18 worked examples28 exercises6 figuresbölüm 4 / 14
04Chain rule, implicit differentiation, and derivatives of trig functions
Drop a stone into still water and the ring it makes widens by about $25$ centimetres every second. Three seconds later somebody asks how fast the wet circle is growing — not how big it is, how fast it is growing. You know how the area depends on the radius, and you know how the radius depends on the time, and you have never had to put two of those together.
By the end of this section you can differentiate a function buried under any number of other functions, find the slope on a curve whose equation nobody can solve for $y$, and convert one measured rate into another with the units and the sign both right.
In 60 seconds
Four skills, one move: when something sits inside something else, differentiating the outside leaves the inside's own derivative behind as a factor.
the equation ties x and y together and will not be solved for y
$\frac{dQ}{dt}=\frac{dQ}{dx}\cdot\frac{dx}{dt}$
two moving quantities are locked together by one equation
Three most common mistakes
Peeling the outer layer and walking away without the inner derivative: $(\sin 3x)'$ is $3\cos 3x$, not $\cos 3x$.
Differentiating a $y$ term as if $y$ were a number: $\frac{d}{dx}y^{2}$ is $2y\,y'$, not $2y$ and certainly not $0$.
Feeding degrees into a trigonometric derivative. Every formula on this page is false in degrees, off by a factor of $\pi/180$.
Quizzes are worth 10% of the grade and the first midterm 28%, and the syllabus assigns four textbook sections to this single week, the joint heaviest week of the term. The compounding matters more than the weight: nothing later in the course can be differentiated without the chain rule, so a gap here keeps charging interest until the final.
How much time do you have?
10 minutes
You leave with the chain rule, the six trig derivatives and the three errors that cost the most marks. Enough to survive a differentiate this question; not enough for a word problem.
The 60 second card, Formula card, The chain rule: one factor per layer, Mistake ledger
45 minutes
You add the two things the formulas cannot give you on their own: how to read an expression before differentiating it, and how to handle a curve that refuses to be solved for y.
The 60 second card, Where the trig derivatives come from, The chain rule: one factor per layer, Reading an expression before differentiating it, Differentiating an equation you cannot solve, Fading ladder, Practice B (computation), Check yourself
Full reading
Everything, in the order it was built: the failure that forces each rule, the rule, the picture, the worked cases, the traps, and the interleaved practice that makes you choose the tool yourself rather than being told which one to use.
Hook, Recall first, Try it yourself first, All six concept blocks, Method boxes, Contrast pairs, Fading ladder, Full exam-style question, Practice A to D, Mistake ledger, Check yourself
By the end of this section
Differentiate the sine and the cosine, and say where the proof used the fact that the angle is in radians.
Derive the remaining four trigonometric derivatives from the quotient rule, and state which three of the six carry a minus sign and what those three have in common.
Apply the chain rule to a two layer composition, naming the outer and before writing anything down.
Decide which rule fires first in an expression that mixes composition with a product or a quotient, and carry the chain rule through three or more layers without dropping a factor.
Compute $dy/dx$ from an equation in $x$ and $y$ that is never solved for $y$, and use it to write the tangent line at a given point on the curve.
Extend an implicit first derivative to a second derivative by substituting the first result back in, and write the at a point.
Translate a word problem about two changing quantities into one equation, differentiate it with respect to time, and report the answer with its sign and its units.
Syllabus coverage
2.4
Derivatives of trigonometric functions
The two limits that make the proof work are restated on this page, so nothing here depends on remembering week 1.
covered
2.5
The chain rule
Split over two blocks: the two layer rule and the picture behind it first, then deep stacks and the question of which rule fires first when a composition sits inside a product.
covered
2.6
Implicit differentiation
Also split over two blocks, because the second derivative of an implicit curve is a different skill from the first and fails for its own reasons.
covered
2.8
Related rates
This is where the chain rule stops being an algebra exercise. The ring of water from the opening paragraph is the first worked case.
covered
2.7
Rates of change in the natural and social sciences
The week's line steps from 2.6 straight to 2.8, so this part of the book is not something you are responsible for. One idea from it is worth two minutes anyway and is used inside the related rates block: a derivative taken with respect to time is a rate, and its units are the units on top divided by the units underneath.
off_syllabus
Recall first
The derivative and its two notations
$f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$, also written $\frac{dy}{dx}$ when $y=f(x)$.
Both notations appear on this page, and the Leibniz one is what makes the chain rule look obvious.
Power rule
$\frac{d}{dx}x^{n}=n\,x^{n-1}$ for every real $n$, so $\sqrt{u}=u^{1/2}$ and $1/u=u^{-1}$ have to be rewritten as powers before they are differentiated.
Half the chain rule problems in this section have a root or a reciprocal as their outer layer.
Product rule
$(fg)'=f'g+fg'$ — two terms, not one.
Every product in this section has a composition inside at least one factor, so the two rules run together.
Quotient rule
$\left(\dfrac{f}{g}\right)'=\dfrac{f'g-fg'}{g^{2}}$, valid where $g\neq0$. The order in the numerator is not negotiable.
The four trigonometric derivatives beyond sine and cosine are all produced by this rule.
The two special trigonometric limits
$\lim_{h\to0}\dfrac{\sin h}{h}=1$ and $\lim_{h\to0}\dfrac{\cos h-1}{h}=0$, both with $h$ in radians.
These are the entire content of the proof that $(\sin x)'=\cos x$; without them the derivative of sine would have to be guessed from a graph.
The difference quotient for sine is unusable until the sum is split apart this way.
Point slope form of a line
The line through $(x_{0},y_{0})$ with slope $m$ is $y-y_{0}=m(x-x_{0})$.
Roughly half the implicit differentiation questions end by asking for a tangent line rather than for a slope.
The
$\sin^{2}\theta+\cos^{2}\theta=1$, and dividing by $\cos^{2}\theta$ gives $1+\tan^{2}\theta=\sec^{2}\theta$.
Expressions collapse dramatically after differentiation when this identity is spotted; several answers on this page look ugly until it is used.
Try it yourself first (3 questions)
1§04.0 — product rule warm up●○○○○
Nothing new yet. This is week 3 machinery, and the point is to find out whether it is automatic before we start stacking things on top of it.
Given
$f(x)=(x^{2}+1)(x^{3}-2x)$
Find
(a) Differentiate $f$ using the product rule.
(b) Expand $f$ first and differentiate again. The two answers must agree.
Hint 1/4
You are asked for $f'(x)$ twice by two different routes, and the real content of the question is whether they agree.
Hint 2/4
Product rule: $(uv)'=u'v+uv'$, with $u=x^{2}+1$ and $v=x^{3}-2x$.
Hint 3/4
With $u=x^{2}+1$ and $v=x^{3}-2x$: $u'=2x$ and $v'=3x^{2}-2$, so $f'=2x(x^{3}-2x)+(x^{2}+1)(3x^{2}-2)$.
Hint 4/4
Both routes give $f'(x)=5x^{4}-3x^{2}-2$.
Show solution
The product rule is the route that survives; expanding is a luxury that disappears as soon as a factor is not a polynomial, which is why it is used here only as the check.
Product rule route
$u=x^{2}+1,\quad v=x^{3}-2x$
naming the factors before differentiating is what stops the two derivatives from getting mixed up
$f'=2x\,(x^{3}-2x)+(x^{2}+1)(3x^{2}-2)$
the rule produces two terms; a single term is the usual sign that one factor was left undifferentiated
$=2x^{4}-4x^{2}+3x^{4}-2x^{2}+3x^{2}-2$
expanding now, not earlier, keeps the two rule applications separate
$=5x^{4}-3x^{2}-2$
collecting like terms
Expand first route
$f(x)=x^{5}-2x^{3}+x^{3}-2x=x^{5}-x^{3}-2x$
for a product of two polynomials expanding is legitimate; it stops being an option the moment a trigonometric factor appears
$f'(x)=5x^{4}-3x^{2}-2$
power rule term by term
Answer $$\boxed{f'(x)=5x^{4}-3x^{2}-2}$$
Check
The two routes were independent of each other and landed on the same polynomial, which is the only check available here.
Whenever a check is this cheap, take it. Later in this section expansion will not be available and this habit is what replaces it.
2§04.0 — a claim about a cube●●○○○
A student differentiates $(2x+1)^{3}$ by looking at the outside only, reasoning that the power rule turns a cube into three times a square. Everything needed to test the claim is already in your hands — the expression can be expanded.
Given
The claim: $\frac{d}{dx}\bigl[(2x+1)^{3}\bigr]=3(2x+1)^{2}$
$(2x+1)^{3}=8x^{3}+12x^{2}+6x+1$
Find
(a) Decide whether the claim is right, using expansion rather than any rule you have not met yet.
Hint 1/4
Do not argue about the claim. Test it: the expanded cube can be differentiated with tools you already have, and then the two answers can be compared.
Hint 2/4
Power rule term by term on a polynomial: $\frac{d}{dx}x^{n}=n\,x^{n-1}$, and constants differentiate to zero.
Hint 3/4
The expansion is $(2x+1)^{3}=8x^{3}+12x^{2}+6x+1$, so its derivative is $24x^{2}+24x+6$. Now factor $6$ out of that.
Hint 4/4
$24x^{2}+24x+6=6(4x^{2}+4x+1)=6(2x+1)^{2}$, so the claim is off by a factor of two.
Numerical check at $x=0$: the expanded derivative gives $6$, the claim gives $3$, and a difference quotient at $x=0$ with $h=0.001$ gives $6.012$, which sides with $6$.
The missing factor was $2$, the derivative of the inside. Hold on to that number — the next block explains why it had to appear and what it becomes when the inside is not linear.
3§04.0 — a limit you will need twice●●○○○
This limit is doing the work inside the proof that opens this section. It is also, on its own, a favourite quiz question, and the obvious answer is not the answer.
Given
$\lim_{h\to0}\dfrac{\sin h}{h}=1$, with $h$ in radians
Compute $\lim_{h\to0}\dfrac{\sin 5h}{h}$
Find
(a) Evaluate the limit.
(b) Say in one sentence why the answer is not $1$.
Hint 1/4
The known limit has the same thing on top and underneath. Here they differ, so the first job is to make them match rather than to compute anything.
Hint 2/4
The known result is $\lim_{t\to0}\frac{\sin t}{t}=1$; it applies to whatever quantity $t$ names, as long as the same quantity appears on top and underneath.
Hint 3/4
Write $\frac{\sin 5h}{h}=5\cdot\frac{\sin 5h}{5h}$. As $h\to0$ the quantity $5h$ also goes to $0$, so the fraction on the right goes to $1$ by the given result.
Hint 4/4
The limit is $5$.
Show solutionForce the denominator to match the angle
$\frac{\sin 5h}{h}=5\cdot\frac{\sin 5h}{5h}$
multiplying and dividing by $5$ costs nothing and is the only move that makes the known result usable
$t=5h\ \Longrightarrow\ t\to0 \text{ as } h\to0$
the substitution is legitimate because $5h$ really does approach $0$; it would fail for something like $\sin(5+h)$
Take the limit
$\lim_{h\to0}5\cdot\frac{\sin t}{t}=5\cdot1=5$
the constant $5$ is outside the limit and simply survives
Answer $$\boxed{5}$$
Check
Numerical check: at $h=0.001$, $\sin(0.005)/0.001=4.99998$, which is closing on $5$ rather than on $1$.
An angle that moves $k$ times as fast as the variable leaves a factor of $k$ behind. That sentence, generalised, is the chain rule.
Notation
symbol
reads as
means
watch out
$f\circ g$
f composed with g
the function $x\mapsto f(g(x))$: run $g$ first, then feed the result to $f$.
$f\circ g$ and $g\circ f$ are different functions. $\sin(x^{2})$ and $(\sin x)^{2}$ are the two orders of the same pair.
$f'(g(x))$
f prime evaluated at g of x
differentiate $f$ first, then substitute $g(x)$ into the result.
Not $f'(x)$, and not $f'$ times $g(x)$. The most common wrong answer in this section comes from evaluating the outer derivative at $x$ instead of at the inner function.
$\frac{dy}{du}\cdot\frac{du}{dx}$
dy by du times du by dx
the Leibniz form of the chain rule, with $u$ the name given to the inner function.
The cancellation of $du$ is a memory aid, not a proof — these are limits, not fractions, even though they behave like fractions here.
$y'$
y prime
$dy/dx$, the derivative of $y$ with respect to $x$, including when $y$ is only known implicitly.
In a related rates problem the independent variable is time, so primes are dropped there in favour of $dx/dt$ and $dy/dt$.
$\sin^{2}x$
sine squared x
$(\sin x)^{2}$ — the sine is taken first and the result is squared.
Not $\sin(x^{2})$. And $\sin^{-1}x$ breaks the pattern completely: it is the inverse function, not $1/\sin x$.
$\sec x,\ \csc x,\ \cot x$
secant, cosecant, cotangent
$1/\cos x$, $1/\sin x$ and $\cos x/\sin x$ respectively.
Cosecant is the reciprocal of sine, not of cosine. The names and the reciprocals are crossed over, which is exactly why the pair gets swapped under exam pressure.
