18 worked examples26 exercises5 figuresbölüm 3 / 14
03Derivatives: definition and basic differentiation rules
An e-scooter pulls away from a red light, and after $t$ seconds it has covered $t^{2}$ metres. Over the second between $t=3$ and $t=4$ it covers $16-9=7$ m, so $7$ m/s; over the tenth of a second after $t=3$ the same arithmetic gives $6.1$ m/s, and over the hundredth, $6.01$ m/s. Every one of those numbers is an average over a stretch of road, and the one number the rider actually sees on the display at the instant $t=3$ is the one we cannot compute: at a single instant the scooter covers $0$ m in $0$ s.
By the end of this section you can produce that instantaneous reading exactly — $6$ m/s — in two independent ways: from the limit that defines it, and in one line from a rule that works just as well on $t^{5}$, $\sqrt{t}$ and $1/t$.
In 60 seconds
A is the limit of an average slope as the interval shrinks to nothing; you compute a handful of them from that limit, and then four rules — power, constant multiple and sum, product, quotient — do the rest for every algebraic function in this course.
Derivative at a point
$f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$
slope of a curve at one point, speed at one instant, and any question whose wording is using the definition
$\frac{d}{dx}x^{n}=n\,x^{n-1}$
every power of $x$, once roots and reciprocals are rewritten as powers
a genuine fraction — one whose denominator does not divide the numerator
Three most common mistakes
Putting $h=0$ into $\frac{f(a+h)-f(a)}{h}$ before cancelling. That is always $\frac{0}{0}$; the algebra that removes $h$ from the denominator is the exercise.
$(fg)'=f'g'$. Test it on $f=g=x$: the product is $x^{2}$ with derivative $2x$, while $f'g'=1\cdot1=1$.
Reading continuous as . $\vert x\vert$ is continuous everywhere and has no derivative at $0$; the implication only runs the other way.
Grade weights this term: Midterm 1 28%, Midterm 2 28%, Final 28%, quizzes 10%, homework 6%. The two midterms also decide who sits the final: a combined score under 40 out of 200 is an FZ.
How much time do you have?
10 minutes
You walk in with the four rules and the three errors that cost the most marks. Enough for a differentiate this question; not enough for a from the definition question.
The 60 second card, Formula card, Powers, constants and sums, Products and quotients, Mistake ledger
45 minutes
You add the one thing the rules cannot give you: the limit the rules came from, plus the classification question that shows up on every quiz — where does a derivative fail to exist.
The 60 second card, From an average to a rate at an instant, The tangent line, Differentiable means continuous, Powers, constants and sums, Products and quotients, Full exam-style question, Practice B (computation), Check yourself
Full read
Everything in order, ending with the mixed set. The mixed problems are the only ones that ask you to decide which tool applies, which is what the midterm actually tests.
Read straight through, Do the four practice tiers in order, Finish with the free recall prompt
By the end of this section
Compute $f'(a)$ straight from the limit definition for a polynomial, for a fraction such as $1/x$ and for an expression containing a square root, and use it to write the tangent line at that point.
Treat $f'$ as a function in its own right: evaluate it, solve equations in it, state its domain, and move between the notations $f'(x)$, $dy/dx$ and $\frac{d}{dx}[\ \cdot\ ]$.
Decide whether $f'(a)$ exists, and name the failure as a , a , a or a discontinuity.
Differentiate any sum of powers — including roots and reciprocals — term by term, using the power, constant multiple and .
Apply the product and quotient rules, and judge when rewriting the expression first is cheaper and safer than either.
Compute higher derivatives, and read $s'$ and $s''$ as and .
Syllabus coverage
2.1
Derivatives and rates of change: tangent lines, velocity, the definition of $f'(a)$
Split over two blocks: the limit itself, then the tangent line built from it. Every rate of change reading — speed, slope, marginal cost — is the same limit.
covered
2.2
The derivative as a function: notation, differentiability, where it fails, higher derivatives
Split over three blocks: the , the continuity theorem with the corner, cusp and vertical tangent failures, and higher derivatives.
Powers and sums first, products and quotients in the block after it.
covered
extra
Differentiability of $x\vert x\vert$ and of $x^{2}\sin(1/x)$ at the origin
Neither is in the week 3 line. They are here because they are the sharpest test of the definition and they appear as bonus parts; you are not responsible for them on this week's quiz.
off_syllabus
Recall first
Substitution is legal only when the denominator survives
$\lim_{x\to a}\frac{p(x)}{q(x)}=\frac{p(a)}{q(a)}$ when $q(a)\neq0$. When numerator and denominator both go to $0$, the expression must be rewritten before any limit can be read off.
Every derivative starts life as a $0/0$ quotient, so for us this is the normal case, not the exception.
Rationalising with the conjugate
$\frac{\sqrt{A}-\sqrt{B}}{C}=\frac{A-B}{C\left(\sqrt{A}+\sqrt{B}\right)}$, obtained by multiplying top and bottom by $\sqrt{A}+\sqrt{B}$.
It is the only way to get $h$ out of the denominator when $f$ contains a square root.
Continuity at a point
$f$ is continuous at $a$ when $f(a)$ is defined, $\lim_{x\to a}f(x)$ exists, and the two are equal.
The first theorem of this section says that having a derivative forces all three conditions at once.
One sided limits and the existence test
$\lim_{x\to a}F(x)=L$ holds exactly when $\lim_{x\to a^{-}}F(x)=\lim_{x\to a^{+}}F(x)=L$.
At a corner the two one sided difference quotients are both perfectly fine and simply disagree; this is the test that catches it.
Expanding a binomial
$(x+h)^{2}=x^{2}+2xh+h^{2}$, $(x+h)^{3}=x^{3}+3x^{2}h+3xh^{2}+h^{3}$, and in general $(x+h)^{n}=x^{n}+n\,x^{n-1}h+(\text{terms carrying }h^{2}\text{ or a higher power})$.
The power rule is nothing but this expansion, minus $x^{n}$, divided by $h$.
Line through a point with a given slope
$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$, and the line through $(a,b)$ with slope $m$ is $y-b=m(x-a)$.
Every tangent line answer in this section is written in that form, and half the marks live in the point, not the slope.
Squeeze theorem
If $g(x)\le F(x)\le h(x)$ near $a$ and $\lim_{x\to a}g(x)=\lim_{x\to a}h(x)=L$, then $\lim_{x\to a}F(x)=L$.
Used once, in the last mixed problem, where the difference quotient oscillates instead of settling.
Try it yourself first (3 questions)
1§03.0 — a quotient that starts as 0 over 0●●○○○
Three warm ups, none of which needs anything from this section. They are the three pieces of algebra that the definition of the derivative will demand of you. Getting one wrong here costs nothing; getting it wrong in the middle of a derivative costs the whole question.
Given
$\displaystyle\lim_{h\to0}\frac{(3+h)^{2}-9}{h}$
Find
(a) Evaluate the limit, or say why it does not exist.
Hint 1/4
Substitution is illegal here, so the expression has to be rewritten first. What is the only algebraic move available on the numerator?
Hint 2/4
Expand $(3+h)^{2}=9+6h+h^{2}$ before doing anything else; the $9$'s then cancel.
Hint 3/4
The quotient becomes $\frac{9+6h+h^{2}-9}{h}=\frac{6h+h^{2}}{h}=6+h$, valid for every $h\neq0$ — which is all the limit ever looks at.
Hint 4/4
So the limit is $6$.
Show solutionMake h a factor of the numerator
$(3+h)^{2}-9=9+6h+h^{2}-9=6h+h^{2}$
expanding is the only move that can produce a factor of $h$; guessing at the answer from the shape of the quotient is what produces the $0/0$ verdict
$\frac{6h+h^{2}}{h}=6+h\quad (h\neq0)$
cancelling is legal precisely because the limit never evaluates at $h=0$
Now the limit is a substitution
$\lim_{h\to0}(6+h)=6$
$6+h$ is a polynomial, so substitution is legal for it even though it was illegal for the original quotient
Answer $$\boxed{6}$$
Check
Independent check with the other form of the same quotient: $\lim_{x\to3}\frac{x^{2}-9}{x-3}=\lim_{x\to3}(x+3)=6$, reached by factoring rather than expanding.
Every quotient in this section behaves this way: it is $0/0$ until you make $h$ a factor of the numerator, and it is a substitution afterwards.
2§03.0 — expanding a cube●●○○○
The same manoeuvre one degree up. This is the algebra behind the derivative of $x^{3}$, and it is the step where most of the arithmetic errors in this section happen.
Given
$\dfrac{(2+h)^{3}-8}{h}$, with $h\neq0$
Find
(a) Simplify the expression as far as it will go.
Hint 1/4
You are not taking a limit yet. The task is to write the expression with no $h$ left in the denominator.
Hint 2/4
$(2+h)^{3}=8+12h+6h^{2}+h^{3}$ — the coefficients are $1,3,3,1$ applied to $2^{3},2^{2}h,2h^{2},h^{3}$.
Hint 3/4
Subtracting $8$ leaves $12h+6h^{2}+h^{3}$, and every term carries an $h$: $\frac{12h+6h^{2}+h^{3}}{h}$.
the denominator tends to $4\neq0$, so the quotient law applies
Answer $$\boxed{\frac{1}{4}}$$
Check
Numerical check at $h=0.01$: $\frac{\sqrt{4.01}-2}{0.01}=\frac{2.0024984-2}{0.01}=0.24984$, closing in on $0.25$ from below.
Whenever a square root sits inside a difference quotient, the conjugate is not one option among several; it is the move.
Notation
symbol
reads as
means
watch out
$f'(a)$
f prime of a
the derivative of $f$ at the single number $a$ — a number, the slope of one tangent line
Differentiate first, substitute second. $\frac{d}{dx}\bigl[f(2)\bigr]=0$, because $f(2)$ is a constant; $f'(2)$ is something else entirely.
$f'(x)$
f prime of x
the derivative function: hand it an $x$, it returns the slope of the graph above that $x$
Its domain can be smaller than the domain of $f$: $\sqrt{x}$ is defined at $0$, its derivative is not.
$\frac{dy}{dx}$
d y d x
the same derivative in , with the variables named
One symbol, not a fraction of two numbers. Do not cancel the $d$'s, and do not read it as $y/x$.
$\left.\frac{dy}{dx}\right\vert_{x=a}$
d y d x evaluated at x equals a
the Leibniz way of writing $f'(a)$
The bar is not decoration: without it you have written a function where the question wanted a number.
$\frac{d}{dx}\bigl[\ \cdot\ \bigr]$
the derivative with respect to x of
an instruction to differentiate whatever is in the bracket
It is a verb, not a quantity. Writing $\frac{d}{dx}$ on its own at the end of a line says nothing.
$f''(x),\ \frac{d^{2}y}{dx^{2}}$
f double prime of x; d two y d x squared
the — the derivative of the derivative
$f''\neq(f')^{2}$. In the Leibniz form the two $2$'s sit in different places on purpose; it is not an exponent on $y$ or on $x$.
$\Delta x,\ \Delta y$
delta x, delta y
finite changes; $\frac{\Delta y}{\Delta x}$ is the over that interval
$h$ and $\Delta x$ are the same thing in this section. Both are allowed to be negative, which is exactly how the left hand side gets tested.
$f'(a)\ \text{DNE}$
f prime of a does not exist
the difference quotient has no finite two sided limit at $a$
Even when both one sided quotients run to $+\infty$, the answer is does not exist plus the words vertical tangent — not $f'(a)=\infty$.
Conventions used here
Continuity and differentiability are only asked at points of the domain
$1/x$ is not discontinuous at $0$: the number $0$ is not in its domain, so continuity there is not a question about the function. Report the behaviour instead — $\lim_{x\to0^{+}}1/x=+\infty$ and $\lim_{x\to0^{-}}1/x=-\infty$.
Does not exist versus infinite
If the difference quotient runs to $+\infty$ or $-\infty$, the derivative does not exist; write that, and add the geometric name (vertical tangent, cusp). $f'(a)=\infty$ is not an acceptable final answer unless the question explicitly asks how the slope behaves.
Endpoints and closed intervals
Differentiable on $[a,b]$ means the two sided derivative exists on $(a,b)$ and the appropriate exists at each endpoint. On an open interval every point gets the two sided test.
Angles are in radians
No trigonometric derivative appears in this section, but from the next one onwards every angle is in radians. The degree versions of the formulas are different, and the difference is a factor of $\pi/180$.
From an average over an interval to a rate at an instant
Where we left off: a $0/0$ quotient is an instruction to rewrite, not a verdict. Here is the $0/0$ quotient that the rest of the course is built out of.
Solvable with what we have
Average speed over $[3,4]$: $\frac{16-9}{4-3}=7$ m/s.
Average over $[3,3.1]$: $\frac{9.61-9}{0.1}=6.1$ m/s.
Average over $[3,3.01]$: $6.01$ m/s.
The slope of the line through any two points of a graph.
Not solvable yet
The speed at the instant $t=3$, with no interval to average over.
The slope of the curve at the single point $(3,9)$: one point does not fix a line.
Anything a speedometer does.
Average over $[3,3+h]$, then set $h=0$: $\dfrac{(3+0)^{2}-9}{0}=\dfrac{0}{0}$.
Why it fails
$0/0$ is not a number, so the calculation stops before it starts. But the averages were already closing in on $6$ while $h$ was still positive. Simplify first — $\frac{(3+h)^{2}-9}{h}=6+h$ for every $h\neq0$ — and let $h\to0$ afterwards.
DefinitionDefinition 3.1: The derivative of f at a
Pin one point of the graph down and slide a second point towards it. Each pair gives a line with a slope, and if those slopes settle on one number, that number is $f'(a)$. The two formulas are the same calculation with the moving point named differently.
The scooter's path $s=t^{2}$, with $P=(3,9)$ fixed and $Q$ sliding towards it. The line through $P$ and $Q$ has the average speed as its slope; as $Q$ slides in, those lines swing onto one limiting line, and $\textcolor{#1f6feb}{6}$ is its slope.
Looks like this, but is not
The quotient at $h=0$ is $\frac{0}{0}$, so the derivative is $0$.
$\frac{0}{0}$ is not $0$; it is the absence of a value. For every $h\neq0$ the same quotient equals $6+h$, and that expression settles on $6$ as $h$ shrinks.
$h$ (seconds)
interval
average speed (m/s)
$1$
$[3,4]$
$7$
$0.5$
$[3,3.5]$
$6.5$
$0.1$
$[3,3.1]$
$6.1$
$0.01$
$[3,3.01]$
$6.01$
$-0.01$
$[2.99,3]$
$5.99$
$-0.1$
$[2.9,3]$
$5.9$
The right hand column is $6+h$ exactly — nothing is rounded — so this is the algebra written out in numbers rather than evidence from an experiment. From the left the averages are too small by $\vert h\vert$, from the right too big by $h$.
The scooter's speed at the instant t = 3
The scooter has covered $s(t)=t^{2}$ metres after $t$ seconds. Find its speed at the instant $t=3$, straight from the definition.
Given
$s(t)=t^{2}$ (metres, with $t$ in seconds)
the instant $a=3$
Find
$s'(3)$, the reading on the display at that instant
SolutionWrite the quotient with h still in it
$s(3+h)=(3+h)^{2}=9+6h+h^{2}$
substitute $3+h$ into the formula, every occurrence; this is where sloppy expansion costs the whole answer
$\frac{s(3+h)-s(3)}{h}=\frac{9+6h+h^{2}-9}{h}$
$s(3)=9$, and the constant terms are the ones that must cancel
Get h out of the denominator
$=\frac{6h+h^{2}}{h}=6+h\qquad(h\neq0)$
legal because the limit only ever looks at $h\neq0$; this is the line that converts a $0/0$ form into something a limit can read
Only now let h go to zero
$s'(3)=\lim_{h\to0}(6+h)=6$
$6+h$ is a polynomial, so substitution is legal here even though it was illegal three lines above
Answer $$\boxed{s'(3)=6\ \text{m/s}}$$
Check
Independent route: the other form of the definition, $\lim_{t\to3}\frac{t^{2}-9}{t-3}=\lim_{t\to3}(t+3)=6$, uses factoring instead of expanding and lands on the same number. Unit check: metres divided by seconds is m/s, which is what a speed should be.
One expansion, one cancellation, one substitution. Every derivative from the definition in this section has exactly this shape.
The $+h$ is the whole story: over an interval of length $h$ the scooter averages $6+h$, which is why every average in the table came out slightly too big. The instant is the case $h=0$, reached by approach rather than by substitution.
Checkpoint
§03.1 — the derivative of a straight line●○○○○
Thirty seconds. A straight line is the one graph whose slope you already know, which makes it a good test of whether the definition is being applied correctly.
Given
$f(x)=5x-7$
$a$ is any real number
Find
(a) Use the definition to compute $f'(a)$.
Hint 1/4
You are being asked for the slope of the graph at a general point $a$, so write the quotient before you think about what the answer ought to be.
Hint 2/4
Definition: $f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$, with $f(a+h)=5(a+h)-7$.
Hint 3/4
$\frac{[5(a+h)-7]-[5a-7]}{h}=\frac{5h}{h}=5$ for every $h\neq0$ — with $f(x)=5x-7$ the quotient does not even depend on $a$.
Hint 4/4
So $f'(a)=5$ for every $a$.
Show solutionWrite and simplify the quotient
$\frac{f(a+h)-f(a)}{h}=\frac{5(a+h)-7-5a+7}{h}$
watch the double negative: subtracting $-7$ adds $7$
$=\frac{5h}{h}=5\quad(h\neq0)$
the $a$ terms cancel, which is what makes a line special
Take the limit of a constant
$f'(a)=\lim_{h\to0}5=5$
the quotient was already constant, so the limit changes nothing
Answer $$\boxed{f'(a)=5}$$
Check
Geometric check: the graph is a straight line of slope $5$, and the tangent to a straight line is the line itself.
⚠ Setting h = 0 before cancelling
the instruction let h go to zero gets read as substitute zero, which worked for every limit met before this one
the notation looks additive, and for a line through the origin nothing goes wrong, so the habit survives
wrong$f(a+h)=a^{2}+h^{2}$
right$f(a+h)=(a+h)^{2}=a^{2}+2ah+h^{2}$
The tangent line: what the number is for
We have a number, $6$. On the graph of the last figure it is the slope of the blue line — and a slope together with a point is a line, which is what exam questions usually ask for.
RuleThe tangent line at a point
Conditions
$f'(a)$ exists
the height $f(a)$ comes from $f$; the slope $f'(a)$ comes from $f'$
$$\boxed{\ y-f(a)=f'(a)\,(x-a)\ }$$
The line through the point of the graph above $a$ whose slope is the derivative there. Two numbers are needed and they come from two different formulas. Supplying only the slope is the commonest way to lose half the marks on a tangent question.
Looks like this, but is not
$\frac{f(3.001)-f(3)}{0.001}=6.001$, so the speed at the instant $t=3$ is $6.001$ m/s.
That is still an average, over an interval a millisecond wide. The derivative is not the quotient for some very small $h$; it is the number those quotients approach. Every average here equals $6+h$ exactly, so each one is wrong by exactly the width of its interval, and only the limit removes the $+h$.
Tangent line to y = 1/x at x = 2
Find $f'(2)$ for $f(x)=\frac{1}{x}$ from the definition, and write the equation of the tangent line at that point.
Given
$f(x)=\dfrac{1}{x}$
$a=2$
Find
$f'(2)$, then the tangent line at $\left(2,\tfrac12\right)$
SolutionCombine the two fractions before touching the h
Sign check: $1/x$ is decreasing for $x>0$, so a negative slope is the only acceptable answer. Numerical check: $\frac{1/2.001-1/2}{0.001}=-0.24988$, closing in on $-0.25$.
Fractions need a common denominator where polynomials needed an expansion. Same goal in both cases: make $h$ a factor of the numerator.
Checkpoint
§03.1 — assembling a tangent line●○○○○
Thirty seconds, and no differentiation at all: both numbers are handed to you. The only question is where each one goes.
Given
$f$ is differentiable at $4$
$f(4)=5$
$f'(4)=3$
Find
(a) Write the equation of the tangent line to $y=f(x)$ at $x=4$.
Hint 1/4
Ask which of the two given numbers is a height and which is a slope, before writing anything down.
Hint 2/4
Point slope form: $y-f(a)=f'(a)(x-a)$, with $a=4$.
Hint 3/4
Here $f(4)=5$ is the height and $f'(4)=3$ is the slope, so $y-5=3(x-4)$.
Hint 4/4
Tidying: $y=3x-7$.