$f^{(n)}(x)$
the n th derivative of f
the result of differentiating $f$ a total of $n$ times.
Parentheses around the $n$: $f^{(4)}$ is the fourth derivative, $f^{4}$ is the fourth power.
$\frac{dV}{dt}$
dV by dt
the rate at which the volume changes per unit of time.
Its units are the units of $V$ divided by the units of $t$. Checking those units catches most related rates errors before the arithmetic does.
Conventions used here
Trigonometric arguments are radians
Every trigonometric derivative in this course assumes the angle is measured in radians. In degrees the formulas are simply false: $\frac{d}{dx}\sin(x^{\circ})=\frac{\pi}{180}\cos(x^{\circ})$. If a problem hands you degrees or degrees per second, convert first and say so in one line.
Where a derivative is evaluated
$f'(g(x))$ means: differentiate $f$, then put $g(x)$ into the result. It does not mean $f'$ multiplied by $g(x)$, and it is not $f'(x)$. The argument travels with the function it belongs to.
An answer may legitimately contain both x and y
An implicit derivative such as $y'=-x/y$ is a finished answer, not an unfinished one. It is a rule for the slope at any point of the curve, and it needs both coordinates because two different points with the same $x$ can sit on the curve with different slopes.
Prime notation names its variable
$y'$ always means $dy/dx$ here. When time is the independent variable we write $dx/dt$, $dV/dt$ and so on in full rather than using primes or dots, because in a related rates problem two different derivatives are in play at once and a prime cannot say which is meant.
Rates carry a sign
A quantity that is shrinking has a negative rate. Report the sign as part of the answer and say what it means physically; an answer of $-0.225$ m/s with no sentence attached is half an answer.
Where the derivatives of sine and cosine come from
Week 3 left you able to differentiate anything assembled out of powers, and stuck the moment a sine walked in.
Solvable with what we have
$x^{7}$, $\sqrt{x}$, $1/x^{3}$ — every power, once it is written as a power
$3x^{2}(x-1)$ and $\dfrac{x}{x^{2}+1}$ — products and quotients of powers
any polynomial, term by term
Not solvable yet
$\sin x$
$\tan x$
$x\sin x$ — the product rule stalls, because it needs $(\sin x)'$ before it can start
Read the slopes off the graph. At $x=0$ the sine climbs at about $45^{\circ}$, so its slope looks like $1$. At $x=\pi/2$ the curve is flat, so $0$. At $x=\pi$ it falls at about $45^{\circ}$, so $-1$. Those three numbers are the heights of the cosine, so guess $(\sin x)'=\cos x$ and move on.
Why it fails
The guess is right, and that is the problem: the picture could not have told you so. A function with slope $0.99\cos x$ everywhere would draw the same graph at this resolution, and that stray $0.99$ would then contaminate every rate for the rest of the course. Redraw the same curve with degrees on the axis and the slopes come out near $0.017$ instead.
TheoremTheorem 4.1: the two seed derivatives
Conditions
$x$ is measured in radians
the two limits $\lim_{h\to0}\frac{\sin h}{h}=1$ and $\lim_{h\to0}\frac{\cos h-1}{h}=0$ are available
The rate at which the sine is climbing at a given angle is the cosine of that same angle; the rate at which the cosine is climbing is minus the sine of it.
Three tangents on $\textcolor{#1f6feb}{\sin x}$, and the height of $\textcolor{#d1690a}{\cos x}$ under each point of tangency. At $0$ the slope is $\textcolor{#128a5a}{1}$ and the cosine is $1$; at $\pi/2$ both are $0$; at $\pi$ both are $-1$, and the theorem says this holds at every angle.
Looks like this, but is not
$\dfrac{d}{dx}\sin(x^{\circ})=\cos(x^{\circ})$ — the same statement with the axis marked in degrees.
False, and off by a factor of about $57$. A degree scale stretches the same wave over a horizontal span $57.3$ times longer and flattens every slope by that factor: $\frac{d}{dx}\sin(x^{\circ})=\frac{\pi}{180}\cos(x^{\circ})$. This is why calculus is done in radians.
The derivative of sin x over 1 + cos x
A quotient where the answer is far simpler than the working, provided you recognise the identity when it shows up.
expand before you look for cancellation; the identity is invisible while the brackets are closed
Use the identity and cancel
$=\frac{\cos x+1}{(1+\cos x)^{2}}$
$\cos^{2}x+\sin^{2}x=1$, so the numerator collapses to $1+\cos x$
$=\frac{1}{1+\cos x}$
one factor of $1+\cos x$ cancels, which is legitimate wherever the original function was defined anyway
Answer $$\boxed{f'(x)=\frac{1}{1+\cos x}}$$
Check
Independent route: the half angle identity gives $f(x)=\tan(x/2)$, whose derivative is $\tfrac12\sec^{2}(x/2)=\dfrac{1}{2\cos^{2}(x/2)}$. Since $1+\cos x=2\cos^{2}(x/2)$, that is the same function — reached without the quotient rule at all.
One quotient rule, one identity, one cancellation.
When $\sin^{2}$ and $\cos^{2}$ both appear after differentiating, reach for the identity before reaching for a common denominator.
Checkpoint
§04.1 — signs on a trig combination●○○○○
Thirty seconds, and the only thing being tested is whether the minus sign landed in the right place.
Given
$f(x)=2\sin x-3\cos x$
Find
(a) Find $f'(x)$.
(b) Evaluate $f'(0)$.
Hint 1/4
Differentiate term by term; the two coefficients just ride along, and the only decision is what happens to each sign.
Hint 2/4
$(\sin x)'=\cos x$ and $(\cos x)'=-\sin x$; a constant multiple survives differentiation untouched.
Hint 3/4
With $f(x)=2\sin x-3\cos x$: the first term gives $2\cos x$, and the second gives $-3\cdot(-\sin x)$.
Hint 4/4
$f'(x)=2\cos x+3\sin x$, so $f'(0)=2$.
Show solutionDifferentiate term by term
$f'(x)=2\cos x-3\cdot(-\sin x)$
the minus in front of the term and the minus inside the derivative of cosine are two different minus signs, and both have to be written before either is simplified
Each of these comes from writing the function as sine over cosine, or one over one of them, and turning the quotient rule once. The three cofunctions come out negative because in each of their cases it is the cosine or the sine downstairs whose minus survives.
Looks like this, but is not
$\dfrac{d}{dx}\sec x=\sec^{2}x$, by analogy with the power rule.
$\sec^{2}x$ is the derivative of the tangent, not of the secant. Test both candidates at $x=0$: the secant has a minimum there, so its derivative must be $0$, and $\sec 0\tan 0=0$ passes while $\sec^{2}0=1$ fails.
function
derivative
a check on the sign
$\sin x$
$\cos x$
at $x=0$ the sine is rising and $\cos0=1>0$
$\cos x$
$-\sin x$
just past $x=0$ the cosine is falling, and $-\sin x<0$ there
$\tan x$
$\sec^{2}x$
never negative, matching a tangent that rises on every branch
$\cot x$
$-\csc^{2}x$
never positive, matching a cotangent that falls on every branch
$\sec x$
$\sec x\tan x$
zero at $x=0$, where $\sec x$ bottoms out at $1$
$\csc x$
$-\csc x\cot x$
zero at $x=\pi/2$, where $\csc x$ bottoms out at $1$
The three cofunctions — cosine, cotangent, cosecant — are exactly the three whose derivatives carry a minus sign. That is the entire mnemonic, and it is worth more than the four formulas above it.
Where sec squared comes from: differentiating tan x
Only two of the six derivatives have to be proved. The other four fall out of the quotient rule, and this is the one worth doing by hand once.
Given
$\tan x=\dfrac{\sin x}{\cos x}$, valid where $\cos x\neq0$
Find
$\dfrac{d}{dx}\tan x$
Solution
The quotient rule is the cheap route because the two seed derivatives are already known. Going back to the difference quotient for the tangent would mean proving the angle addition formula for tangent first.
the numerator's second term picks up a minus from the quotient rule and another from the derivative of cosine, so the two cancel and the term arrives positive
$=\frac{\cos^{2}x+\sin^{2}x}{\cos^{2}x}$
a sum of two squares is the shape that always signals the Pythagorean identity
Collapse with the identity
$=\frac{1}{\cos^{2}x}=\sec^{2}x$
the identity turns the whole numerator into $1$, which is why this derivative is so much simpler than its construction
Answer $$\boxed{\frac{d}{dx}\tan x=\sec^{2}x}$$
Check
Independent check on the sign and the size: $\sec^{2}x\geq1$ for every $x$, and the graph of $\tan x$ never has a slope below $1$ — at the origin it crosses at exactly $45^{\circ}$, and $\sec^{2}0=1$.
One quotient rule and one identity.
The same two lines with $\cos x/\sin x$ produce $-\csc^{2}x$; the minus arrives because the derivative of the denominator is now the one that keeps its sign.
Checkpoint
§04.2 — a secant and a cotangent together●●○○○
Thirty seconds, and it tests the two formulas that get swapped most often.
Given
$f(x)=\sec x+\cot x$
Find
(a) Find $f'(x)$.
(b) Evaluate $f'\left(\tfrac{\pi}{4}\right)$.
Hint 1/4
Two separate formulas, one per term, and the only judgement needed is which formula belongs to which name.
Hint 2/4
$(\sec x)'=\sec x\tan x$ and $(\cot x)'=-\csc^{2}x$; the cotangent is a cofunction, so its derivative is negative.
Hint 3/4
$f'(x)=\sec x\tan x-\csc^{2}x$, and at $\tfrac{\pi}{4}$ we have $\sec\tfrac{\pi}{4}=\sqrt2$, $\tan\tfrac{\pi}{4}=1$ and $\csc^{2}\tfrac{\pi}{4}=2$.
Sign check by inspection: at $\pi/4$ the cotangent is falling steeply while the secant is rising only gently, so a negative total is what the graph predicts. Numerically, $f(0.785)\approx 2.41445$ and $f(0.786)\approx 2.41386$, giving a difference quotient of about $-0.5849$, within $0.001$ of $\sqrt2-2\approx-0.5858$ — the gap is the step size, which is what a one sided quotient costs.
Two of the four formulas carry a minus, and they are exactly the two whose names start with co.
⚠ The two secant formulas swapped
$\sec^{2}x$ and $\sec x\tan x$ both start with a secant, and the one that belongs to the tangent gets attached to the secant instead
wrong$\frac{d}{dx}\sec x=\sec^{2}x$
right$\frac{d}{dx}\sec x=\sec x\tan x$
⚠ A cofunction derivative written without its minus
the four derived formulas are memorised as a list, and a sign is the easiest part of a list item to lose
wrong$\frac{d}{dx}\cot x=\csc^{2}x$
right$\frac{d}{dx}\cot x=-\csc^{2}x$
The chain rule: one factor per layer
The warm up left a factor of $2$ unaccounted for. Here is where it came from, and what it turns into when the inside is not a straight line.
Differentiate the and leave its input exactly as it was, then multiply by the derivative of that input.
Why the two derivatives multiply rather than add. A step of $\textcolor{#1f6feb}{1}$ in $x$ opens into a step of $\textcolor{#d1690a}{3}$ in $u$ because $g$ stretches by $3$; each of those units opens into $2$ units of $y$ because $f$ stretches by $2$. The stretches compound: $\textcolor{#128a5a}{6}=2\times3$.
Looks like this, but is not
$\dfrac{d}{dx}\sin(x^{2})=\cos(x^{2})$, or its close relative $\dfrac{d}{dx}\sin(x^{2})=\cos(x^{2})\cdot x^{2}$.
The first keeps only the outer factor and throws the inner one away. The second remembers that something multiplies but multiplies by the inner function instead of by its derivative. The rule asks for $\cos(x^{2})\cdot2x$, and a quick numerical test at $x=1$ settles it: the true slope there is $2\cos1\approx1.081$, while the two wrong versions predict $0.540$.
The derivative of (2x+1) cubed, without expanding
The warm up produced $6(2x+1)^{2}$ by brute force. The rule should reproduce it in one line, and if it does not, the rule is wrong.
naming them is the whole job; once they are named the rule is mechanical
$f'(u)=3u^{2},\qquad u'=2$
the inner derivative is the constant $2$ because the inside is a straight line of slope $2$
Assemble
$\frac{dy}{dx}=3(2x+1)^{2}\cdot 2$
the outer derivative keeps the inside untouched inside its own brackets; only the factor at the end knows about the $2$
$=6(2x+1)^{2}$
multiplying the two constants
Answer $$\boxed{\frac{dy}{dx}=6(2x+1)^{2}}$$
Check
This is exactly the expression the expansion produced in the warm up, reached by a completely different route.
Chain rule applied once.
The mysterious factor of $2$ was the slope of the inside. When the inside is not a line, that factor stops being constant — which is the only thing that changes in the next example.
The derivative of the square root of x squared plus nine
A root on the outside is the most common shape in this section, and it is also where the outer derivative is most often evaluated at the wrong place.