Show solutionPut each number where it belongs
$y-5=3(x-4)$
the value of $f$ fixes the point the line passes through; the value of $f'$ fixes its slope — swapping them gives $y-3=5(x-4)$, which is a different line entirely
$y=3x-7$
expand and tidy
Answer $$\boxed{y=3x-7}$$
Check
Check the point: at $x=4$ the line gives $12-7=5$, which is $f(4)$, so it does touch the graph where it should.
⚠ Writing the tangent line without the point
the slope was the part that took work, so it gets written down and the height is forgotten; the line then has the right direction and misses the curve
wrong$y=f'(a)(x-a)$
right$y-f(a)=f'(a)(x-a)$
⚠ Swapping the roles of f(a) and f prime of a
both numbers are attached to the same point, and in a hurry the one written first becomes the slope
wrong$y-f'(a)=f(a)(x-a)$
right$y-f(a)=f'(a)(x-a)$
The derivative as a function you can graph
One slope at one point is not much of a return for half a page of algebra. Notice, though, that nothing in that calculation used the number $3$ — replace it by a letter and the same work hands back every slope at once.
DefinitionDefinition 3.2: The derivative function
Conditions
the same limit as Definition 3.1, with the fixed number $a$ replaced by a variable $x$
the domain of $f'$ is the set of $x$ at which the limit exists, and it can be smaller than the domain of $f$
Feed it a value of $x$ and it returns the slope of the graph above that value. Written as $dy/dx$ it says the same thing in the language of changes: the rate at which $y$ responds to a change in $x$.
The same $x$ runs along both pictures. The height of the lower graph is the slope of the upper one: where the upper curve levels off, at $x=-1$ and $x=1$, the lower curve crosses zero.
Looks like this, but is not
The graph of $f'$ is a shifted copy of the graph of $f$ — both curves in the picture wave up and down, after all.
Compare heights against slopes, not shape against shape. At $x=0$ the upper curve has height $0$ but is falling steeply, and the lower curve sits at $-3$, not at $0$. The zeros of $f'$ are where $f$ is flat, not where $f$ is zero: $f(x)=x^{3}-3x$ vanishes at $x=0$ and $x=\pm\sqrt3$, while $f'$ vanishes at $x=\pm1$, and $f'(\sqrt3)=6$ is nowhere near zero.
The derivative of √x, and the one point where it fails
Find $f'(x)$ for $f(x)=\sqrt{x}$ from the definition, and state the domain of $f'$.
Numerical check at $x=4$: the formula gives $\tfrac14=0.25$, and $\frac{\sqrt{4.001}-\sqrt{4}}{0.001}=0.24999$. Shape check: $\sqrt{x}$ flattens as $x$ grows, and $\frac{1}{2\sqrt{x}}$ does shrink towards $0$.
$f$ is defined at $0$ and $f'$ is not: the domain of a derivative is something you report, not something you inherit.
Where does y = x² + 3x have slope 1?
For $f(x)=x^{2}+3x$, find $f'(x)$ from the definition and then find every point of the graph at which the tangent line has slope $1$.
Given
$f(x)=x^{2}+3x$
target slope $=1$
Find
$f'(x)$, and all points where $f'(x)=1$
SolutionDifferentiate once, for all x at once
$f(x+h)=(x+h)^{2}+3(x+h)=x^{2}+2xh+h^{2}+3x+3h$
substitute $x+h$ into every occurrence of $x$, including the linear term
Independent check on the slope at $x=-1$: a centred numerical quotient, $\frac{f(-0.999)-f(-1.001)}{0.002}=\frac{-1.998999-(-2.000999)}{0.002}=1.000$. And $2x+3=1$ has a single root, so one point is the right count for a parabola.
Two different questions were answered by the same formula: what is the slope here is an evaluation, where is the slope this is an equation. Exams ask both.
Checkpoint
§03.2 — reading f prime as a function●○○○○
Thirty seconds, and the point is the difference between a function and one of its values.
Given
$f'(x)=2x+3$ for a certain function $f$
Find
(a) What does $f'(0)=3$ tell you about the graph of $f$?
Hint 1/4
Ask what the number $f'(0)$ measures before asking what it equals.
Hint 2/4
$f'$ reports slope: $f'(c)$ is the slope of the tangent line to $y=f(x)$ at the point where $x=c$.
Hint 3/4
Here $f'(0)=2(0)+3=3$, so the slope at the point of the graph above $x=0$ is $3$.
Hint 4/4
The tangent line at that point rises three units for each unit to the right.
Show solutionEvaluate, then translate
$f'(0)=2\cdot0+3=3$
substituting into $f'$, not into $f$
$\text{slope of the tangent at }x=0\ \text{is}\ 3$
$f'$ is a slope reporter; the height of the graph is the job of $f$, and $f$ is not even determined by this information
Answer $$\boxed{\text{slope }3\ \text{at}\ x=0}$$
Check
Consistency check: $f'(x)=2x+3$ is negative for $x<-1.5$ and positive after it, so the graph falls then rises — a parabola opening upwards, which is the only shape compatible with a linear $f'$.
⚠ Substituting the point before differentiating
the number in the question is written first, so it gets used first
Differentiable forces continuous, and the three ways a graph refuses a slope
The last example produced a point where the limit refused to exist: $\sqrt{x}$ at $0$. It is worth knowing exactly which points can do that, because exam questions are built out of them.
TheoremTheorem 3.3: Differentiable at a implies continuous at a
A graph that has a tangent line at a point cannot have a hole or a jump there. The contrapositive is the version you will actually use: a break in the graph kills the derivative on the spot, with no calculation needed.
Proof, in one line
For $x\neq a$ write $f(x)-f(a)=\frac{f(x)-f(a)}{x-a}\cdot(x-a)$ — legal, since we multiplied and divided by the same nonzero number. Let $x\to a$. The first factor tends to $f'(a)$, which exists by hypothesis; the second tends to $0$. So $\lim_{x\to a}\bigl[f(x)-f(a)\bigr]=f'(a)\cdot0=0$, which says $\lim_{x\to a}f(x)=f(a)$: continuity. The hypothesis was used exactly once, in the claim that the first factor has a finite limit.
Every line joining the origin to a point on the right branch has slope $\textcolor{#1f6feb}{+1}$, and every line joining it to a point on the left branch has slope $\textcolor{#d1690a}{-1}$. Shrinking $h$ changes neither number, so the two sides never agree on a tangent.
Looks like this, but is not
The converse: $f$ is continuous at $a$, so $f$ must have a slope at $a$. After all, $f(x)=\vert x\vert$ can be drawn without lifting the pen.
At $0$ the quotient is $\frac{\vert h\vert}{h}$, which is $+1$ for every $h>0$ and $-1$ for every $h<0$. Both one sided limits are finite and they disagree, so $f'(0)$ does not exist. The implication runs one way only: differentiability buys continuity, continuity buys nothing.
$h$
$\dfrac{\vert h\vert}{h}$
$\dfrac{h^{2/3}}{h}$
$\dfrac{h^{1/3}}{h}$
$0.1$
$1$
$2.15$
$4.64$
$0.01$
$1$
$4.64$
$21.5$
$0.001$
$1$
$10$
$100$
$-0.001$
$-1$
$-10$
$100$
$-0.01$
$-1$
$-4.64$
$21.5$
$-0.1$
$-1$
$-2.15$
$4.64$
Three different refusals. Column two settles on two different finite numbers — a corner. Column three runs to $+\infty$ from the right and $-\infty$ from the left — a cusp. Column four runs to $+\infty$ from both sides — a vertical tangent. In each case $f'(0)$ fails to exist while the graph stays unbroken, and the exam question is usually which one.
Is f(x) = |x − 2| differentiable at x = 2?
Decide whether $f'(2)$ exists for $f(x)=\vert x-2\vert$, and give $f'(x)$ wherever it does exist.
Given
$f(x)=\vert x-2\vert$
the suspicious point $a=2$
Find
whether $f'(2)$ exists, and a formula for $f'$ elsewhere
SolutionContinuity first — otherwise there is nothing to test
$\lim_{x\to2}\vert x-2\vert=0=f(2)$
if this had failed, Theorem 3.3 would end the question immediately
The two one sided quotients
$h>0:\ \frac{\vert h\vert}{h}=\frac{h}{h}=1$
for $h>0$ the point $2+h$ lies to the right, where the graph is the line of slope $+1$
$h<0:\ \frac{\vert h\vert}{h}=\frac{-h}{h}=-1$
for $h<0$, $\vert h\vert=-h$; this sign flip is the entire content of the example
Verdict, and the derivative away from the corner
$1\neq-1\ \Rightarrow\ f'(2)\ \text{does not exist}$
the two sided limit needs both sides to agree, and shrinking $h$ never brings $+1$ and $-1$ closer
away from the corner the graph is locally a straight line, so its slope is that line's slope
Answer $$\boxed{f'(2)\ \text{does not exist (corner)};\quad f'(x)=\pm1\ \text{for}\ x\neq2}$$
Check
Geometric check: the graph is two straight pieces of slopes $-1$ and $+1$ meeting at $(2,0)$, and a single point cannot be given two slopes at once. Continuity check confirms this is a corner and not a jump.
A corner is not an accident of the picture; it is two one sided answers that will not merge no matter how far you zoom in.
A piecewise function that does have a derivative at the joint
Let $f(x)=x^{2}$ for $x\le1$ and $f(x)=2x-1$ for $x>1$. Decide whether $f'(1)$ exists.
same trick on the other side; the factor $x-1$ cancels exactly
Verdict
$2=2\ \Rightarrow\ f'(1)=2$
equal, finite one sided quotients are precisely the condition for the two sided limit to exist
Answer $$\boxed{f'(1)=2}$$
Check
Independent check: the line $2x-1$ is the tangent to $y=x^{2}$ at $x=1$ (it has slope $2$ and passes through $(1,1)$), so the two pieces join with matching slope by construction — the graph has no visible kink.
Two limits, plus one continuity check that most students skip. The check is what stops you handing a slope to a graph that jumps.
The recipe never changes: continuity first, then the two one sided quotients, then compare.
Checkpoint
§03.3 — the implication and its direction●●○○○
Thirty seconds. Getting the direction of this implication wrong is worth several marks a term.
Given
Statement: if $f$ is not continuous at $a$, then $f'(a)$ cannot exist.
Find
(a) True or false?
Hint 1/4
Do not test the statement on examples yet. Write down the theorem and see which way its arrow points.
Hint 2/4
Theorem 3.3 says $f'(a)$ exists $\Rightarrow$ $f$ continuous at $a$. The contrapositive of $P\Rightarrow Q$ is not $Q$ $\Rightarrow$ not $P$, and it is always as true as the original.