Given
$y=\sqrt{x^{2}+9}$
Find
$dy/dx$, and its value at $x=4$
SolutionRewrite the root as a power
$y=\left(x^{2}+9\right)^{1/2}$
the power rule is the only version of the outer derivative we have; a root has to be written as a power before it can be used
Independent check by geometry: $\sqrt{x^{2}+9}$ is the length of the hypotenuse of a right triangle with legs $x$ and $3$, so its rate of growth per unit of $x$ is the cosine of the angle at the far vertex, which is $x$ over the hypotenuse and can never exceed $1$. The formula agrees, and at $x=4$ the triangle is the $3,4,5$ one, giving $4/5$.
Chain rule applied once, plus one rewrite.
Every root in this section follows the same two moves: write it as a power, then evaluate the outer derivative at the whole inside.
Checkpoint
§04.3 — sine squared against sine of a square●●○○○
Thirty seconds. These two expressions are written almost identically and differentiate completely differently, which is why exams keep putting them next to each other.
Given
$p(x)=\sin\left(x^{2}\right)$
$s(x)=\sin^{2}x=(\sin x)^{2}$
Find
(a) Differentiate $p$.
(b) Differentiate $s$.
Hint 1/4
For each one, decide which operation happens last when you evaluate it by hand at, say, $x=2$. That operation is the outer function.
In $p$ the last operation is taking a sine, so outer is $\sin$ and inner is $x^{2}$. In $s$ the last operation is squaring, so outer is $u^{2}$ and inner is $\sin x$.
Hint 4/4
$p'(x)=2x\cos\left(x^{2}\right)$ and $s'(x)=2\sin x\cos x$.
Show solutionSine on the outside
$p'(x)=\cos\left(x^{2}\right)\cdot 2x$
the outer derivative of sine is cosine, evaluated at the whole inside; the inner derivative of $x^{2}$ is $2x$
$=2x\cos\left(x^{2}\right)$
writing the polynomial factor first, as is conventional
Square on the outside
$s'(x)=2(\sin x)^{1}\cdot\cos x$
the outer derivative of $u^{2}$ is $2u$ with $u=\sin x$; the inner derivative is $\cos x$
Check at $x=0$: $p$ has a there because $p'(0)=0$, and indeed $\sin(x^{2})\approx x^{2}$ near zero, a parabola. And $s(x)=\sin^{2}x\approx x^{2}$ near zero too, so $s'(0)=0$ as well — both formulas return $0$.
Deciding which function is on the outside is the entire difficulty. Once that is settled the algebra is automatic.
⚠ The inner derivative left behind
the outer rule is the visible one; the inner factor has to be asked for, and under time pressure nobody asks
wrong$\frac{d}{dx}\sin 3x=\cos 3x$
right$\frac{d}{dx}\sin 3x=3\cos 3x$
⚠ The outer derivative evaluated at x instead of at the inside
the power rule was learned as a statement about $x$, so the hand writes $x$ where the inside belongs
One factor for each layer, and each factor is evaluated at everything that was still sitting inside it.
The question that decides which rule fires first: if you evaluated the expression by hand at $x=2$, what would the last keystroke be? That operation is the outermost one, and its rule goes first. Everything still not elementary goes back through the same question.
Looks like this, but is not
$h(x)=x^{2}\sin(3x)$ looks like a chain, because it has three moving parts and one of them is clearly inside another.
Evaluate it by hand at $x=2$: you would compute $4$, compute $\sin 6$, and multiply last. The outermost operation is a multiplication, so the product rule fires first and the chain rule appears only later, inside the second factor. Starting with the chain rule here produces an answer with one term where there should be two.
layer
what it does
its input
the factor it contributes
outer
takes the sine
$\sqrt{x^{2}+1}$
$\cos\!\left(\sqrt{x^{2}+1}\right)$
middle
takes the square root
$x^{2}+1$
$\dfrac{1}{2\sqrt{x^{2}+1}}$
inner
squares and adds one
$x$
$2x$
The product of the three factors is $\dfrac{x\cos\!\left(\sqrt{x^{2}+1}\right)}{\sqrt{x^{2}+1}}$, after the $2$ from the bottom row cancels the $2$ in the middle row. Notice that each input column is a copy of the row below it, wrapped in one more function — that pattern is what tells you no layer has been skipped.
The derivative of x squared times sin 3x
Two rules in one expression, and the only thing that matters is which one goes first.
Given
$h(x)=x^{2}\sin(3x)$
Find
$h'(x)$
Solution
The product rule has to go first because the multiplication is the outermost operation. Expanding is not an option here — there is nothing to expand once a sine is involved.
Independent check near the origin: for small $x$, $\sin(3x)\approx3x$, so $h(x)\approx3x^{3}$ and its slope should be about $9x^{2}$. The formula at small $x$ gives $2x(3x)+3x^{2}(1)=9x^{2}$, which matches.
One product rule and one chain rule.
Whenever the answer comes out with fewer terms than the product rule promises, a factor was left undifferentiated.
The derivative of cos cubed of 5x
Three layers, none of them optional, and a natural place to lose the middle one.
Given
$y=\cos^{3}(5x)$, which means $\bigl(\cos(5x)\bigr)^{3}$
Find
$dy/dx$
SolutionName all three layers before differentiating
Independent check at two special angles. At $x=0$ the function sits at its maximum value $1$, so the slope must be $0$ — the formula gives $-15\cdot1\cdot0=0$. At $5x=\pi/2$ the cosine crosses zero, and cubing a number near zero flattens it, so the slope must be $0$ there too — the formula gives $-15\cdot0\cdot1=0$.
Chain rule applied twice, inside one expression.
A power written on the function name, as in $\cos^{3}$, is still an outer layer. Rewriting it as a bracket before you start costs three seconds and saves the whole factor.
Checkpoint
§04.4 — which rule fires first●●○○○
Thirty seconds, and no differentiating at all. For each expression, say which rule you would apply first if you had to differentiate it.
Given
$A(x)=\dfrac{\tan(2x)}{x^{3}+1}$
$B(x)=\sqrt{x\cos x}$
$C(x)=x^{4}\sec(x^{2})$
Find
(a) Name the first rule for $A$.
(b) Name the first rule for $B$.
(c) Name the first rule for $C$.
Hint 1/4
You are not being asked for derivatives. For each one, imagine evaluating it by hand at $x=1$ and ask what the final keystroke would be.
Hint 2/4
The outermost operation decides: a final multiplication means the product rule, a final division means the quotient rule, a final apply a function to what I have means the chain rule.
Hint 3/4
For $A$ the last act is dividing $\tan 2$ by $2$. For $B$ the last act is taking a square root of the number $\cos 1$. For $C$ the last act is multiplying $1$ by $\sec 1$.
Hint 4/4
Quotient rule for $A$, chain rule for $B$, product rule for $C$.
Show solutionEvaluate each one mentally and watch the last keystroke
Cross check by counting terms: $A$ and $C$ must end with two terms, because both the quotient and the product rule produce two. $B$ must end as a single fraction with a product rule inside its numerator.
Answering this question first is worth more than any speed gained later; almost every lost mark in this section comes from starting at the wrong level.
⚠ The chain rule applied where the top level is a product
a composition sitting inside is more visually striking than the multiplication holding the whole thing together
Every time a $y$ gets differentiated it leaves a $dy/dx$ behind, because $y$ is quietly a function of $x$ and the chain rule charges for that.
The , $\textcolor{#1f6feb}{x^{3}+y^{3}=6xy}$. No formula for $y$ in terms of $x$ exists in elementary terms, and for most of the loop a single $x$ has two different $y$ values above it. The point $\textcolor{#128a5a}{P=(3,3)}$ sits on the mirror line $y=x$, and the $\textcolor{#d1690a}{\text{tangent}}$ there has slope $-1$.
Looks like this, but is not
Solve for $y$ first — surely that is simpler than learning a new method.
For $x^{3}+y^{3}=6xy$ there is no elementary formula to solve for, so the plan fails outright. And even when solving works it can cost you: $x^{2}+y^{2}=25$ splits into $y=\sqrt{25-x^{2}}$ and $y=-\sqrt{25-x^{2}}$, two separate functions, and the derivative of the branch you picked is the wrong answer at the mirrored point. The implicit result $y'=-x/y$ covers both branches at once and reports the difference through the sign of $y$.
The tangent to the folium of Descartes at (3, 3)
The showpiece for this method. No formula for $y$ exists, the point is on the curve, and the tangent still comes out in four lines.
Given
$x^{3}+y^{3}=6xy$
the point $(3,3)$, which is on the curve since $27+27=54=6\cdot 9$
Find
$dy/dx$ in terms of $x$ and $y$, then the tangent line at $(3,3)$
Solution
Implicit differentiation is not merely the faster route here, it is the only one: the cubic cannot be solved for $y$ in elementary terms at all.
Differentiate both sides with respect to x
$3x^{2}+3y^{2}\frac{dy}{dx}=6y+6x\frac{dy}{dx}$
the left side uses the chain rule on $y^{3}$; the right side is a product $6xy$, so it needs the product rule and produces two terms
$3y^{2}\frac{dy}{dx}-6x\frac{dy}{dx}=6y-3x^{2}$
everything carrying $\frac{dy}{dx}$ moves to one side — the equation is linear in it, so this always works
Solve for the derivative
$\frac{dy}{dx}\bigl(3y^{2}-6x\bigr)=6y-3x^{2}$
factoring $\frac{dy}{dx}$ out is the step that finishes the method
substituting only now, after the general formula is finished; the denominator is $3$, not zero, so the tangent is not vertical
$y-3=-1\,(x-3)\ \Longrightarrow\ x+y=6$
point slope form with the point that was given
Answer $$\boxed{\frac{dy}{dx}=\frac{2y-x^{2}}{y^{2}-2x},\qquad \text{tangent at }(3,3):\ x+y=6}$$
Check
Independent check by symmetry: swapping $x$ and $y$ leaves the equation $x^{3}+y^{3}=6xy$ unchanged, so the curve is its own mirror image in the line $y=x$. The point $(3,3)$ lies on that mirror line, so the tangent there must be perpendicular to it — slope $-1$. No differentiation was involved in that argument.
One chain rule, one product rule, one division.
The derivative contains both $x$ and $y$, and that is normal rather than a loose end. The curve has two different points above most $x$ values, and they genuinely have different slopes.
Finding where the folium has a horizontal tangent
A question the picture can answer approximately and the formula can answer exactly. Both are done here, which is what makes it a check rather than a claim.
Given
$x^{3}+y^{3}=6xy$
from the previous example, $\dfrac{dy}{dx}=\dfrac{2y-x^{2}}{y^{2}-2x}$
Find
the point on the loop where the tangent is horizontal
SolutionSet the numerator to zero
$2y-x^{2}=0\ \Longrightarrow\ y=\frac{x^{2}}{2}$
a fraction is zero exactly when its top is zero and its bottom is not, so the condition splits into an equation plus a check
and the denominator there is $y^{2}-2x\approx10.08-5.04\approx5.04\neq0$, so the tangent really is horizontal rather than undefined
Answer $$\boxed{(x,y)\approx(2.52,\;3.17)}$$
Check
Independent check against the figure: the top of the drawn loop sits at about $x=2.5$, $y=3.2$, which is where a horizontal tangent has to touch. The picture and the algebra were produced separately and agree to the precision the picture can offer.
One extra equation on top of the previous example.
A horizontal tangent needs the numerator to vanish; a vertical one needs the denominator to vanish. By the mirror symmetry, the vertical tangent of this loop is at $(3.17,\,2.52)$, the same pair of numbers reversed.
Checkpoint
§04.5 — differentiating three mixed terms●●○○○
Thirty seconds. No equation to solve, just the three shapes that appear in every implicit problem.
Given
$y$ is a function of $x$, written implicitly
Find
(a) Find $\dfrac{d}{dx}\bigl[y^{4}\bigr]$.
(b) Find $\dfrac{d}{dx}\bigl[xy\bigr]$.
(c) Find $\dfrac{d}{dx}\bigl[x^{2}y^{3}\bigr]$.
Hint 1/4
Each of these is an ordinary rule you already know; the only new thing is that $y$ is a function whose formula is missing.
Hint 2/4
Chain rule for a power of $y$: $\frac{d}{dx}y^{n}=n\,y^{n-1}y'$. Product rule whenever $x$ and $y$ are multiplied together.
Hint 3/4
For (c) the two factors are $x^{2}$ and $y^{3}$, so the product rule gives $2x\cdot y^{3}+x^{2}\cdot\frac{d}{dx}\bigl[y^{3}\bigr]$, and the second piece needs the chain rule.
Hint 4/4
$4y^{3}y'$, then $y+xy'$, then $2xy^{3}+3x^{2}y^{2}y'$.
Show solutionA pure power of y
$\frac{d}{dx}y^{4}=4y^{3}\,y'$
chain rule: the power is the outer function and the unnamed $y(x)$ is the inner one
Consistency check: put $y=x$, so that $y'=1$ and the three expressions are really $x^{4}$, $x^{2}$ and $x^{5}$. The answers become $4x^{3}$, then $x+x=2x$, then $2x\cdot x^{3}+3x^{2}\cdot x^{2}=5x^{4}$ — exactly what the power rule gives for each one.
Count the $y'$ factors before you go on: one per $y$ that was differentiated. A term with a $y$ in it and no $y'$ after it is almost always a mistake.