Hint 3/4
Here not $Q$ is $f$ is not continuous at $a$ and not $P$ is $f'(a)$ does not exist, which is exactly the statement being tested.
Hint 4/4
So the statement is true — it is the theorem itself, read backwards.
a contrapositive is logically identical to the original; a converse is not
$\text{not continuous at }a\ \Rightarrow\ f'(a)\ \text{does not exist}$
which is the statement, so it is true
Answer $$\boxed{\text{True}}$$
Check
Check with an example: $f(x)=\lfloor x\rfloor$ jumps at $x=1$, and its difference quotient there is $\frac{\pm1}{h}$, which blows up. No derivative, as predicted.
⚠ Reading the implication backwards
continuous and smooth are the same word in everyday speech, and every graph drawn in class before this week was both
right$0\notin\operatorname{dom}f,\ \text{so continuity at}\ 0\ \text{is not a question};\ \lim_{x\to0^{\pm}}\tfrac1x=\pm\infty$
⚠ Writing f prime of a equals infinity
the difference quotient really does run to infinity, so writing that down feels like reporting what happened
wrong$f(x)=x^{1/3}\ \Rightarrow\ f'(0)=+\infty$
right$f'(0)\ \text{does not exist; the graph has a vertical tangent at}\ 0$
Powers, constants and sums: the rules that retire the limit
Every derivative so far has cost half a page. The algebra was the same shape every time — substitute, expand, cancel one $h$, throw away what still carries an $h$ — so we do it once with a general exponent and never do it again.
RuleRule 3.4: Constant, power, constant multiple and sum
Conditions
$f$ and $g$ are differentiable at $x$, and $c$ is a constant
$n$ is any real number — the derivation below covers positive integers, and the roots and reciprocals of this course obey the same formula
A constant has no slope. For a power, bring the exponent down in front and knock one off it. A constant factor rides along untouched, and a sum is differentiated one term at a time.
Where the power rule comes from
$(x+h)^{n}=x^{n}+n\,x^{n-1}h+(\text{terms carrying }h^{2}\text{ or more})$. Subtract $x^{n}$ and every surviving term has an $h$ in it; divide by $h$ and you are left with $n\,x^{n-1}+(\text{terms that still carry at least one }h)$. Let $h\to0$ and those terms die, leaving $n\,x^{n-1}$. That is the whole content of the rule: after one division by $h$, the $h^{2}$ terms are still too small to matter and the $h^{1}$ term is not.
Why the exponent drops in front. Growing a square of side $x$ by $h$ adds $\textcolor{#1f6feb}{2xh}$ in strips and $\textcolor{#d1690a}{h^{2}}$ in the corner, so $\frac{(x+h)^{2}-x^{2}}{h}=\textcolor{#1f6feb}{2x}+\textcolor{#d1690a}{h}$ — and only the corner term still carries an $h$.
Looks like this, but is not
$\frac{d}{dx}2^{x}=x\,2^{x-1}$ — it has a base, it has an exponent, so the power rule should apply.
The power rule is about a variable base with a constant exponent; here the variable is upstairs, which is a different animal and a later section. Test the claim where it is cheapest: at $x=0$ it predicts a slope of $0\cdot2^{-1}=0$, but $2^{x}$ increases everywhere, and the numerical quotient $\frac{2^{0.001}-1}{0.001}=0.693$ agrees with the objection.
$f(x)$ as written
as a power
$f'(x)$
tidied
$x$
$x^{1}$
$1\cdot x^{0}$
$1$
$x^{2}$
$x^{2}$
$2x^{1}$
$2x$
$\sqrt{x}$
$x^{1/2}$
$\tfrac12x^{-1/2}$
$\dfrac{1}{2\sqrt{x}}$
$\dfrac{1}{x}$
$x^{-1}$
$-1\cdot x^{-2}$
$-\dfrac{1}{x^{2}}$
$\dfrac{1}{x^{2}}$
$x^{-2}$
$-2x^{-3}$
$-\dfrac{2}{x^{3}}$
$\dfrac{1}{\sqrt{x}}$
$x^{-1/2}$
$-\tfrac12x^{-3/2}$
$-\dfrac{1}{2x\sqrt{x}}$
Every row is the same rule; the only work is the middle column. The two rows that cost marks are the negative exponents: $-1$ becomes $-2$, and $-2$ becomes $-3$ — you subtract one from the exponent, you do not subtract one from its size.
Differentiating 4x⁵ − 7x³ + 2x − 9 term by term
Find $p'(x)$ for $p(x)=4x^{5}-7x^{3}+2x-9$, and evaluate the slope at $x=1$.
Numerical check at $x=1$: $\frac{p(1.001)-p(0.999)}{0.002}=\frac{-9.998981-(-10.000981)}{0.002}=1.000$, matching $p'(1)=1$. Degree check: a degree $5$ polynomial must have a degree $4$ derivative.
Four terms, four applications of one rule, no limits. Compare with the two page version from the definition.
Numerical check at $x=1$: the formula gives $1.5+6-5=2.5$, and $\frac{g(1.001)-g(0.999)}{0.002}=\frac{6.0024926-5.9974925}{0.002}=2.50005$.
The rewriting step is not decoration. Once every term is $c\,x^{n}$, the rest of the problem is mechanical — and the exam's difficulty is usually hidden in exactly that first line.
Checkpoint
§03.4 — one negative exponent●○○○○
Thirty seconds. The rewriting step is the only step.
Given
$y=\dfrac{1}{x^{2}}$
Find
(a) Find $\dfrac{dy}{dx}$.
Hint 1/4
The power rule needs a power. What does this expression look like once it is written as one?
Hint 2/4
$\frac{1}{x^{2}}=x^{-2}$, and $\frac{d}{dx}x^{n}=n\,x^{n-1}$.
Hint 3/4
With $n=-2$: the exponent comes down in front as $-2$, and the new exponent is $-2-1=-3$.
Hint 4/4
So $\frac{dy}{dx}=-2x^{-3}=-\frac{2}{x^{3}}$.
Show solutionRewrite and apply the rule
$y=x^{-2}$
the fraction hides a power; the rule only sees powers
For a product: differentiate the first factor and keep the second, then keep the first and differentiate the second, and add. For a quotient: the same two products, but subtracted in that order — derivative of the top times the bottom, minus the top times the derivative of the bottom — all over the bottom squared. Order matters only in the quotient, because subtraction is not symmetric.
Where the product rule comes from
Add and subtract the same thing: $f(x+h)g(x+h)-f(x)g(x)=f(x+h)\bigl[g(x+h)-g(x)\bigr]+g(x)\bigl[f(x+h)-f(x)\bigr]$. Divide by $h$ and let $h\to0$. The two brackets become $g'(x)$ and $f'(x)$, $g(x)$ is just a number, and $f(x+h)\to f(x)$ — which is legal because $f$, being differentiable, is continuous by Theorem 3.3. The result is $f(x)g'(x)+g(x)f'(x)$. The one clever step is the adding and subtracting, and the rectangle picture is that step drawn.
The same picture as the square, except that the two sides now move at different rates. Growing $f$ by $\Delta f$ and $g$ by $\Delta g$ adds $\textcolor{#1f6feb}{f\,\Delta g}$ and $\textcolor{#1f6feb}{g\,\Delta f}$ in strips and $\textcolor{#d1690a}{\Delta f\,\Delta g}$ in the corner: two strips survive, which is why $(fg)'$ has two terms and is not $f'g'$.
Looks like this, but is not
Sums differentiate term by term, so products should differentiate factor by factor: $\left(x^{2}\cdot x^{3}\right)'=2x\cdot3x^{2}=6x^{3}$.
The left hand side is $\left(x^{5}\right)'=5x^{4}$, and $6x^{3}\neq5x^{4}$ — at $x=1$ one says $5$ and the other says $6$. The rectangle says why: growing both sides at once adds two strips, and their areas are $f\,\Delta g$ and $g\,\Delta f$, not a product of the two increments. The product rule agrees with the power rule here: $2x\cdot x^{3}+x^{2}\cdot3x^{2}=2x^{4}+3x^{4}=5x^{4}$.
Derivative of (x³ + 2x)(x² − 5), two ways
Differentiate $u(x)=\left(x^{3}+2x\right)\left(x^{2}-5\right)$ with the product rule, then check by expanding first.
derivative of the first times the second, plus the first times the derivative of the second
$=3x^{4}-15x^{2}+2x^{2}-10+2x^{4}+4x^{2}$
expand both products fully before collecting; this is where signs get lost
$=5x^{4}-9x^{2}-10$
collect like terms
Answer $$\boxed{u'(x)=5x^{4}-9x^{2}-10}$$
Check
Independent route: expand first, $u(x)=x^{5}-3x^{3}-10x$, then differentiate term by term to get $5x^{4}-9x^{2}-10$. Two genuinely different calculations, one answer.
Product rule: about four lines. Expanding first: about three. For polynomials this small, expanding is not a bad choice — the product rule earns its keep when the factors are roots, fractions or, from next week, trigonometric.
Slope of y = (x² − 4)/(x + 3) at x = 1, and the rewrite that avoids the quotient rule
Find $v'(x)$ for $v(x)=\dfrac{x^{2}-4}{x+3}$ and evaluate $v'(1)$.
Independent route by long division: $x^{2}-4=(x+3)(x-3)+5$, so $v(x)=x-3+\frac{5}{x+3}$. Differentiating that (the last term by the quotient rule with a constant numerator) gives $1-\frac{5}{(x+3)^{2}}=\frac{(x+3)^{2}-5}{(x+3)^{2}}=\frac{x^{2}+6x+4}{(x+3)^{2}}$, and at $x=1$, $1-\frac{5}{16}=\frac{11}{16}$.
When the division comes out clean, the rewritten form is shorter and has no sign trap. Always spend three seconds checking whether the denominator divides the numerator before committing to the quotient rule.
Checkpoint
§03.5 — product rule against expanding●○○○○
Thirty seconds, and a free consistency check: this function can be differentiated both ways, so the two answers must agree.
Given
$y=x^{2}(x+1)$
Find
(a) Differentiate with the product rule.
(b) Differentiate by expanding first, and compare.
Hint 1/4
Two routes are available. Decide what each one needs before starting either.
Hint 2/4
Product rule: $(fg)'=f'g+fg'$, with $f=x^{2}$ and $g=x+1$. Expansion: $x^{2}(x+1)=x^{3}+x^{2}$.