⚠ Treating y as a constant
$y$ looks like a letter rather than a function, and letters that are not $x$ have behaved like constants all term
wrong$\frac{d}{dx}y^{2}=0$
right$\frac{d}{dx}y^{2}=2y\,y'$
⚠ Differentiating a mixed term without the product rule
$xy$ reads as one object rather than as two functions multiplied together
wrong$\frac{d}{dx}\bigl[6xy\bigr]=6xy'$
right$\frac{d}{dx}\bigl[6xy\bigr]=6y+6xy'$
⚠ Leaving y prime on both sides of the equation
the aim was to find $y'$, so the collecting step feels like bookkeeping and gets skipped when time is short
wrong$y'=\frac{6y-3x^{2}+6xy'}{3y^{2}}$
right$y'=\frac{6y-3x^{2}}{3y^{2}-6x}$
Going back through an implicit answer: second derivatives and normals
An implicit first derivative still contains $y$, and that single fact is what makes the second derivative a different problem rather than a repetition.
MethodMethod 4.5: the second derivative of an implicit curve
Conditions
$y'$ has already been found and its denominator is not zero at the point
the substitution at the end is done before, not after, simplifying
$$\boxed{\;y''=\frac{d}{dx}\Bigl[\,y'\,\Bigr]\ \text{treating }y\text{ as a function},\ \text{then replace every }y'\text{ that appears by its known expression}\;}$$
Differentiate the formula you already have, remembering that it contains $y$ and will therefore hand you another $y'$, and then put the known $y'$ back in so that the answer mentions only $x$ and $y$.
On $\textcolor{#1f6feb}{x^{2}+y^{2}=25}$ the implicit slope $y'=-x/y$ gives $-3/4$ at $(3,4)$, so the $\textcolor{#d1690a}{\text{tangent}}$ falls three units for every four across. The $\textcolor{#128a5a}{\text{normal}}$ turns that over to $4/3$ — and a line of slope $4/3$ through $(3,4)$ passes exactly through the centre, which is the check that the two slopes were reciprocals with the right sign.
Looks like this, but is not
From $y'=-\dfrac{x}{y}$, differentiate the top and the bottom separately to get $y''=-\dfrac{1}{y'}$, or treat $y$ as a constant and get $y''=-\dfrac{1}{y}$.
Both skip the quotient rule, and both forget that the $y$ downstairs is a function of $x$. The correct calculation produces $y''=-\dfrac{x^{2}+y^{2}}{y^{3}}$, which on this circle is $-25/y^{3}$. A quick test at $(0,5)$, the top of the circle: the true value is $-25/125=-0.2$, while $-1/y=-0.2$ agrees by coincidence and $-1/y'$ is undefined because $y'=0$ there. Coincidences at one point are exactly why the shortcut survives so long.
The second derivative on a circle of radius 5
The standard exercise, and the one where the substitution step either happens or the answer never closes.
Given
$x^{2}+y^{2}=25$
Find
$y'$, then $y''$ in terms of $y$ alone
Solution
Substituting $y'$ back in before simplifying is the cheaper order. Simplifying first leaves a compound fraction containing $y'$ and the substitution then has to be done inside it.
First derivative
$2x+2y\,y'=0$
chain rule on $y^{2}$; the constant $25$ contributes nothing
$y'=-\frac{x}{y}$
solving; this is already a complete answer and it needs both coordinates, because the upper and lower halves of the circle have opposite slopes at the same $x$
Differentiate the answer, quotient rule
$y''=-\frac{(1)(y)-x\,y'}{y^{2}}$
the numerator is $x$ and the denominator is $y$, and $y$ is a function of $x$, so its derivative $y'$ appears rather than $0$
$=-\frac{y-x\left(-\frac{x}{y}\right)}{y^{2}}$
substituting the known $y'$ back in now, while the expression is still small enough to handle
Independent check on the upper half, where the circle can be solved explicitly: $y=\sqrt{25-x^{2}}$ gives $y'=-x(25-x^{2})^{-1/2}$ and then $y''=-25(25-x^{2})^{-3/2}=-25/y^{3}$ — the same expression, reached without implicit differentiation. The sign also passes inspection: on the upper half $y>0$, so $y''<0$ and the arc is concave down, which is what the picture shows.
Chain rule once, quotient rule once, one substitution.
Using the original equation at the end is the move that turns an ugly answer into a clean one. Look for the original left hand side inside your expression before you start expanding.
The normal line to the circle at (3, 4)
The tangent's perpendicular partner. It appears in exams because it costs one extra line and catches a missing minus sign.
Given
$x^{2}+y^{2}=25$
the point $(3,4)$, on the circle since $9+16=25$
from the previous example, $y'=-x/y$
Find
the tangent slope, the normal slope, and the equation of the normal
SolutionSlopes
$y'\big|_{(3,4)}=-\frac{3}{4}$
substituting both coordinates, since the implicit formula needs both
$m_{\text{normal}}=-\frac{1}{-3/4}=\frac{4}{3}$
perpendicular means negative reciprocal — flip it and change the sign, and here the sign change turns a negative slope positive
Independent check from geometry: the normal to a circle at any point is the radius through that point, so it must pass through the centre $(0,0)$. The line $y=\frac43x$ does, and this was never used in the calculation.
One reciprocal and one line.
The constant term vanishing was not luck; it is the algebraic trace of the normal passing through the centre. When a geometric fact is available, use it as the check rather than as the method.
Checkpoint
§04.6 — the normal on an ellipse●●○○○
Thirty seconds, and the whole test is whether the minus sign survives the flip.
Given
$x^{2}+4y^{2}=8$
the point $(2,1)$, on the curve since $4+4=8$
Find
(a) Find the slope of the tangent at $(2,1)$.
(b) Find the slope of the normal there.
Hint 1/4
Two slopes, and the second is built from the first by one operation. Get the implicit derivative first and do not evaluate anything before it is finished.
Hint 2/4
Differentiate both sides: the $y^{2}$ term gives $8y\,y'$ by the chain rule. Perpendicular slopes multiply to $-1$.
Hint 3/4
$2x+8y\,y'=0$ gives $y'=-\dfrac{x}{4y}$, and at $(2,1)$ that is $-\dfrac{2}{4}$.
Hint 4/4
Tangent slope $-\tfrac12$, normal slope $2$.
Show solutionImplicit derivative
$2x+8y\,y'=0$
the $4y^{2}$ term gives $4\cdot 2y\,y'=8y\,y'$
$y'=-\frac{x}{4y}$
solving for the derivative
Evaluate and flip
$y'\big|_{(2,1)}=-\frac{2}{4}=-\frac{1}{2}$
substituting both coordinates
$m_{\text{normal}}=-\frac{1}{-1/2}=2$
negative reciprocal; the two minus signs cancel, which is precisely the step that gets dropped
How fast the answer moves equals how much the answer changes per unit of the middle quantity, times how fast the middle quantity is itself moving.
The instant that the ladder problem asks about: the foot is $\textcolor{#1f6feb}{3}$ m out, the top is $\textcolor{#128a5a}{4}$ m up, and the $\textcolor{#d1690a}{\text{ladder}}$ is $5$ m long and stays that way. Both $x$ and $y$ are moving; the length is the only thing that is not, and that is why it is the constant in the equation.
Looks like this, but is not
The radius is $75$ cm, so put that in first: $A=\pi(75)^{2}=5625\pi$, and now differentiate.
$5625\pi$ is a number, and the derivative of a number is zero. Substituting the instant's value froze the very quantity whose motion the question was about. The rule is mechanical: differentiate while everything is still a symbol, and let the numbers in only at the last line.
How fast the wet circle grows, three seconds in
The question from the opening paragraph, now answerable. A stone has been dropped, the ring is spreading, and the area is wanted as a rate rather than as a size.
Given
the radius grows at a steady $\dfrac{dr}{dt}=25$ cm/s
the instant asked about is $t=3$ s, when $r=75$ cm
$A=\pi r^{2}$
Find
$\dfrac{dA}{dt}$ at that instant, in cm$^{2}$/s
Solution
The area formula is the relation to differentiate because it connects the quantity asked about to the quantity whose rate was given. Any relation that does not contain both of those is the wrong one to pick.
Write the relation with everything still symbolic
$A(t)=\pi\bigl(r(t)\bigr)^{2}$
writing $r(t)$ rather than $r$ once, at the start, is what stops the radius from being treated as a constant three lines later
$\frac{dA}{dt}=2\pi r\cdot\frac{dr}{dt}$
chain rule: the outer derivative of the square is $2\pi r$, and the inner derivative is the radius' own speed
Now, and only now, put the numbers in
$r=25\cdot 3=75\ \text{cm}$
the radius at the instant asked about; it was never needed before this line
Independent check by geometry, with no calculus at all: in one instant the wet region gains a thin ring of area equal to its circumference times its extra width. The circumference at $r=75$ is $2\pi(75)\approx471.2$ cm and the width added each second is $25$ cm, so the ring gains about $471.2\times25\approx11781$ cm$^{2}$ per second, the same number with no calculus in it.
One chain rule.
The answer grows with $r$: the same $25$ cm of extra radius buys more area when the ring is already wide. That is the sentence that explains why a rate question is not a size question.
The ladder sliding down the wall
The standard word problem of this section, and the one that teaches the sign.
Given
a ladder $5$ m long, foot on the ground and top against a vertical wall
the foot slides away from the wall at $\dfrac{dx}{dt}=0.3$ m/s
the instant asked about is when the foot is $x=3$ m from the wall
Find
$\dfrac{dy}{dt}$ at that instant, and what its sign means
SolutionFind the relation
$x^{2}+y^{2}=25$
Pythagoras on the triangle formed by wall, ground and ladder; the $25$ is constant because the ladder does not stretch, and that is the only reason a constant appears
$y=\sqrt{25-9}=4\ \text{m at the instant in question}$
computed and set aside — it is needed at the end, not now
Differentiate with respect to time
$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$
both $x$ and $y$ are functions of $t$, so each square contributes a chain rule factor; the right hand side is a constant and gives zero
$\frac{dy}{dt}=-\frac{x}{y}\cdot\frac{dx}{dt}$
solving; the minus sign is already telling you that the two motions must be in opposite directions
Independent check by solving explicitly: on this configuration $y=\sqrt{25-x^{2}}$, so $\frac{dy}{dt}=\frac{-x}{\sqrt{25-x^{2}}}\cdot\frac{dx}{dt}=-\frac34(0.3)=-0.225$ m/s — the same number without ever differentiating implicitly. The size also passes inspection: at this instant $x
One implicit differentiation with respect to time.
The negative sign is the answer's most informative part: the top of the ladder is going down at $0.225$ m/s. Report the sentence, not just the number. And notice what happens as $y\to0$: the formula sends the speed to infinity, which is the model breaking down rather than the ladder doing something remarkable.
Checkpoint
§04.7 — a growing cube●●○○○
Thirty seconds. One relation, one chain rule, and a unit check at the end.
Given
a cube whose edge length $s$ grows at $\dfrac{ds}{dt}=2$ cm/s
the instant asked about is when $s=5$ cm
the volume of a cube of edge $s$ is $V=s^{3}$
Find
(a) Find $\dfrac{dV}{dt}$ at that instant, with units.
Hint 1/4
You want a rate of volume, and you are given a rate of length. Something has to connect the two, and it is the volume formula.
Hint 2/4
Chain rule with time underneath: $\dfrac{dV}{dt}=\dfrac{dV}{ds}\cdot\dfrac{ds}{dt}$, and $\dfrac{dV}{ds}=3s^{2}$ from the power rule.
Hint 3/4
$\dfrac{dV}{dt}=3s^{2}\dfrac{ds}{dt}$, and at the instant asked about $s=5$ cm with $\dfrac{ds}{dt}=2$ cm/s.
Hint 4/4
$\dfrac{dV}{dt}=3(25)(2)=150$ cm$^{3}$/s.
Show solutionDifferentiate the relation, symbols only
$\frac{dV}{dt}=3s^{2}\frac{ds}{dt}$
the chain rule with $s$ as the middle quantity; substituting $s=5$ before this line would have turned $V$ into the constant $125$
Substitute the instant
$=3(5)^{2}(2)=150$
the units are cm$^{2}$ times cm/s, that is cm$^{3}$/s, which is a volume per unit time as required
Independent check by counting slabs, no calculus: an edge growing at $2$ cm/s means each of the six faces moves outward at $1$ cm/s, and each face has area $5\times5=25$ cm$^{2}$. Six faces times $25$ cm$^{2}$ times $1$ cm/s is $150$ cm$^{3}$/s.
Every related rates answer should be read back with its units. If they do not come out as the quantity asked for, per unit time, something was differentiated with respect to the wrong thing.
⚠ Substituting the instant's numbers before differentiating
the numbers are given, so they feel like the first thing to use; and the expression does get shorter
one function has been fed into another, at any depth, and nothing is being multiplied or divided at the top level.
Name the layers, outside in
Write down what the outer function is and what its input is. If the input is itself built from another function, keep going. Notation like $\sin^{3}$ or $\sqrt{\ }$ hides a layer, so rewrite it as a bracket first.
One factor per layer
Write the derivative of each layer as a separate factor, each one evaluated at everything that was still inside it. Write all the factors down before simplifying any of them.
Stop at x
The last layer is the one whose input is $x$ itself. Its derivative is the final factor, and it is the one that goes missing most often.
Simplify last
Only now collect constants and use identities. Simplifying halfway through is how a middle factor gets lost.
Where it goes wrong
The outer derivative evaluated at $x$ rather than at the inner function.
The innermost factor left off, which is invisible when the inside is $x+c$ because that factor happens to be $1$.
A power written on the function name, as in $\tan^{2}x$, not recognised as an outer layer.
The four step implicit recipe
an equation ties $x$ and $y$ together and either cannot be solved for $y$ or would be unpleasant to solve.
Differentiate both sides with respect to x
Treat $y$ as a function of $x$ throughout. Every $y$ that gets differentiated leaves a $y'$ behind, and every product of $x$ and $y$ needs the product rule.
Collect the y prime terms
Move every term containing $y'$ to one side and everything else to the other. The equation is linear in $y'$, so this always succeeds.
Factor y prime out
One bracket, one factor. If $y'$ is still stuck inside a fraction or a product with itself, step 1 went wrong.
Divide, then check the denominator
Solve for $y'$. Say where the answer is valid: the denominator vanishing means a vertical tangent, not an error.
Where it goes wrong
A $y$ differentiated without leaving a $y'$ behind.
A term such as $6xy$ differentiated as if it were a single symbol.
The point substituted before the general formula is finished, which freezes both variables and produces zero.
The five step related rates recipe
one rate is given, another is wanted, and the two quantities are locked together by geometry or by a physical law.
Name everything as a function of time
Write $r(t)$, $x(t)$, $V(t)$ once, explicitly. This single habit prevents the commonest error in the whole method.
Write the equation that holds at every instant
Geometry, Pythagoras, a similar triangle ratio, a volume formula. If a quantity in your equation is only true at the one instant being asked about, the equation is wrong.
Differentiate with respect to t
Every variable contributes a chain rule factor. Constants — and only genuine constants such as a ladder's length or a cone's shape ratio — contribute nothing.
Substitute the instant, last
Now put in the numbers for the particular moment, including any quantity you had to compute from the others.
Report sign and units
Say what the sign means physically and check that the units read as the quantity asked for, per unit time.
Where it goes wrong
Numbers substituted at step 2 instead of step 4, which turns a moving quantity into a constant.
A second moving quantity treated as fixed because the problem only mentioned one rate out loud.
An angular rate left in degrees per second.
Sine on the outside: differentiating sin of x squared
The square happens first, the sine second.
Given
$p(x)=\sin\left(x^{2}\right)$
Find
$p'(x)$
SolutionIdentify and differentiate
$\text{outer}=\sin,\quad \text{inner}=x^{2}$
evaluating by hand, the sine is the last keystroke
the outer derivative keeps $x^{2}$ inside it untouched
Answer $$\boxed{p'(x)=2x\cos\left(x^{2}\right)}$$
Check
At $x=0$ the answer is $0$; near zero $\sin(x^{2})\approx x^{2}$, a parabola with a horizontal tangent at the origin.
Chain rule once.
Square on the outside: differentiating sine squared
The sine happens first, the square second.
Given
$s(x)=\sin^{2}x=(\sin x)^{2}$
Find
$s'(x)$
SolutionIdentify and differentiate
$\text{outer}=u^{2},\quad \text{inner}=\sin x$
evaluating by hand, the squaring is the last keystroke
$s'(x)=2\sin x\cdot\cos x=\sin 2x$
the outer derivative $2u$ is evaluated at $\sin x$, and the double angle identity tidies the result
Answer $$\boxed{s'(x)=\sin 2x}$$
Check
At $x=\pi/2$ the answer is $\sin\pi=0$, matching the fact that $\sin^{2}x$ reaches its maximum value $1$ there.
Chain rule once.
Identical symbols, opposite order, and nothing in common between the two answers: one carries a polynomial factor from the inside, the other carries a trigonometric one.
How to tell them apart
Read the expression as a sequence of keystrokes on a calculator. Whichever operation you would press last is the outer function. In $\sin(x^{2})$ that is the sine key; in $\sin^{2}x$ it is the squaring key.
The circle solved for y, then differentiated
The explicit route works here, and it costs a branch.
Given
$x^{2}+y^{2}=25$, upper half only
Find
$dy/dx$ on the upper half
SolutionSolve, then differentiate
$y=\sqrt{25-x^{2}}=(25-x^{2})^{1/2}$
choosing the upper branch; the lower one is a different function and has to be handled separately
At $x=3$ this gives $-3/4$, and the point on the upper half is $(3,4)$.
One rewrite plus one chain rule, valid on one branch.
The same circle differentiated implicitly
No solving, no branches, and both halves covered by one formula.
Given
$x^{2}+y^{2}=25$
Find
$dy/dx$ anywhere on the circle
SolutionDifferentiate as it stands
$2x+2y\,y'=0$
chain rule on $y^{2}$
$y'=-\frac{x}{y}$
the answer keeps $y$, and that is what lets it describe both halves at once
Answer $$\boxed{y'=-\frac{x}{y}}$$
Check
At $(3,4)$ it gives $-3/4$, matching the explicit route; at $(3,-4)$ on the lower half it gives $+3/4$, which the explicit formula above cannot produce at all.
One chain rule.
The two answers are the same function in disguise — substituting $y=\sqrt{25-x^{2}}$ into $-x/y$ reproduces the explicit formula exactly — but the implicit one also covers the half of the circle the explicit one had to throw away.
How to tell them apart
Solve for $y$ only when the solving is free and the curve has one branch above each $x$. Otherwise differentiate as it stands and accept an answer containing $y$.
Scaffolding comes off
The common skeleton
Rewrite any hidden layer as a bracket, so that roots and powers on function names become visible.
Name the layers from the outside in, and stop when the input is $x$ itself.
Write one derivative factor per layer, each evaluated at everything that was inside it, before simplifying anything.
Multiply the factors together.
Simplify only at the end, and check the result at one convenient value of $x$.
1 · fully worked
Fully worked: the derivative of sin(3x squared + 1)
Every line has its reason written next to it. Read the reasons, not just the algebra — the reasons are what you will need on the next rung.
Check at $x=0$: the formula gives $0$, and the function $\sin(3x^{2}+1)$ is even, so it must have a horizontal tangent at the origin.
Chain rule once.
The inner derivative $6x$ is not a constant, so unlike the linear case it changes the shape of the answer rather than just its scale.
2 · you write the reasoning
Easier algebra this time, and the algebra is already done for you. Your job is the column that is missing: write down why each line is allowed, in your own words, before opening the model answers. Writing the reason is the part that transfers to the next problem; copying the algebra is not.
$y=(4x+7)^{5}$
reasoning
Line 1: nothing to rewrite. The bracket is already visible, so the two layers can be read straight off.
outer $u^{5}$, inner $4x+7$
reasoning
Line 2: the last operation performed by hand would be raising to the fifth power, so that is the outer function and everything under it is the inner one.
$\frac{dy}{dx}=5(4x+7)^{4}\cdot 4$
reasoning
Line 3: the outer derivative $5u^{4}$ is evaluated at $u=4x+7$, and the inner derivative of a straight line is its slope, $4$. Two factors, one per layer.
$=20(4x+7)^{4}$
reasoning
Line 4: $5\times4=20$. Worth a check: at $x=0$ the formula gives $20\cdot 7^{4}=48020$, and expanding $(4x+7)^{5}$ would give a linear coefficient of $5\cdot 7^{4}\cdot 4$, the same number.
3 · find the buried error
Harder than the rung above, and nobody is checking your work now. The solution below differentiates $y=\sqrt{\cos 2x}$, reaches an answer and then evaluates it, and the work contains exactly two errors. Each line follows correctly from the one before it, so you cannot find them by looking for a jump.
Step 1. $y=(\cos 2x)^{1/2}$, so $y'=\tfrac12(\cos 2x)^{-1/2}\cdot(\cos 2x)'$.
Step 2. The middle layer: $(\cos 2x)'=-\sin 2x$.
Step 3. So $y'=\dfrac{-\sin 2x}{2\sqrt{\cos 2x}}$.
Step 4. At $x=\tfrac{\pi}{2}$: $y'=\dfrac{-\sin\pi}{2\sqrt{\cos\pi}}=\dfrac{0}{2\sqrt{-1}}=0$, so the graph has a horizontal tangent there.
the two buried errors (2)
⚠ step 2
the innermost layer was never differentiated. $\cos 2x$ is itself a composition, so its derivative is $-\sin(2x)\cdot 2=-2\sin 2x$, twice what is written.
the first chain rule was applied consciously and the second one silently was not; an inner factor of $2$ is small enough that the answer still looks right
right
With the missing factor, $y'=\dfrac{-\sin 2x}{\sqrt{\cos 2x}}$ — the $2$ from the inside cancels the $\tfrac12$ from the outside, so the denominator loses its $2$.
⚠ step 4
the point $x=\pi/2$ is not in the domain. There $\cos\pi=-1$, so $\sqrt{\cos 2x}$ is not a real number and the function itself does not exist, let alone its derivative. The line writes $\sqrt{-1}$ and then quietly stops looking at it because the numerator is zero.
substituting into a finished formula feels like arithmetic rather than like a step that can fail, and a zero on top hides whatever is underneath
right
Ask first where the answer is valid: we need $\cos 2x>0$, that is $2x$ within $\pi/2$ of a multiple of $2\pi$. At a legitimate point such as $x=\pi/4^{-}$ the slope runs to $-\infty$; at $x=0$ it is $0$, and there the horizontal tangent claim would actually be true.
4 · the bare problem
§04.4 — three layers, no scaffolding●●●●○
No hints written into the question, no layers named for you. The skeleton from this ladder is the whole method.
Given
$y=\tan^{3}\left(\sqrt{x}\right)$, which means $\bigl(\tan(\sqrt{x})\bigr)^{3}$
$x>0$, and $\sqrt{x}$ is not an odd multiple of $\pi/2$
Find
(a) Find $dy/dx$.
Hint 1/4
Before differentiating, count the layers by asking what you would do first, second and third if you evaluated this at $x=4$ by hand.
Hint 2/4
Chain rule through three layers: $\frac{d}{dx}f(g(h(x)))=f'(g(h(x)))\,g'(h(x))\,h'(x)$, with $(\tan u)'=\sec^{2}u$ and $(\sqrt{x})'=\frac{1}{2\sqrt{x}}$.
Hint 3/4
For $y=\bigl(\tan(\sqrt{x})\bigr)^{3}$ the three factors are $3\tan^{2}(\sqrt{x})$, then $\sec^{2}(\sqrt{x})$, then $\dfrac{1}{2\sqrt{x}}$.
Hint 4/4
Multiplying them gives $\dfrac{3\tan^{2}(\sqrt{x})\sec^{2}(\sqrt{x})}{2\sqrt{x}}$.
Show solutionMake the hidden layer visible
$y=\bigl(\tan(\sqrt{x})\bigr)^{3}$
the cube written on the function name is an outer layer; rewriting it as a bracket is what makes it countable
Sign check: for small $x>0$ the angle $\sqrt{x}$ is small and positive, so $\tan^{2}$ and $\sec^{2}$ are both positive and the whole expression is positive — matching a function that is increasing just to the right of the origin. Numerical check at $x=1$: the formula gives $3(2.4255)(3.4255)/2\approx 12.46$, and the difference quotient $[y(1.001)-y(1)]/0.001\approx 12.48$.
Three layers, three factors, no cancellation. The only way to get this wrong is to miscount the layers, which is why they are counted before anything is differentiated.
Full exam-style question
Exam format: an implicit curve, its tangent, and a point moving along itexam format
A midterm question in the shape they are actually set: three parts that build on each other, where part (c) is unreachable if part (a) went wrong. Twelve minutes is a realistic budget.
Given
the curve $x^{2}+xy+y^{2}=7$
the point $(1,2)$, which is on the curve since $1+2+4=7$
at that instant a particle moving along the curve has $\dfrac{dx}{dt}=3$ units per second
Find
(a) $dy/dx$ in terms of $x$ and $y$; (b) the tangent line at $(1,2)$; (c) $dy/dt$ at that instant
Solution
Part (a) is done implicitly rather than by solving the quadratic for $y$: solving is possible here but produces a square root whose derivative is far uglier, and it also forces a choice of branch that part (c) would then have to justify.