Differentiating again: acceleration and the second derivative
$f'$ is a function, and functions can be differentiated. Nothing new is required — but the second derivative answers a physical question that the first one cannot.
DefinitionDefinition 3.6: Higher derivatives
Conditions
$f'$ is itself differentiable at $x$ (which can fail — see the counterexample below)
Differentiate, then differentiate the answer, and keep going. If $s(t)$ is position, then $s'$ is velocity — how fast the position is changing — and $s''$ is acceleration — how fast the velocity is changing.
Looks like this, but is not
The double prime means squaring: $f(x)=x^{3}$ has $f'(x)=3x^{2}$, so $f''(x)=\left(3x^{2}\right)^{2}=9x^{4}$.
It means differentiate twice, not square once: $f''(x)=\left(3x^{2}\right)'=6x$. A size check settles it without any calculation — each differentiation lowers the degree of a polynomial by one, so the second derivative of a cubic has to be linear. Degree $4$ was going the wrong way.
$n$
$\dfrac{d^{n}}{dx^{n}}x^{5}$
$\dfrac{d^{n}}{dx^{n}}\dfrac{1}{x}$
$1$
$5x^{4}$
$-x^{-2}$
$2$
$20x^{3}$
$2x^{-3}$
$3$
$60x^{2}$
$-6x^{-4}$
$4$
$120x$
$24x^{-5}$
$5$
$120$
$-120x^{-6}$
$6$
$0$
$720x^{-7}$
The left column dies: each differentiation drops the degree by one, so a polynomial of degree $n$ has $f^{(n+1)}=0$ and everything after it is zero too. The right column never dies — it only alternates sign and grows a factorial, $\frac{d^{n}}{dx^{n}}\frac{1}{x}=\frac{(-1)^{n}\,n!}{x^{n+1}}$.
The scooter's acceleration, and why its speed grew evenly
With $s(t)=t^{2}$ metres, find the velocity and the acceleration, and check them against the opening numbers.
Given
$s(t)=t^{2}$ metres, $t$ in seconds
Find
$s'(t)$, $s''(t)$, and their meaning
SolutionTwo applications of the power rule
$s'(t)=2t\ \text{m/s}$
velocity is the rate at which position changes
$s''(t)=2\ \text{m/s}^{2}$
acceleration is the rate at which velocity changes, and here it is the same at every instant
Check against the instant we computed by hand
$s'(3)=6\ \text{m/s}$
the same $6$ that the definition produced at the start of this section, now in one line
$s'(4)-s'(3)=8-6=2$
which is what a constant acceleration of $2$ m/s$^{2}$ has to mean over one second
Unit check: metres per second per second is the unit of acceleration, and the numbers agree with the table — the averages there rose by exactly $0.1$ when the interval grew by $0.1$, which is the signature of a constant acceleration of $2$.
Constant acceleration is why the velocity graph is a straight line while the distance graph bends. The bending of a graph is a second derivative statement, which is the thread picked up later in the course.
Test the general formula on a case computed by hand: $n=2$ gives $(-1)^{2}2!\,x^{-3}=2x^{-3}$, which is exactly what differentiating $-x^{-2}$ produces. A formula that survives a case you can check is worth trusting on a case you cannot.
No differentiation was performed a hundred times; four were, and the fifth step was reading a pattern. Exam questions with an absurd order are always pattern questions.
Checkpoint
§03.6 — acceleration from a position formula●●○○○
Thirty seconds. The only trap is stopping one derivative early.
Given
$s(t)=t^{3}-6t^{2}$ metres, with $t$ in seconds
Find
(a) Find the acceleration at $t=1$.
Hint 1/4
Decide which derivative the word acceleration is asking for before differentiating anything.
Hint 2/4
Velocity is $s'$ and acceleration is $s''$, so the position formula has to be differentiated twice.
Hint 3/4
$s'(t)=3t^{2}-12t$, then $s''(t)=6t-12$; the question asks for $t=1$.
Hint 4/4
$s''(1)=6-12=-6$ m/s$^{2}$.
Show solutionDifferentiate twice
$s'(t)=3t^{2}-12t$
velocity
$s''(t)=6t-12$
acceleration; each step is the power rule on two terms
Evaluate
$s''(1)=6-12=-6$
negative, so the velocity is decreasing at that instant
Answer $$\boxed{-6\ \text{m/s}^{2}}$$
Check
Consistency check on the velocity: $s'(1)=3-12=-9$ and $s'(1.1)=3.63-13.2=-9.57$, so the velocity is indeed falling near $t=1$, as a negative acceleration requires.
⚠ Reading the Leibniz second derivative as a square
the two $2$'s look like exponents, and one of them is written where an exponent normally goes
⚠ Answering acceleration with the first derivative
rate of change is the phrase attached to both velocity and acceleration, so the count of derivatives gets lost
wrong$a(t)=s'(t)$
right$a(t)=v'(t)=s''(t)$
Differentiating straight from the definition
the question contains the words using the definition of the derivative — in which case the rules earn zero marks — or the function is new and no rule reaches it yet.
Substitute
Write $f(x+h)$ by putting $x+h$ into every occurrence of $x$ in the formula.
Subtract
Form $f(x+h)-f(x)$ and simplify it to a single expression. The constant terms must cancel; if they do not, the substitution is wrong.
Manufacture a factor of h
Expand a power, put fractions over a common denominator, or multiply by the conjugate for a root. One of those three always works at this level.
Cancel
Divide the factor of $h$ out of numerator and denominator. This is the step that turns a $0/0$ form into something a limit can read.
Now take the limit
Substitute $h=0$ into what is left — legal at last, because the denominator no longer vanishes.
Test
Check the answer at one convenient value against a numerical quotient such as $\frac{f(a+0.001)-f(a)}{0.001}$.
Where it goes wrong
Setting $h=0$ while $h$ is still in the denominator.
Writing $f(x+h)=f(x)+f(h)$.
Cancelling $h$ from only one term of the numerator.
Stopping at the simplified quotient and forgetting that the limit is still to be taken.
Choosing the cheapest route for an algebraic function
any differentiate this question that does not force the definition on you.
Rewrite
Turn every root and every reciprocal into a power of $x$: $\sqrt[3]{x^{2}}=x^{2/3}$, $\frac{5}{x^{4}}=5x^{-4}$.
Sum?
If the expression is now a sum of terms $c\,x^{n}$, differentiate term by term and stop. Most exam questions end here.
Product?
If it is a product of two things you cannot cheaply multiply out, use $(fg)'=f'g+fg'$. If they are two short polynomials, expanding is often faster.
Fraction, or only fraction shaped?
Before the quotient rule, ask whether the denominator divides the numerator, or is a single power of $x$. If it does, split the fraction — the rewritten form is shorter and carries no sign trap.
Genuine quotient
Otherwise use $\left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^{2}}$, writing the numerator in that order and squaring the whole denominator.
Simplify only as far as the question needs
If it asks for a value, substitute now. If it asks for a formula, match the notation the question used.
Where it goes wrong
Reaching for the quotient rule on $\frac{2x+1}{x}$, which is $2+x^{-1}$.
Expanding a product of two long polynomials when the product rule is three lines.
Leaving negative exponents when the question was written with radicals, or the reverse.
Testing whether f prime of a exists at a suspicious point
piecewise definitions, absolute values, roots and fractional powers — anywhere the formula changes at $a$ or the graph comes to a point.
Continuity first
Check $\lim_{x\to a^{-}}f=\lim_{x\to a^{+}}f=f(a)$. If this fails, $f'(a)$ does not exist by Theorem 3.3 and you are finished.
Left quotient
Compute $\lim_{x\to a^{-}}\frac{f(x)-f(a)}{x-a}$ using the branch that applies on the left.
Right quotient
Compute $\lim_{x\to a^{+}}\frac{f(x)-f(a)}{x-a}$ using the branch on the right.
Compare and name
Equal and finite gives $f'(a)$. Different and finite is a corner. Both infinite with opposite signs is a cusp. Both infinite with the same sign is a vertical tangent.
Where it goes wrong
Differentiating the two branch formulas and comparing those, without checking continuity — that hands a slope to a graph that jumps.
Deciding from a plot: $x^{1/3}$ looks smooth at $0$ and has no derivative there.
Reporting $f'(0)=\infty$ instead of does not exist, vertical tangent.
Quotient rule on y = (2x + 1)/x
Differentiate $y=\dfrac{2x+1}{x}$ with the quotient rule.
Given
$y=\dfrac{2x+1}{x}$
Find
$y'$
SolutionApply the rule in the fixed order
$y'=\frac{(2)(x)-(2x+1)(1)}{x^{2}}$
top derivative times bottom, minus top times bottom derivative
$=\frac{2x-2x-1}{x^{2}}=-\frac{1}{x^{2}}$
the minus sign has to reach both terms of $2x+1$; forgetting that is how this route goes wrong
Answer $$\boxed{y'=-\frac{1}{x^{2}}}$$
Check
Sign check: for large $x$ the function flattens towards $2$ from above, so the slope should be small and negative — and $-1/x^{2}$ is.
Four lines, one sign trap.
Same function, split first: y = 2 + 1/x
Differentiate $y=\dfrac{2x+1}{x}$ after dividing through.
Given
$y=\dfrac{2x+1}{x}=2+x^{-1}$
Find
$y'$
SolutionSplit, then use the power rule
$y=\frac{2x}{x}+\frac{1}{x}=2+x^{-1}$
legal for every $x\neq0$, which is the whole domain anyway
$y'=0+(-1)x^{-2}=-\frac{1}{x^{2}}$
a constant contributes nothing and the reciprocal is a single power rule
Answer $$\boxed{y'=-\frac{1}{x^{2}}}$$
Check
Same answer as the quotient rule route, reached without a subtraction — so nothing was there to lose a sign in.
Two lines, no sign trap.
Same function, same answer, four lines against two — and the longer route is the one with a subtraction in it, which is where marks are lost.
How to tell them apart
If the denominator is a single power of $x$, or divides the numerator exactly, split the fraction first. Keep the quotient rule for denominators that genuinely do not divide, such as $\frac{2x+1}{x^{2}+x+1}$.
f(x) = |x| at the origin: no derivative
Decide whether $f'(0)$ exists for $f(x)=\vert x\vert$.