(a) Differentiate the equation as it stands
$2x+\bigl(y+x\,y'\bigr)+2y\,y'=0$
the middle term $xy$ is a product of two functions of $x$, so it contributes two pieces; the $y^{2}$ term contributes a chain rule factor
$y'\,(x+2y)=-(2x+y)$
collecting the two terms carrying $y'$ on one side and everything else on the other
$y'=-\frac{2x+y}{x+2y}$
dividing; the symmetry of the answer under swapping $x$ and $y$ is inherited from the same symmetry in the original equation, and is a free check that no term was mishandled
(b) Evaluate and write the line
$y'\big|_{(1,2)}=-\frac{2+2}{1+4}=-\frac{4}{5}$
substituting both coordinates; the denominator is $5$, so the tangent is not vertical here
$y-2=-\tfrac45\,(x-1)\ \Longrightarrow\ 4x+5y=14$
point slope form, cleared of fractions
(c) Convert the slope into a rate
$\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}$
the chain rule with time underneath is the only link between a slope and a speed; the particle's path supplies the first factor and its motion the second
$=-\tfrac45\cdot 3=-\tfrac{12}{5}=-2.4\ \text{units per second}$
the sign says the particle is descending as it moves, which matches a negative slope with $x$ increasing
Independent check on part (c), taking the whole equation to time directly: $2x\frac{dx}{dt}+\left(y\frac{dx}{dt}+x\frac{dy}{dt}\right)+2y\frac{dy}{dt}=0$. At $(1,2)$ with $\frac{dx}{dt}=3$ this is $6+\left(6+\frac{dy}{dt}\right)+4\frac{dy}{dt}=0$, so $5\frac{dy}{dt}=-12$ and $\frac{dy}{dt}=-2.4$. The two routes never shared a line, and they agree.
One implicit differentiation, one line, one chain rule with respect to time.
Part (c) is the whole point of the question: a slope is a ratio of rates, so once you have $dy/dx$ any one of the two rates gives you the other.
Practice
A · concept 4 questions
1§04.3 — a plausible looking chain rule●●○○○
A claim about compositions that copies the shape of the product rule. Decide whether it holds, and be ready to justify the decision with a single example rather than with a feeling.
Given
$f$ and $g$ are both differentiable everywhere
the claim: $(f\circ g)'(x)=f'(x)\cdot g'(x)$
Find
(a) True or false?
(b) Support your answer with one concrete pair of functions.
Hint 1/4
A claim about all functions is destroyed by one example, so look for the simplest pair you can compute both sides of.
Hint 2/4
The chain rule says $(f\circ g)'(x)=f'(g(x))\cdot g'(x)$; compare where each factor is evaluated.
Hint 3/4
Take $f(u)=u^{2}$ and $g(x)=x+1$. Then $(f\circ g)(x)=(x+1)^{2}$, whose derivative is $2(x+1)$. The claim would give $f'(x)g'(x)=2x\cdot1=2x$.
Hint 4/4
$2(x+1)\neq 2x$, so the claim is false.
Show solution
One counterexample settles a universal claim faster than any general argument, and $f(u)=u^{2}$ with $g(x)=x+1$ is the pair to pick because the composite can be expanded, so its derivative needs no rule that is itself under dispute.
the composite can be expanded here, so its derivative needs no rule under dispute
$f'(x)\,g'(x)=2x\cdot 1=2x$
this is what the claim predicts
Compare
$2(x+1)-2x=2\neq 0$
the gap is a constant $2$, so the two never agree anywhere, not even at a single lucky point
Answer $$\boxed{\text{False}}$$
Check
The chain rule's version, $f'(g(x))g'(x)=2(x+1)\cdot1$, reproduces the expanded answer exactly, which also confirms that the error is in where $f'$ gets evaluated rather than in the multiplication.
The product of derivatives is a different object from the derivative of a composition. The word times in the chain rule refers to the two factors, not to the two functions.
2§04.5 — differentiating an equation●○○○○
The first line of an implicit differentiation, written by a student in a hurry. Everything about it looks like the original equation, which is exactly the problem.
Given
the equation $x^{2}+y^{2}=25$, with $y$ a function of $x$
the claim: differentiating both sides with respect to $x$ gives $2x+2y=0$
Find
(a) True or false?
(b) If false, write the correct line.
Hint 1/4
Ask what kind of object $y$ is in this equation. If it is a function of $x$, then differentiating anything built from it is a chain rule problem.
Hint 2/4
$\frac{d}{dx}\bigl[y^{n}\bigr]=n\,y^{n-1}\frac{dy}{dx}$ whenever $y$ depends on $x$.
Hint 3/4
With $n=2$ the term $y^{2}$ gives $2y\frac{dy}{dx}$, so the line should read $2x+2y\frac{dy}{dx}=0$.
Hint 4/4
False; the missing factor is $\frac{dy}{dx}$.
Show solution
We differentiate the equation as it stands instead of solving $x^{2}+y^{2}=25$ for $y$ first: solving splits the circle into two branches and buries the very factor the claim is missing.
Differentiate term by term
$\frac{d}{dx}x^{2}=2x$
an ordinary power of the independent variable
$\frac{d}{dx}y^{2}=2y\frac{dy}{dx}$
a power of a function of $x$, so the chain rule contributes the inner derivative
without the missing factor the equation would say $x=-y$, which is a line rather than a slope
Answer $$\boxed{2x+2y\frac{dy}{dx}=0}$$
Check
Test the claimed version at the point $(3,4)$: it would say $6+8=0$, which is false, so the line cannot be an identity on the circle. The corrected version says $6+8\frac{dy}{dx}=0$, giving the slope $-3/4$ that the picture shows.
Count the $y$ symbols that got differentiated and then count the $dy/dx$ factors in your line. The two numbers must match.
3§04.4 — which rule fires first●●○○○
Four expressions, all of them mixtures. Only one of them has a multiplication as its outermost operation, and the question is which — not what the derivative is.
Given
$P=\sqrt{x\sin x}$
$Q=x^{3}\cos(2x)$
$R=\dfrac{\sin(3x)}{x^{2}+1}$
$S=\tan^{4}(x+1)$
Find
(a) Which expression needs the product rule as its very first step?
Hint 1/4
Do not differentiate anything. For each one, imagine evaluating it by hand at $x=1$ and watch which operation you perform last.
Hint 2/4
The outermost operation names the first rule: a final multiplication means the product rule, a final division means the quotient rule, and applying a function to what you already have means the chain rule.
Hint 3/4
For $P$ the last act is a square root; for $Q$ it is multiplying $1$ by $\cos 2$; for $R$ it is a division; for $S$ it is raising to the fourth power.
Hint 4/4
$Q$ is the only one whose last operation is a multiplication.
Show solution
We read each expression from the outside in rather than differentiating all four, because the question is which rule fires first and that is decided by the outermost operation alone.
a power written on the function name is still an outer layer
Answer $$\boxed{Q}$$
Check
Cross check by counting terms: only $Q$ and $R$ can produce two terms, and $R$'s two terms sit inside a single fraction. $P$ and $S$ must come out as one term each.
This question is worth answering silently before every derivative you take for the rest of the course.
4§04.1 — units inside a trig derivative●●○○○
A statement that is either harmless or disastrous depending on one word that the formula does not mention out loud.
Given
the claim: $\dfrac{d}{dx}\sin x=\cos x$ holds whether $x$ is measured in degrees or in radians
Find
(a) True or false?
(b) If false, give the correct derivative in the other unit.
Hint 1/4
Ask where the proof of the formula used the unit at all, and whether anything in it would survive a change of scale.
Hint 2/4
The proof runs through $\lim_{h\to0}\frac{\sin h}{h}=1$, and that limit is a statement about $h$ in radians.
Hint 3/4
In degrees the same limit becomes $\lim_{h\to0}\frac{\sin(h^{\circ})}{h}=\frac{\pi}{180}\approx0.0175$, so the constant that came out as $1$ now comes out as $\pi/180$.
Hint 4/4
False: in degrees the derivative is $\frac{\pi}{180}\cos(x^{\circ})$.
Show solution
We go back to the difference quotient instead of arguing about the formula, because the unit enters through one limit and nowhere else, and only the limit makes it visible.
Numerical check: $\sin(1^{\circ})=0.017452$ and $\sin(0^{\circ})=0$, so the slope near zero on a degree scale is about $0.0175$ per degree, not $1$.
This is the only place in the section where the unit changes an answer rather than merely its presentation, which is why the convention is stated at the top of the page.
B · computation 6 questions
1§04.3 — a polynomial inside a cosine●●○○○
The standard two layer quiz pattern: a polynomial buried inside a trigonometric function. The only decisions are where the minus sign goes and what gets left behind.
Given
$f(x)=\cos\left(x^{3}-2x+5\right)$
Find
(a) Find $f'(x)$.
(b) Evaluate $f'(1)$ to three decimal places, with the angle in radians.
(c) Rewrite part (a) in Leibniz form using $u=x^{3}-2x+5$.
Hint 1/4
Two layers only. Decide which function is on the outside before you write a single symbol.
Hint 2/4
Chain rule with $(\cos u)'=-\sin u$: the derivative is $-\sin(u)\cdot u'$.
Hint 3/4
With $u=x^{3}-2x+5$ we get $u'=3x^{2}-2$, so $f'(x)=-\sin\left(x^{3}-2x+5\right)\cdot\left(3x^{2}-2\right)$. At $x=1$: $u=1-2+5=4$ and $u'=3-2=1$.
Hint 4/4
$f'(1)=-\sin 4\approx 0.757$, positive because $\sin 4$ is negative.
Show solution
The inner function is named before anything is differentiated, because the outer derivative has to be evaluated at that whole expression; there is nothing to expand here, so the layers are the only route.
Name and differentiate the layers
$u=x^{3}-2x+5,\qquad f=\cos u$
the cosine is the last operation performed by hand, so it is the outer function
Numerical check: $f(1)=\cos 4\approx-0.65364$ and $f(1.001)\approx-0.65288$, so the difference quotient is about $0.76$, matching the sign and the size.
The sign of the final number came from the angle being $4$ radians rather than from anything in the differentiation. Keep the two sources of sign separate when you check your work.
2§04.1 — a trig quotient that collapses●●●○○
A quotient of trigonometric functions whose derivative looks unpleasant for exactly one line, and then does not.
Given
$h(\theta)=\dfrac{1-\cos\theta}{\sin\theta}$
$\theta\in(0,\pi)$
Find
(a) Apply the quotient rule and expand the numerator fully.
(b) Simplify using the Pythagorean identity.
(c) Show that the result can be written as $\dfrac{1}{1+\cos\theta}$.
Hint 1/4
Expand everything before you look for cancellation. The identity is invisible while the brackets are still closed.
Hint 2/4
Quotient rule $\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^{2}}$, with $u'=\sin\theta$ and $v'=\cos\theta$.
Hint 3/4
With $u=1-\cos\theta$ and $v=\sin\theta$: the numerator is $\sin\theta\cdot\sin\theta-(1-\cos\theta)\cos\theta=\sin^{2}\theta-\cos\theta+\cos^{2}\theta$.
Hint 4/4
That numerator is $1-\cos\theta$, so $h'(\theta)=\dfrac{1-\cos\theta}{\sin^{2}\theta}=\dfrac{1}{1+\cos\theta}$.
Show solution
We differentiate first and use the identity afterwards: the Pythagorean identity is applied to the finished numerator and then, in reverse, to the denominator, because it is the factoring $\sin^{2}\theta=(1-\cos\theta)(1+\cos\theta)$ that produces the cancellation. Rewriting $h$ with an identity before differentiating only trades one quotient for another.
Independent route: the half angle identity gives $h(\theta)=\tan(\theta/2)$, whose derivative is $\tfrac12\sec^{2}(\theta/2)=\frac{1}{2\cos^{2}(\theta/2)}$, and $2\cos^{2}(\theta/2)=1+\cos\theta$. Same function, no quotient rule used.
The identity earns its keep in both directions: forwards to collapse a sum of squares, backwards to factor a denominator.
3§04.4 — a root over a trig function●●●○○
A root on the outside and a composition on the inside: two chain rules, one inside the other, and one place where a factor of $2$ cancels.
Given
$y=\sqrt{1+\sin(2x)}$
restrict to $x$ where $1+\sin(2x)>0$
Find
(a) Name the three layers.
(b) Find $dy/dx$.
(c) Evaluate $dy/dx$ at $x=0$.
Hint 1/4
Write the root as a power first. Then count the layers before differentiating any of them.
Hint 2/4
Chain rule twice: $\frac{d}{dx}u^{1/2}=\frac12 u^{-1/2}u'$, and $\frac{d}{dx}\sin(2x)=2\cos(2x)$.
Hint 3/4
With $u=1+\sin(2x)$: $u'=2\cos(2x)$, so $\frac{dy}{dx}=\frac{2\cos(2x)}{2\sqrt{1+\sin(2x)}}$.
Hint 4/4
The twos cancel: $\frac{dy}{dx}=\frac{\cos(2x)}{\sqrt{1+\sin(2x)}}$, which is $1$ at $x=0$.
Show solution
The root is rewritten as a power before anything else, since without $u^{1/2}$ there is no outer derivative to apply, and all three layers are named before any is used so that the innermost factor $2$ cannot be dropped.
Numerical check at $x=0$: $y(0)=1$ and $y(0.001)=\sqrt{1+0.002}\approx1.0009995$, so the difference quotient is about $1.0$.
When the twos cancel like this, the answer no longer displays the layer that produced them. Do not read that as evidence that the layer was not needed.
4§04.5 — a tangent line on an ellipse●●●○○
Implicit differentiation on a curve that could be solved for $y$ but is much cheaper not to. The constants underneath are where the errors hide.