Given
$f(x)=\vert x\vert$
$a=0$
Find
whether $f'(0)$ exists
SolutionBoth one sided quotients
$h>0:\ \frac{\vert h\vert-0}{h}=1$
on the right the graph is the line $y=x$
$h<0:\ \frac{\vert h\vert-0}{h}=-1$
on the left it is the line $y=-x$
$1\neq-1$
so the two sided limit does not exist
Answer $$\boxed{f'(0)\ \text{does not exist — corner}}$$
Check
The function is continuous at $0$, so this is a genuine corner and not a jump; Theorem 3.3 is not contradicted, since it never promised the converse.
f(x) = x|x| at the origin: derivative 0
Decide whether $f'(0)$ exists for $f(x)=x\vert x\vert$.
Given
$f(x)=x\vert x\vert$
$a=0$
Find
whether $f'(0)$ exists
SolutionBoth one sided quotients again
$\frac{h\vert h\vert-0}{h}=\vert h\vert$
the $h$ cancels, and what is left is the same expression on both sides
$\lim_{h\to0}\vert h\vert=0$
from the left and from the right alike
Answer $$\boxed{f'(0)=0}$$
Check
Independent check: $f(x)=x^{2}$ for $x\ge0$ and $-x^{2}$ for $x<0$, and both pieces have slope $0$ at the origin. So the graph really does have a horizontal tangent there.
Its derivative is $f'(x)=2\vert x\vert$, which is continuous but has a corner at $0$ — so $f''(0)$ does not exist. Differentiable once does not mean differentiable twice.
Both functions are built by gluing two formulas at the origin, and both graphs are continuous there — yet one has no tangent line and the other has a perfectly good horizontal one.
How to tell them apart
Never classify by how sharp the picture looks; compute both one sided difference quotients. If they agree, the point is fine no matter how the graph is drawn.
Scaffolding comes off
The common skeleton
Rewrite. Turn every root and reciprocal into a power of $x$ so that the sum rule can reach each term.
Differentiate. Apply the power rule term by term, using the product or quotient rule only where a term forces it.
Evaluate. Substitute the given $x$ into $f'$ — never into $f$ first.
Assemble. If a line is wanted, write $y-f(a)=f'(a)(x-a)$, with $f(a)$ computed from the original function.
Test. Check the slope's sign against the shape of the graph, or the value against a numerical difference quotient.
1 · fully worked
Tangent line to y = x³ − 4x at x = 1
Find the equation of the tangent line to $y=x^{3}-4x$ at $x=1$.
Given
$f(x)=x^{3}-4x$
$a=1$
Find
the equation of the tangent line
SolutionRewrite and differentiate
$f(x)=x^{3}-4x$
already a sum of powers, so the rewriting step costs nothing here
$f'(x)=3x^{2}-4$
power rule on each term; the coefficient $-4$ rides along
Evaluate both f and f prime at the point
$f(1)=1-4=-3$
the point of tangency is $(1,-3)$ — this comes from $f$, not from $f'$
$f'(1)=3-4=-1$
the slope comes from $f'$; keeping the two calculations apart is what stops the classic mix up
Assemble the line
$y-(-3)=-1(x-1)$
point slope form with the point and the slope just computed
$y=-x-2$
tidy up
Answer $$\boxed{y=-x-2}$$
Check
Numerical check on the slope: $\frac{f(1.001)-f(0.999)}{0.002}=\frac{-3.000997-(-2.998997)}{0.002}=-1.000$. Point check: the line at $x=1$ gives $-3$, which is $f(1)$, so it does pass through the point of tangency.
Two numbers, two sources: the height always comes from $f$ and the slope always from $f'$.
2 · you write the reasoning
The same skeleton on an easier function: find the tangent line to $y=x^{2}+5$ at $x=2$. The algebra is done for you — your job is to write down why each line is allowed. Writing the reason in your own words is the part that transfers to the next problem.
$f(x)=x^{2}+5$
reasoning
Step 1: nothing to rewrite — the function is already a sum of powers, and $5$ is a constant term rather than a factor.
$f'(x)=2x$
reasoning
Step 2: the power rule gives $2x$ for $x^{2}$, and the constant $5$ contributes $0$ because a horizontal line has no slope.
$f(2)=9$ and $f'(2)=4$
reasoning
Step 3: the height of the graph comes from $f$ and the slope from $f'$; they are computed from different formulas even though both use $x=2$.
$y-9=4(x-2)$, that is $y=4x+1$
reasoning
Step 4: point slope form through $(2,9)$ with slope $4$. Test: at $x=2$ the line gives $9$, matching the point of tangency, and its slope $4$ is positive, which fits a parabola rising to the right of its vertex.
3 · find the buried error
Harder, and now nobody is checking your work but you. This solution finds the tangent line to $y=\dfrac{x}{x^{2}+1}$ at $x=2$ and reaches an answer, and the answer is wrong. Exactly two of the four steps contain an error. Each step is consistent with the one before it, so you cannot find them by looking for a jump.
Step 1. $f(2)=\frac{2}{2^{2}+1}=\frac{2}{5}$, so the point of tangency is $\left(2,\frac{2}{5}\right)$.
Step 2. By the quotient rule, $f'(x)=\frac{x\cdot 2x-\left(x^{2}+1\right)\cdot 1}{\left(x^{2}+1\right)^{2}}=\frac{x^{2}-1}{\left(x^{2}+1\right)^{2}}$.
Step 4. So the tangent line is $y=\frac{3}{25}(x-2)$.
the two buried errors (2)
⚠ step 2
the quotient rule numerator is written in the wrong order. It should be $\frac{(1)\left(x^{2}+1\right)-x(2x)}{\left(x^{2}+1\right)^{2}}=\frac{1-x^{2}}{\left(x^{2}+1\right)^{2}}$, the negative of what is written.
the product rule is symmetric, so the order feels unimportant; the error survives because the answer keeps its size and only loses its sign
right
With the correct order, $f'(x)=\frac{1-x^{2}}{\left(x^{2}+1\right)^{2}}$ and $f'(2)=-\frac{3}{25}$.
⚠ step 4
the point of tangency has been dropped: point slope form is $y-f(2)=f'(2)(x-2)$, and $f(2)=\frac{2}{5}$ was computed in step 1 and then never used.
the slope is the part that felt like work, so it gets all the attention; the line is then written through the origin without anyone noticing
right
$y-\frac{2}{5}=-\frac{3}{25}(x-2)$, that is $y=\frac{-3x+16}{25}$. Test: at $x=2$ this gives $\frac{10}{25}=\frac{2}{5}$, which is $f(2)$.
4 · the bare problem
§03.5 — the bare problem●●●○○
No scaffolding. The function is fraction shaped, which is the decision the skeleton asks you to make first.
Given
$y=\dfrac{x^{2}+1}{x}$
the point $x=2$
Find
(a) Find the equation of the tangent line at that point.
Hint 1/4
Before differentiating, decide whether this is really a quotient or a fraction that can be split.
Hint 2/4
$\frac{x^{2}+1}{x}=x+x^{-1}$, so the power and sum rules are enough; then $y-f(a)=f'(a)(x-a)$.
Hint 3/4
$y'=1-x^{-2}$, so at $x=2$: $y'=1-\frac14=\frac34$, and $y(2)=\frac{4+1}{2}=\frac52$.
Hint 4/4
So the tangent line is $y-\frac52=\frac34(x-2)$, that is $y=\frac34x+1$.
Show solutionSplit the fraction rather than reaching for the quotient rule
$y=\frac{x^{2}}{x}+\frac{1}{x}=x+x^{-1}$
the denominator is a single power of $x$, so division is exact and the quotient rule is not needed
Independent check by the quotient rule: $y'=\frac{2x\cdot x-\left(x^{2}+1\right)}{x^{2}}=\frac{x^{2}-1}{x^{2}}$, which at $x=2$ is $\frac{3}{4}$ — the same slope by a route that does need the rule. Point check: the line at $x=2$ gives $\frac{3}{2}+1=\frac{5}{2}$.
The rewrite cost one line and removed the only place a sign could have been lost.
Full exam-style question
Full exam-style question on f(x) = x³ − 3xexam format
Let $f(x)=x^{3}-3x$. (a) Find $f'(x)$ from the definition of the derivative. (b) Find the equation of the tangent line at $x=2$. (c) Find every point of the graph where the tangent is horizontal. (d) Show that no tangent to this curve has slope less than $-3$, and say where that smallest slope occurs.
Given
$f(x)=x^{3}-3x$
part (a) must use the definition, not the rules
parts (b) to (d) may use the rules
Find
$f'(x)$; the tangent at $x=2$; all points with horizontal tangent; the smallest possible slope
Numerical check of part (b): $\frac{f(2.001)-f(1.999)}{0.002}=\frac{2.009006-1.991006}{0.002}=9.000$. Consistency check of (c) against (d): the slope is $0$ at $x=\pm1$ and $-3$ at $x=0$, so the curve falls between the two horizontal tangents and is steepest downhill exactly in the middle — which matches the shape of the graph in the earlier figure.
One derivative, asked about four different ways. That is the standard construction of a midterm question: the algebra is done once and then interrogated.
Notice that (c) and (d) never touch $f$ until the last line. Questions about slope are questions about $f'$; $f$ only reappears when a point on the graph is wanted.
Practice
A · concept 3 questions
1§03.3 — continuity is not enough●●○○○
One sentence, and it is the sentence that costs the most marks in this section. Decide it by finding either a proof or a single counterexample.
Given
Claim: every function that is continuous at $x=0$ has a derivative at $x=0$.
Find
(a) True or false, with a reason of one line.
Hint 1/4
A claim of the form every falls to one counterexample. Ask which continuous graph you have already seen that misbehaves at a point.
Hint 2/4
Differentiability at $0$ needs $\lim_{h\to0}\frac{f(h)-f(0)}{h}$ to exist as a finite number; continuity only needs $\lim_{x\to0}f(x)=f(0)$.
Hint 3/4
Take $f(x)=\vert x\vert$: it is continuous at $0$, and its quotient there is $\frac{\vert h\vert}{h}$, which is $+1$ for $h>0$ and $-1$ for $h<0$.
Hint 4/4
So the claim is false, and $\vert x\vert$ at $0$ is the counterexample.
so the conclusion of the claim fails, which is all a counterexample has to do
Answer $$\boxed{\text{False}}$$
Check
Cross check with Theorem 3.3: it states the implication in the other direction only, and a theorem never promises its converse. Two more counterexamples sit in the table of difference quotients: $x^{2/3}$ and $x^{1/3}$ at $0$.
Continuity is necessary for differentiability, not sufficient. Necessary and sufficient are not interchangeable words on an exam script.
2§03.5 — products do not differentiate factor by factor●●○○○
One product, four opinions about it. Only one of them survives the check that the same function can also be differentiated by the power rule after multiplying out.