Given
$\dfrac{x^{2}}{16}+\dfrac{y^{2}}{9}=1$
the point $\left(2,\ \tfrac{3\sqrt{3}}{2}\right)$, on the curve since $\tfrac{4}{16}+\tfrac{27/4}{9}=\tfrac14+\tfrac34=1$
Find
(a) Differentiate both sides with respect to $x$.
(b) Solve for $y'$ in terms of $x$ and $y$.
(c) Write the tangent line at the given point.
Hint 1/4
The denominators $16$ and $9$ are constants, so they simply ride along. Differentiate as if they were not there and then put them back.
Hint 2/4
$\frac{d}{dx}\left[\frac{y^{2}}{9}\right]=\frac{2y}{9}y'$ by the chain rule, and $\frac{d}{dx}\left[\frac{x^{2}}{16}\right]=\frac{x}{8}$.
Hint 3/4
$\frac{x}{8}+\frac{2y}{9}y'=0$, so $y'=-\frac{9x}{16y}$. At $x=2$, $y=\frac{3\sqrt{3}}{2}$ this is $-\frac{18}{24\sqrt{3}}$.
Hint 4/4
$y'=-\frac{\sqrt{3}}{4}\approx-0.433$, so the tangent is $y-\frac{3\sqrt{3}}{2}=-\frac{\sqrt{3}}{4}(x-2)$.
Show solution
We differentiate the ellipse as it stands rather than solving for $y=\tfrac34\sqrt{16-x^{2}}$: solving forces a choice of branch and a square root whose derivative needs its own chain rule, while the implicit route keeps the two constants in plain sight.
Differentiate, constants and all
$\frac{2x}{16}+\frac{2y}{9}y'=0$
each term is a constant multiple of a square, so the constant survives untouched and the chain rule acts only on $y^{2}$
Check by the explicit route on the upper half: $y=\frac34\sqrt{16-x^{2}}$ gives $y'=\frac{-3x}{4\sqrt{16-x^{2}}}$, and at $x=2$ that is $\frac{-6}{4\sqrt{12}}=-\frac{\sqrt3}{4}$ — the same slope. And a sanity check on the line: at $x=2$ it gives $-\frac{\sqrt3}{2}+2\sqrt3=\frac{3\sqrt3}{2}$, which is the point we started from.
The general formula $y'=-\frac{b^{2}x}{a^{2}y}$ for the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ falls out of the same two lines, and the circle's $-x/y$ is the case $a=b$.
5§04.6 — a second derivative on an ellipse●●●●○
The second derivative of an implicit curve, where the whole difficulty is the substitution in the middle and the use of the original equation at the end.
Given
$x^{2}+4y^{2}=8$
Find
(a) Show that $y'=-\dfrac{x}{4y}$.
(b) Find $y''$ in terms of $x$ and $y$.
(c) Simplify $y''$ using the original equation.
Hint 1/4
Part (b) differentiates a fraction whose denominator contains $y$, and $y$ is a function of $x$. That is the only thing being tested.
Hint 2/4
Quotient rule on $-\frac{x}{4y}$, remembering that $\frac{d}{dx}[4y]=4y'$, then substitute the known $y'$ back in.
Hint 3/4
$y''=-\frac{(1)(4y)-x(4y')}{16y^{2}}$, and putting $y'=-\frac{x}{4y}$ in gives $y''=-\frac{4y+\frac{x^{2}}{y}}{16y^{2}}$.
Hint 4/4
Multiplying top and bottom by $y$ gives $y''=-\frac{4y^{2}+x^{2}}{16y^{3}}=-\frac{8}{16y^{3}}=-\frac{1}{2y^{3}}$.
Show solution
The known $y'$ is substituted into $y''$ while the expression is still small, not at the end: carrying $y'$ through the simplification leaves a compound fraction containing $y'$ inside it, and the substitution then has to be done in the worst possible place.
First derivative
$2x+8y\,y'=0\ \Longrightarrow\ y'=-\frac{x}{4y}$
the $4y^{2}$ term gives $8y\,y'$; the factor $4$ stays because it multiplies the whole square
Differentiate again
$y''=-\frac{(1)(4y)-x\,(4y')}{16y^{2}}$
quotient rule on $-\frac{x}{4y}$; the denominator's derivative is $4y'$, not $4$, because $y$ moves with $x$
Check on the upper half, where the curve can be solved: $y=\tfrac12\sqrt{8-x^{2}}$ gives $y'=\frac{-x}{2\sqrt{8-x^{2}}}$ and then $y''=\frac{-8}{2\left(8-x^{2}\right)^{3/2}}$. Since $8-x^{2}=4y^{2}$, that is $\frac{-8}{2\cdot 8y^{3}}=-\frac{1}{2y^{3}}$ — the same. Sign check: on the upper half $y>0$ so $y''<0$, and the arc is concave down.
Look for the original equation inside your expression before expanding anything. It is the difference between $-\frac{x^{2}+4y^{2}}{16y^{3}}$ and $-\frac{1}{2y^{3}}$.
6§04.7 — inflating a spherical balloon●●●○○
A rate of volume is given and a rate of length is wanted, which is the reverse of the usual direction and is where the division goes wrong.
Given
air is pumped in at $\dfrac{dV}{dt}=100$ cm$^{3}$/s
the instant asked about is when the radius is $r=10$ cm
the volume of a sphere of radius $r$ is $V=\tfrac43\pi r^{3}$
Find
(a) Find $\dfrac{dr}{dt}$ at that instant.
(b) Say in one sentence what happens to that rate as the balloon gets bigger, and why.
Hint 1/4
Both quantities move with time. Write the relation between them first, with the radius kept as a symbol.
Hint 2/4
Chain rule with time underneath: $\frac{dV}{dt}=\frac{dV}{dr}\cdot\frac{dr}{dt}$, and $\frac{dV}{dr}=4\pi r^{2}$.
Hint 3/4
$100=4\pi r^{2}\frac{dr}{dt}$, and at the instant asked about $r=10$, so $4\pi(100)=400\pi$.
We differentiate the volume formula rather than first solving it for $r$: $r=\left(\tfrac{3V}{4\pi}\right)^{1/3}$ is available but brings a cube root back into the derivative, whereas differentiating $V$ directly leaves the surface area as the factor and keeps the units readable.
Relate the two rates
$\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}$
differentiating $\tfrac43\pi r^{3}$ with respect to $t$; the factor $4\pi r^{2}$ is the sphere's surface area, which is not a coincidence
Independent check by geometry: air arriving at $100$ cm$^{3}$/s has to spread over the whole surface, $4\pi(10)^{2}\approx1257$ cm$^{2}$, so the skin advances at $100/1257\approx0.0796$ cm/s. Same number, no calculus.
Because $\frac{dr}{dt}=\frac{dV/dt}{4\pi r^{2}}$, a steady supply of air moves the surface ever more slowly as the balloon grows: doubling the radius quarters the speed.
C · exam level 4 questions
1§04.3 — the chain rule from a table●●●○○
No formulas anywhere, only a table of values. This is the exam's way of checking whether you know where each factor is evaluated, because guessing is impossible without a formula to lean on.
Given
The table of values, valid at the three listed inputs only:
For each one, write the general rule with the number $2$ in it before you look up a single table entry. Which row you need is decided by the rule, not the other way round.
Hint 2/4
$(f\circ g)'(a)=f'(g(a))\cdot g'(a)$, and $(fg)'(a)=f'(a)g(a)+f(a)g'(a)$.
Hint 3/4
At $a=2$ the table gives $f(2)=1$, $f'(2)=-2$, $g(2)=3$, $g'(2)=6$, and the composition also needs $f'(3)=8$ and $g'(1)=5$.
Each general rule is written out before a single number is looked up, because the whole question is which input each factor is evaluated at, and a table invites reading the row for $x=2$ when the rule asks for the row for $g(2)$.
f composed with g
$h'(2)=f'\bigl(g(2)\bigr)\cdot g'(2)$
the general rule first, so that the table lookups are forced rather than chosen
$=f'(3)\cdot 6=8\cdot 6=48$
$g(2)=3$, so the outer derivative is needed at $3$ and not at $2$ — this single lookup is what the question is testing
g composed with f
$k'(2)=g'\bigl(f(2)\bigr)\cdot f'(2)$
the same rule with the roles swapped
$=g'(1)\cdot(-2)=5\cdot(-2)=-10$
$f(2)=1$, so this time the outer derivative is needed at $1$
f times g
$p'(2)=f'(2)g(2)+f(2)g'(2)$
the product rule needs no composition, so every entry comes from the row $x=2$
$=(-2)(3)+(1)(6)=-6+6=0$
the two terms cancel, which is a legitimate answer rather than a sign that something was lost
Structural check: $h'$ and $k'$ came out different, as they must, since $f\circ g$ and $g\circ f$ are different functions. And $p'(2)=0$ can be read as a statement about the product $f\cdot g$ having a flat spot at $x=2$, which is consistent with one factor rising while the other falls at matching rates.
If you find yourself reading only the row $x=2$ for parts (a) and (b), the outer derivative is being evaluated at the wrong place. The composition always sends you to a different row.
2§04.5 — an implicit cubic and its flat point●●●●○
A full midterm-shaped implicit question: the derivative, a tangent, and then the point where the curve levels off. Part (c) is unreachable if part (a) contains an error.
Given
the curve $x^{2}y+y^{3}=10$
the point $(1,2)$, on the curve since $1\cdot 2+8=10$
Find
(a) Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$.
(b) Find the tangent line at $(1,2)$.
(c) Find every point on the curve where the tangent is horizontal.
Hint 1/4
The first term is a product of two functions of $x$, so it needs the product rule; the second is a power of $y$, so it needs the chain rule. Handle them separately before combining.
Hint 2/4
$\frac{d}{dx}\bigl[x^{2}y\bigr]=2xy+x^{2}y'$ and $\frac{d}{dx}\bigl[y^{3}\bigr]=3y^{2}y'$. A horizontal tangent needs the numerator of $y'$ to vanish while the denominator does not.
Hint 3/4
$2xy+x^{2}y'+3y^{2}y'=0$ gives $y'=-\dfrac{2xy}{x^{2}+3y^{2}}$; at $(1,2)$ that is $-\dfrac{4}{1+12}$.
Hint 4/4
$y'(1,2)=-\tfrac{4}{13}$, and $y'=0$ needs $2xy=0$, which on this curve happens only at $x=0$, $y=10^{1/3}$.
Show solution
Part (a) is done implicitly rather than by solving $x^{2}y+y^{3}=10$ for $y$: solving means a cubic in $y$ whose formula is far worse to differentiate than the two lines the implicit route costs. Part (c) is then a matter of setting the numerator to zero, which is only possible because $y'$ was left as one fraction.
(a) Differentiate both sides
$2xy+x^{2}y'+3y^{2}y'=0$
the first term is a product of $x^{2}$ and $y$, so it splits in two; the second is a power of $y$, so it contributes one $y'$
$y'\left(x^{2}+3y^{2}\right)=-2xy$
collecting; the bracket can only vanish at the origin, which is not on this curve
Check part (c) against part (a) from the other side: at $x=0$ the curve reduces to $y^{3}=10$, a single point, and near it $y$ depends on $x$ only through the term $x^{2}y$, which is flat in $x$ at $x=0$. A horizontal tangent there is exactly what an even dependence predicts. Check the tangent line numerically: at $x=1.01$ the curve gives $y\approx1.9969$ and the line gives $y=30/13-4(1.01)/13\approx1.9969$.
A horizontal tangent kills the numerator, a vertical one kills the denominator, and every candidate has to be tested against the curve's own equation before it counts.
3§04.7 — water rising in a conical tank●●●●○
The related rates problem that appears in some form on almost every calculus midterm. The trap is that the tank has two dimensions that both change, and only one of them can be allowed to survive.
Given
an inverted cone, $10$ m tall, with top radius $4$ m
water enters at $\dfrac{dV}{dt}=2$ m$^{3}$/min
the instant asked about is when the depth is $h=5$ m
the volume of a cone of radius $r$ and height $h$ is $V=\tfrac13\pi r^{2}h$
Find
(a) Express $r$ in terms of $h$ using similar triangles.
(b) Find $\dfrac{dh}{dt}$ at the instant in question.
(c) Say whether the water rises faster or slower as the tank fills, and why.
Hint 1/4
The volume formula contains two variables and you only want one. Something about the shape of the cone ties them together at every depth.
Hint 2/4
Similar triangles give $\frac{r}{h}=\frac{4}{10}$; substitute before differentiating, then use $\frac{dV}{dt}=\frac{dV}{dh}\cdot\frac{dh}{dt}$.
Hint 3/4
With $r=0.4h$: $V=\frac13\pi(0.16h^{2})h=\frac{0.16\pi}{3}h^{3}$, so $\frac{dV}{dt}=0.16\pi h^{2}\frac{dh}{dt}$. At $h=5$ that coefficient is $0.16\pi(25)=4\pi$.
Hint 4/4
$2=4\pi\frac{dh}{dt}$, so $\frac{dh}{dt}=\frac{1}{2\pi}\approx0.159$ m/min.
Show solution
The radius is eliminated before differentiating rather than after, because $r$ and $h$ are both moving and keeping both would leave two unknown rates in one equation.
the water's surface and the tank's rim are similar triangles, and this holds at every depth rather than only at $h=5$ — which is precisely why it may be substituted before differentiating
$V=\tfrac13\pi(0.4h)^{2}h=\frac{0.16\pi}{3}h^{3}$
with one variable left, the differentiation has one term
Differentiate with respect to time
$\frac{dV}{dt}=0.16\pi h^{2}\frac{dh}{dt}$
chain rule; the coefficient $0.16\pi h^{2}$ is exactly the area of the water's surface, which is the sanity check for this line
Independent check with no calculus: at depth $5$ m the surface is a circle of radius $0.4(5)=2$ m, area $4\pi\approx12.57$ m$^{2}$. Two cubic metres per minute spread over that area raises the level by $2/12.57\approx0.159$ m per minute. The same number, and it explains part (c) at the same time.
The rate of rise is the inflow divided by the surface area, always. As the tank fills the surface widens, so the same inflow lifts the level more slowly — at $h=10$ the rate is four times smaller than at $h=5$.
4§04.4 — a secant with a quadratic inside●●●○○
Four candidate derivatives for one expression. Three of them are what you get by making one specific mistake, so identifying the mistake is faster than checking the algebra four times.
Given
$y=\sec\left(3x^{2}\right)$
recall $\dfrac{d}{du}\sec u=\sec u\tan u$
Find
(a) Which expression is $\dfrac{dy}{dx}$?
Hint 1/4
Two layers: a secant on the outside and a quadratic on the inside. Decide what each layer contributes before comparing the options.
Hint 2/4
Chain rule: the outer factor is $\sec\left(3x^{2}\right)\tan\left(3x^{2}\right)$ and the inner factor is the derivative of $3x^{2}$.
Hint 3/4
With $u=3x^{2}$ we have $u'=6x$, so the answer is $\sec(u)\tan(u)\cdot 6x$ with $u=3x^{2}$ put back.
We build the derivative from the layers instead of testing the four options one at a time: naming the outer and the inner factor takes two lines and identifies the mistake behind each wrong option for free.
Check at $x=0$: the formula gives $0$, and $\sec(3x^{2})$ is an even function with a minimum at the origin, so a horizontal tangent there is required. Numerical check at $x=0.5$: the formula gives $3\sec(0.75)\tan(0.75)\approx 3(1.3667)(0.9316)\approx 3.82$, and the difference quotient of $\sec(3x^{2})$ at $0.5$ with step $0.001$ is $3.84$.
Two derivatives start with a secant and they are not interchangeable: $\sec^{2}$ belongs to the tangent, $\sec\tan$ belongs to the secant.
D · interleaved 3 questions
1§04.1 — a limit with no obvious method●●●○○
This one is deliberately not labelled. Before reaching for any technique, look at the shape of the expression and ask what it reminds you of. There is a route that takes one line and a route that takes twenty.
Look at the shape rather than at the trigonometry: something is evaluated at a point plus $h$, something is evaluated at the point itself, they are subtracted, and the result is divided by $h$.
Hint 2/4
That shape is the definition $f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$, here with $f=\sin$ and $a=\frac{\pi}{6}$.
Hint 3/4
With $f(x)=\sin x$ and $a=\frac{\pi}{6}$, note that $f(a)=\sin\frac{\pi}{6}=\frac12$, exactly the constant being subtracted. So the limit is $f'\left(\frac{\pi}{6}\right)$.
Hint 4/4
$f'(x)=\cos x$, so the limit is $\cos\frac{\pi}{6}=\frac{\sqrt3}{2}\approx0.866$.
Show solution
Recognising the difference quotient is the one line route. The angle addition expansion also works and is used above only as the check, because it takes five lines rather than one.
the constant $\tfrac12$ is not decoration: it is $\sin\tfrac{\pi}{6}$, which is what makes the expression a difference quotient rather than merely similar to one
$\text{so the limit is } f'\!\left(\tfrac{\pi}{6}\right)$
by the definition of the derivative, read from right to left
the derivative of sine, evaluated at the base point
Answer $$\boxed{\frac{\sqrt3}{2}}$$
Check
Numerical check: with $h=0.001$ the quotient is $\frac{\sin(0.5245988)-0.5}{0.001}\approx0.8658$, closing on $0.8660$. Expanding with the angle addition formula instead gives $\frac12\cdot\frac{\cos h-1}{h}+\frac{\sqrt3}{2}\cdot\frac{\sin h}{h}\to\frac12(0)+\frac{\sqrt3}{2}(1)$ — the same answer by the long route.
Any limit of the form something at a plus h, minus the same thing at a, over h is a derivative in disguise, and naming the function is the whole job.
2§04.3 — joining two pieces smoothly●●●●○
A function defined in two pieces, with two constants free. The question asks for the values that make the join invisible to a derivative, which is a stronger demand than making it invisible to the eye.
(a) Find the condition on $b$ that makes $f$ continuous at $0$.
(b) Find the condition on $a$ that makes $f$ differentiable at $0$.
(c) State both values and say why the order of the two parts matters.
Hint 1/4
Two separate demands are being made at the same point: the two pieces must meet, and they must meet at the same slope. Deal with meeting first.
Hint 2/4
Continuity at $0$ needs the two one sided values to agree; differentiability needs the two one sided derivatives to agree, and $\frac{d}{dx}\sin(2x)=2\cos(2x)$ by the chain rule.
Hint 3/4
From the left, $f\to\sin 0=0$ and $f'\to 2\cos 0=2$. From the right, $f\to b$ and $f'\to a$.
Hint 4/4
$b=0$ and $a=2$.
Show solution
Continuity is settled before differentiability, because the slope condition is written at a point where the two pieces already meet; matching slopes first would fix $a$ from a join that might not exist.
Make the pieces meet
$\lim_{x\to0^{-}}\sin(2x)=0,\qquad f(0)=b$
the left hand limit and the value at the join must agree for continuity
$b=0$
which pins the second piece down to $ax$
Make the slopes meet
$\text{from the left: } \frac{d}{dx}\sin(2x)=2\cos(2x)\to 2\ \text{as } x\to0^{-}$
the chain rule supplies the factor $2$; without it the answer would come out as $a=1$
$\text{from the right: } \frac{d}{dx}(ax)=a$
the slope of a line is the same everywhere, including at its endpoint
$a=2$
the two one sided derivatives must be equal
Say why the order matters
$\text{continuity}\ \Leftarrow\ \text{differentiability, not the reverse}$
if the pieces did not meet, the difference quotient from one side would blow up and no choice of $a$ could rescue it
Answer $$\boxed{a=2,\quad b=0}$$
Check
Check by comparing the two formulas near zero: $\sin(2x)\approx 2x$ for small $x$, and the right hand piece with $a=2$, $b=0$ is exactly $2x$. The two pieces agree to first order, which is what differentiability at the join means.
The factor of $2$ is the chain rule's entire contribution here, and getting it wrong changes the answer rather than just its appearance.
3§04.1 — where a wave levels off●●●○○
A short question that needs one derivative, one equation and one interval restriction. Deciding which of those three is the hard part is left to you.
Given
$y=\sin x+\cos x$ on the interval $[0,2\pi]$
Find
(a) Find every point of the interval where the tangent is horizontal.
(b) Find the value of $y$ at each of those points.
(c) Say which one is the maximum and which the minimum, using the values rather than a second derivative.
Hint 1/4
A horizontal tangent is a statement about the derivative, so the first move is to have the derivative rather than the function.
Hint 2/4
Set $y'=0$ and solve on the given interval; $(\sin x)'=\cos x$ and $(\cos x)'=-\sin x$.
Hint 3/4
$y'=\cos x-\sin x$, so $y'=0$ means $\cos x=\sin x$, that is $\tan x=1$, and on $[0,2\pi]$ that happens at $x=\frac{\pi}{4}$ and $x=\frac{5\pi}{4}$.
Hint 4/4
At $\frac{\pi}{4}$, $y=\sqrt2$; at $\frac{5\pi}{4}$, $y=-\sqrt2$.
Show solution
Setting $y'=0$ is turned into $\tan x=1$ rather than solved graphically, because one equation with a known period gives both solutions on $[0,2\pi]$ at once, and the maximum and minimum are then read off the two values, which is cheaper than a second derivative.
Differentiate and set to zero
$y'=\cos x-\sin x$
the minus comes from the derivative of the cosine, not from anything in the original expression
Independent check by rewriting: $\sin x+\cos x=\sqrt2\sin\!\left(x+\tfrac{\pi}{4}\right)$, a single wave of amplitude $\sqrt2$. Its peaks and troughs are at $x+\tfrac{\pi}{4}=\tfrac{\pi}{2}$ and $\tfrac{3\pi}{2}$, that is $x=\tfrac{\pi}{4}$ and $\tfrac{5\pi}{4}$, with values $\pm\sqrt2$ — the same answer without differentiating at all.
Comparing values is enough to sort a maximum from a minimum when there are only two candidates on a closed interval. The machinery for doing it in general arrives in a later section.
Mistake ledger (19 entries)
⚠ The cofunction minus sign goes missing
sine and cosine feel like mirror images of each other, so the minus reads as decoration rather than as content
wrong$\frac{d}{dx}\cos x=\sin x$
right$\frac{d}{dx}\cos x=-\sin x$
⚠ Degrees fed into a trigonometric derivative
the calculator was left in degree mode, or the problem quoted an angle in degrees and nobody converted it
Close the page. On one sheet, from memory: the six trigonometric derivatives with the three that carry a minus sign marked; the chain rule in both notations; the four steps of the implicit recipe; and the one sentence that says what goes wrong if you substitute the instant's numbers too early in a related rates problem. Then open the formula card and mark what was missing. What you missed is your revision list, and the rest of this section is not.
Prove that $(\sin x)'=\cos x$ from the difference quotient, and point at the line where radians were needed?
c-trig-derivatives
Write all six trigonometric derivatives and say which three carry a minus sign, and what those three have in common?
c-trig-family
Differentiate $\sqrt{x^{2}+9}$ and say out loud where the outer derivative is being evaluated?
c-chain-rule
Look at $x^{3}\cos(2x)$ and $\sqrt{x\sin x}$ and name the first rule for each without differentiating either?
c-chain-structure
Differentiate $x^{3}+y^{3}=6xy$ and reach $y'$ without leaving a $y'$ on the wrong side?
c-implicit
Take $y'=-x/y$ and produce $y''=-25/y^{3}$, including the substitution step in the middle?
c-implicit-second
State why substituting $r=75$ before differentiating turns the ripple problem into the answer zero?
c-related-rates
Glossary (14 terms)
bileşke fonksiyon
A function built by feeding one function into another, written $f\circ g$ and evaluated as $f$ applied to $g(x)$.
chain rulezincir kuralı
The rule for differentiating a composition: differentiate the outer function at the inner one, then multiply by the derivative of the inner one.
outer functiondış fonksiyon
In a composition, the function applied last; it is the one whose derivative keeps the whole inside untouched.
inner functioniç fonksiyon
In a composition, the function applied first; its derivative is the factor the chain rule leaves behind.
implicit differentiationkapalı türev alma
Differentiating an equation relating $x$ and $y$ without solving it for $y$, treating $y$ as a function of $x$ throughout.
açık fonksiyon
A function given by a formula that already isolates $y$ on one side, such as $y=\sqrt{25-x^{2}}$.
related rates
A problem in which two quantities change with time and are tied together by one equation, so that one rate determines the other.
radianradyan
The angle unit in which the derivative formulas of this course are true; one radian is the angle whose arc equals the radius.
cofunctionkofonksiyon
Cosine, cotangent and cosecant; the three whose derivatives carry a minus sign.
Pythagorean identityPisagor özdeşliği
The relation $\sin^{2}\theta+\cos^{2}\theta=1$, together with $1+\tan^{2}\theta=\sec^{2}\theta$, used to collapse a sum of two squares of trigonometric function values into the number one.
folium of DescartesDescartes yaprağı
The curve $x^{3}+y^{3}=6xy$, a closed loop through the origin that cannot be solved for $y$ in elementary terms.
normal linenormal doğru
The line through a point of a curve perpendicular to the tangent there; its slope is the negative reciprocal of the tangent's.
horizontal tangentyatay teğet
A point where the derivative is zero, so the curve levels off; on an implicit curve it is where the numerator of $y'$ vanishes and the denominator does not.
açısal hız
The rate at which an angle changes with time, which must be in radians per unit time before any trigonometric derivative is applied to it.
What comes next
§05 · Linear approximation, differentials, extreme values, and the Mean Value Theorem
This section could find the slope anywhere on a curve, but it never asked what that slope is good for beyond drawing a line. Next comes the answer: a tangent line is the cheapest possible model of a function, accurate enough to estimate values by hand, and the same derivative that built it also decides where a function reaches its largest and smallest values.
Sources
James Stewart, Calculus, Metric Version, Ninth Edition — sections 2.4, 2.5, 2.6 and 2.8 The topic split follows the week 4 line of the course syllabus, and section numbers are quoted only as they appear there. Section 2.7 is not on that line and is not covered.
Conventions used on this page Trigonometric arguments are radians throughout; an implicit derivative containing both $x$ and $y$ is a finished answer; rates are reported with their sign and their units.