Given
$y=x^{2}\cdot x^{3}$
Find
(a) Which statement about $\dfrac{dy}{dx}$ is correct?
Hint 1/4
This function can be differentiated two ways. Find the route that needs no new rule at all, and use it as the referee.
Hint 2/4
$x^{2}\cdot x^{3}=x^{5}$, and the power rule gives $\frac{d}{dx}x^{5}=5x^{4}$. The product rule must agree: $(fg)'=f'g+fg'$.
Hint 3/4
Product rule with $f=x^{2}$, $g=x^{3}$: $2x\cdot x^{3}+x^{2}\cdot3x^{2}=2x^{4}+3x^{4}$.
Hint 4/4
Both routes give $5x^{4}$.
Show solutionReferee route: multiply out first
$y=x^{5}\ \Rightarrow\ y'=5x^{4}$
exponents add when powers of the same base are multiplied, so no new rule is needed
derivative of the first times the second, plus the first times the derivative of the second
$=2x^{4}+3x^{4}=5x^{4}$
the two terms are the two strips in the rectangle picture, and their sum is the whole added area
Answer $$\boxed{5x^{4}}$$
Check
Check at $x=1$: both routes give $5$, while the plausible looking $f'g'=6x^{3}$ would give $6$. One test value is enough to expose it.
Whenever two rules both apply, run both. Agreement is the cheapest verification available in this chapter.
3§03.6 — one derivative does not buy the next●●●●○
A trap with a short answer and a specific counterexample. It is worth knowing because questions about $f''$ silently assume the answer.
Given
Claim: if $f'(a)$ exists, then $f''(a)$ exists.
Find
(a) True or false? If false, produce a counterexample.
Hint 1/4
$f''(a)$ is a statement about the function $f'$, not about $f$. So ask what $f'$ has to be like near $a$, not what $f$ is like.
Hint 2/4
$f''(a)$ exists exactly when $f'$ is differentiable at $a$, and a function is differentiable only where it has no corner. So the question is whether $f'$ can have a corner.
Hint 3/4
Take $f(x)=x\vert x\vert$, which is $x^{2}$ for $x\ge0$ and $-x^{2}$ for $x<0$. Then $f'(x)=2\vert x\vert$ — a corner at $0$.
Hint 4/4
So the claim is false: $f'(0)=0$ exists, but $f''(0)$ does not.
Show solutionWrite the function without the absolute value
the two sided quotient at the origin exists, so the branches join smoothly
$f'(x)=2\vert x\vert$
the two branch formulas plus the value at $0$ collapse into one absolute value
Second derivative: it fails at one point
$f''(0)\ \text{is the derivative of}\ 2\vert x\vert\ \text{at}\ 0$
which is the standard corner, twice as tall
$\lim_{h\to0^{\pm}}\frac{2\vert h\vert}{h}=\pm2\ \Rightarrow\ f''(0)\ \text{does not exist}$
two finite one sided answers that disagree
Answer $$\boxed{\text{False}}$$
Check
Independent check on $f'(0)=0$: the graph of $x\vert x\vert$ is a parabola for $x\ge0$ and an upside down parabola for $x<0$, and both have slope $0$ at the origin, so a horizontal tangent is right. Meanwhile $f''$ jumps from $-2$ to $2$ there.
Each new derivative is a fresh question about a new function. Twice differentiable is a stronger assumption than differentiable, which is why theorems state it separately.
B · computation 5 questions
1§03.1 — three derivatives from the definition●●●○○
Three functions of increasing algebraic difficulty. The three parts use the three standard ways of manufacturing a factor of $h$: expanding, rationalising, and putting fractions over a common denominator. Rules earn no marks here.
Given
$f_{1}(x)=x^{2}-4x+1$
$f_{2}(x)=\sqrt{x+3}$
$f_{3}(x)=\dfrac{1}{x+1}$
Find
(a) Compute $f_{1}'(3)$ from the definition.
(b) Compute $f_{2}'(1)$ from the definition.
(c) Compute $f_{3}'(x)$ for general $x\neq-1$ from the definition.
Hint 1/4
Each part is the same three moves: substitute, make $h$ a factor of the numerator, cancel. Decide in advance which move will do it for each function.
Hint 2/4
(a) expand $(3+h)^{2}$; (b) multiply by the conjugate $\sqrt{4+h}+2$; (c) put $\frac{1}{x+h+1}-\frac{1}{x+1}$ over one denominator. Then $f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$ in each case.
Rule check (allowed as verification, not as the solution): $f_{1}'(x)=2x-4$ gives $2$ at $x=3$; $\left(\sqrt{x+3}\right)'$ at $x=1$ should be $\frac{1}{2\sqrt{4}}=\frac14$; and $\left((x+1)^{-1}\right)'=-(x+1)^{-2}$. All three agree. Sign check on (c): $\frac{1}{x+1}$ decreases wherever it is defined, and the answer is negative everywhere.
Which of the three moves you need is decided by the shape of the function, not by the difficulty of the question. Polynomial to expand, root to rationalise, fraction to combine.
2§03.4 — rewrite, then differentiate●●○○○
A drill in the step everyone skips. Two of these three need to be rewritten before any rule can be applied, and one of them can be done two ways.
Given
$p(x)=6x^{4}-\tfrac12x^{3}+7x-11$
$q(x)=3\sqrt{x}-\dfrac{2}{x^{3}}$
$r(x)=(2x-1)^{2}$
Find
(a) Find $p'(x)$ and $p''(x)$.
(b) Find $q'(x)$, written with radicals and positive exponents.
(c) Find $r'(x)$ twice: by expanding first, and by the product rule on $(2x-1)(2x-1)$.
Hint 1/4
Look at each expression and ask whether the power rule can even see it. If it cannot, the first line of your answer is a rewrite.
Hint 2/4
$\frac{d}{dx}x^{n}=nx^{n-1}$ with $\sqrt{x}=x^{1/2}$ and $\frac{2}{x^{3}}=2x^{-3}$; and $(fg)'=f'g+fg'$ for part (c).
Hint 3/4
(a) $p'=24x^{3}-\tfrac32x^{2}+7$, then differentiate again. (b) $q=3x^{1/2}-2x^{-3}$, so $q'=\tfrac32x^{-1/2}+6x^{-4}$. (c) expanded: $r=4x^{2}-4x+1$.
Hint 4/4
(a) $p''=72x^{2}-3x$; (b) $q'=\frac{3}{2\sqrt{x}}+\frac{6}{x^{4}}$; (c) $r'=8x-4$ by both routes.
Show solution(a) Straight sum of powers
$p'(x)=24x^{3}-\tfrac{3}{2}x^{2}+7$
term by term; the constant $-11$ contributes nothing
$p''(x)=72x^{2}-3x$
differentiate the answer, not the original
(b) Rewrite before differentiating
$q(x)=3x^{1/2}-2x^{-3}$
the rule cannot see a root or a fraction
$q'(x)=\tfrac{3}{2}x^{-1/2}+6x^{-4}$
note the sign: $-2\cdot(-3)=+6$
$=\frac{3}{2\sqrt{x}}+\frac{6}{x^{4}}$
back into the notation the question used
(c) Two routes that must agree
$r(x)=4x^{2}-4x+1\ \Rightarrow\ r'(x)=8x-4$
expanding first turns it into a sum, which is the cheaper route here
$r'=(2)(2x-1)+(2x-1)(2)=4(2x-1)=8x-4$
product rule with both factors equal to $2x-1$; the two terms are identical, which is why the answer carries a factor of $2$
Numerical check on (b) at $x=1$: the formula gives $1.5+6=7.5$, and $\frac{q(1.001)-q(0.999)}{0.002}=7.5001$. Degree check on (a): a quartic has a cubic derivative and a quadratic second derivative. Route check on (c): the two methods agreed, which is the point of asking for both.
If you cannot see a rewrite, look for a bracket: $(2x-1)^{2}$ is the same function as $4x^{2}-4x+1$, and one of the two forms is always easier.
3§03.5 — quotient rule, then solve for zero slope●●●○○
A genuine quotient: the denominator does not divide the numerator and is not a single power of $x$, so there is no rewrite to escape into. The second part is the standard follow up.
Given
$v(x)=\dfrac{2x+1}{x^{2}+x+1}$
Find
(a) Find $v'(x)$.
(b) Find every $x$ at which the tangent to $y=v(x)$ is horizontal.
Hint 1/4
Check first whether the fraction can be split — if it cannot, the quotient rule is unavoidable. Then remember what horizontal tangent says about $v'$.
Hint 2/4
$\left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^{2}}$ with $f=2x+1$, $f'=2$, $g=x^{2}+x+1$, $g'=2x+1$. Horizontal tangent means $v'(x)=0$, and a fraction is zero exactly when its numerator is.
Hint 3/4
$v'=\frac{2\left(x^{2}+x+1\right)-(2x+1)(2x+1)}{\left(x^{2}+x+1\right)^{2}}=\frac{2x^{2}+2x+2-4x^{2}-4x-1}{\left(x^{2}+x+1\right)^{2}}$. Setting the numerator to zero: $-2x^{2}-2x+1=0$.
Hint 4/4
$v'(x)=\frac{1-2x-2x^{2}}{\left(x^{2}+x+1\right)^{2}}$, which is zero at $x=\frac{-1\pm\sqrt3}{2}$.
Show solutionCheck for an escape route first
$x^{2}+x+1\nmid 2x+1$
the denominator has higher degree, so no division simplifies this; the quotient rule it is
$x^{2}+x+1>0\ \text{for all}\ x$
its discriminant is $1-4=-3<0$, so the denominator never vanishes and $v$ is defined everywhere
Sanity check at $x=0$: the formula gives $v'(0)=\frac{1}{1}=1$, and numerically $\frac{v(0.001)-v(0)}{0.001}=\frac{1.000999-1}{0.001}=0.999$. Count check: $v$ rises then falls, so two horizontal tangents is the expected number, and the two roots straddle the interval where the numerator is positive.
Horizontal tangent, stationary point and $f'=0$ are three names for the same equation. Whenever a question uses one of them, write the other two down.
4§03.6 — higher derivatives and two patterns●●●○○
Repeated differentiation, and the two patterns worth memorising: the one that terminates and the one that never does.
Given
$f(x)=x^{5}+4x^{3}-6$
$g(x)=\dfrac{1}{x}$
$k(x)=x^{4}-2x^{3}$
Find
(a) Compute $f',f'',f''',f^{(4)},f^{(5)},f^{(6)}$. At what order does it become identically zero, and why?
(b) Compute the first four derivatives of $g$ and give a formula for $g^{(n)}(x)$.
(c) Compute $k''(1)$.
Hint 1/4
For (a) and (b) do not try to see the pattern before computing; compute three or four, then look.
Hint 2/4
Each differentiation is the power rule again. For (b) write $g(x)=x^{-1}$ so that every derivative is a single power.
(a) $f^{(6)}=0$, because each derivative drops the degree by one and the degree was $5$. (b) $g^{(n)}(x)=\frac{(-1)^{n}n!}{x^{n+1}}$. (c) $k''(x)=12x^{2}-12x$, so $k''(1)=0$.
Show solution(a) Differentiate until nothing is left
Test the general formula in (b) against a case computed by hand: $n=3$ gives $(-1)^{3}3!\,x^{-4}=-6x^{-4}$, matching $g'''$. Degree check in (a): $f^{(5)}$ must be a constant, since $5-5=0$.
An absurd order in the question — the twelfth derivative, the hundredth — is never a request to differentiate twelve times. It is a request to find a pattern and justify it.
5§03.4 — tangent lines and the slopes you are asked to hit●●●○○
One cubic, three questions about its tangents. This is the single most repeated exam pattern in the section, and all three parts are equations in $f'$.
Given
$y=x^{3}-3x^{2}+2$
Find
(a) Find the tangent line at $x=1$.
(b) Find all points where the tangent is horizontal.
(c) Find all points where the tangent has slope $9$.
Hint 1/4
All three parts ask about slope, so the first line of the answer is the same in each: differentiate once and keep the result in front of you.
Hint 2/4
$y'=3x^{2}-6x$. A tangent line needs $y-f(a)=f'(a)(x-a)$; horizontal means $y'=0$; slope $9$ means $y'=9$.
Hint 3/4
(a) $f(1)=0$ and $f'(1)=3-6=-3$. (b) $3x^{2}-6x=3x(x-2)=0$. (c) $3x^{2}-6x=9\Rightarrow x^{2}-2x-3=0\Rightarrow(x-3)(x+1)=0$.
Hint 4/4
(a) $y=-3x+3$; (b) $(0,2)$ and $(2,-2)$; (c) $(3,2)$ and $(-1,-2)$.
Show solutionDifferentiate once, use it three times
$f'(x)=3x^{2}-6x$
power rule, term by term
(a) The line at x = 1
$f(1)=1-3+2=0,\qquad f'(1)=3-6=-3$
height from $f$, slope from $f'$
$y-0=-3(x-1)\ \Rightarrow\ y=-3x+3$
point slope form through $(1,0)$
(b) Horizontal tangents
$3x^{2}-6x=3x(x-2)=0\ \Rightarrow\ x=0,\ x=2$
factoring beats the quadratic formula whenever the constant term is missing
$f(0)=2,\quad f(2)=8-12+2=-2$
the question asked for points, so both coordinates are needed
Numerical check of (a): $f(1.001)=-0.003000$ and $f(0.999)=0.003000$, so $\frac{f(1.001)-f(0.999)}{0.002}=-3.000$, matching $f'(1)=-3$. Consistency check between (b) and (c): the slope $-3$ found at $x=1$ sits between the two zeros of $f'$, which is exactly where the curve should be running downhill.
Every one of these parts converted an English sentence about tangents into an equation about $f'$. That translation is the skill being examined, not the algebra.
C · exam level 4 questions
1§03.3 — making a piecewise function differentiable●●●●○
The classic midterm question on this material. Two unknowns means two conditions are needed, and the whole question is knowing which two.
(a) Find $a$ and $b$ so that $f$ is differentiable at $x=1$ (and hence everywhere).
Hint 1/4
Two unknowns need two equations. Ask what would go wrong at $x=1$ if only the slopes matched, and what would go wrong if only the values matched.
Hint 2/4
Differentiability at $1$ requires continuity at $1$ (Theorem 3.3) and equal one sided difference quotients. That is the pair of equations.
Hint 3/4
Continuity: $\lim_{x\to1^{-}}x^{2}=1$ must equal $\lim_{x\to1^{+}}(ax+b)=a+b$, so $a+b=1$. Slopes: the left branch has slope $2x\big\vert_{x=1}=2$, the right branch has slope $a$, so $a=2$.
the left hand difference quotient, computed with the branch that applies on the left
$\lim_{x\to1^{+}}\frac{(ax+b)-1}{x-1}=a$
using $a+b=1$ from the first condition, the numerator is $a(x-1)$, and the quotient is the constant $a$
$a=2$
equal one sided quotients
Solve the pair
$a=2,\ a+b=1\ \Rightarrow\ b=-1$
the second equation only becomes usable once the first is in hand
Answer $$\boxed{a=2,\ b=-1}$$
Check
Check the answer geometrically: with $a=2,b=-1$ the right branch is $y=2x-1$, which is precisely the tangent line to $y=x^{2}$ at $x=1$. A curve continued by its own tangent has no visible kink — exactly what differentiability at the joint means.
Order matters: solve continuity first, because the slope condition is only meaningful once the graph is unbroken. Matching slopes across a jump is a mark losing move.
2§03.5 — rules applied to values from a table●●●●○
No formulas at all, only four numbers, so the only thing being tested is whether each rule is known in the right order. Read every entry of the table before starting.
Given
$f(2)=3$ and $f'(2)=-1$
$g(2)=4$ and $g'(2)=5$
Both $f$ and $g$ are differentiable at $2$
Find
(a) $(fg)'(2)$
(b) $\left(\dfrac{f}{g}\right)'(2)$
(c) $(5f-2g)'(2)$
Hint 1/4
Write down the three rules you will use before you look at any numbers. Then check that each rule's hypothesis is met by the table.
Hint 2/4
$(fg)'=f'g+fg'$; $\left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^{2}}$ (legal here since $g(2)=4\neq0$); $(cf\pm dg)'=cf'\pm dg'$.
Structural check: in (a) and (b) both a value and a derivative of each function appear, while in (c) only derivatives do — that is the fingerprint of a linear combination against a product. A quick model also confirms (a): take $f(x)=-x+5$ and $g(x)=5x-6$, which match the table; then $(fg)'(2)=11$ by direct expansion.
Table questions are pure rule recall. The numbers are chosen so that a misremembered rule gives a different answer — which is exactly why they appear on midterms.
3§03.1 — a limit that is a derivative in disguise●●●●○
A limit with no obvious factoring route. Questions of this shape are marked on whether you recognise the difference quotient, and the recognition is worth more than the algebra.
(a) This limit is the derivative of which function, at which point, and what is its value?
Hint 1/4
Compare the expression with $\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$ and ask what $f$ and $a$ would have to be for the two to match.
Hint 2/4
Matching term by term: $f(a+h)=\sqrt[3]{8+h}$ and $f(a)=2$ force $f(x)=x^{1/3}$ and $a=8$, since $\sqrt[3]{8}=2$. Then use $\frac{d}{dx}x^{n}=nx^{n-1}$.
Hint 3/4
$f'(x)=\frac13x^{-2/3}$, and at $x=8$: $8^{2/3}=\left(\sqrt[3]{8}\right)^{2}=4$.
Hint 4/4
So the limit is $f'(8)=\frac{1}{3\cdot4}=\frac{1}{12}$.
$8^{2/3}=4$, since the cube root of $8$ is $2$ and $2^{2}=4$
Answer $$\boxed{\frac{1}{12}}$$
Check
Numerical check: $\frac{\sqrt[3]{8.001}-2}{0.001}=\frac{2.0000833-2}{0.001}=0.0833$, and $\frac{1}{12}=0.0833$. Size check: the cube root grows slowly at $8$, so a small positive slope is what we should expect.
Any limit of the form $\lim_{h\to0}\frac{(\text{something in }h)-(\text{constant})}{h}$ should trigger the question whose derivative is this. Rationalising a cube root by hand is possible and slow; recognising it takes seconds.
4§03.2 — a derivative that is never negative●●●●○
A cubic whose derivative factors as a perfect square. That single algebraic fact answers two questions that would otherwise need machinery from later in the course.
Given
$y=x^{3}+3x^{2}+3x+3$
Find
(a) Find $y'$ and factor it.
(b) Show that no tangent to this curve has negative slope.
(c) Find every point where the tangent has slope $3$.
Hint 1/4
Differentiate first and look hard at the result before answering (b) and (c). The shape of $y'$ is the entire question.
Hint 2/4
$\frac{d}{dx}x^{n}=nx^{n-1}$ term by term; then recall that a squared real quantity is never negative, and that slope $3$ means $y'=3$.
Hint 3/4
$y'=3x^{2}+6x+3=3\left(x^{2}+2x+1\right)=3(x+1)^{2}$, which is $3$ times a square. For (c): $3(x+1)^{2}=3$.
Hint 4/4
(b) $y'\ge0$ always, with $y'=0$ only at $x=-1$; (c) $(x+1)^{2}=1$ gives $x=0$ and $x=-2$, so the points are $(0,3)$ and $(-2,1)$.
Show solution(a) Differentiate and factor
$y'=3x^{2}+6x+3$
power rule term by term; the constant $3$ contributes nothing
$=3\left(x^{2}+2x+1\right)=3(x+1)^{2}$
factoring is not decoration here — the next two parts read their answers straight off this form
(b) The sign argument
$(x+1)^{2}\ge0\ \text{for every real}\ x$
a square of a real number is never negative; no calculus is involved in this step
Numerical check at $x=-1$: $\frac{y(-0.999)-y(-1.001)}{0.002}=\frac{2.000000-2.000000}{0.002}\approx0$, agreeing with the single horizontal tangent. Symmetry check on (c): the two solutions $0$ and $-2$ sit one unit either side of $x=-1$, which is what $(x+1)^{2}=1$ has to give.
When a derivative factors into a square, questions about can the slope ever be negative stop needing any theory. Always factor $f'$ before answering anything about it.
D · interleaved 3 questions
1§03.1 — a limit with no obvious factoring●●●○○
Nothing in this problem announces which chapter it belongs to; that decision is the first part of the work. Two routes exist, and one of them takes about twenty seconds.
Given
$\displaystyle\lim_{x\to2}\frac{x^{5}-32}{x-2}$
Find
(a) Evaluate the limit.
Hint 1/4
Substitution gives $\frac{0}{0}$, so something has to be rewritten — but before grinding, look at the shape of the expression and ask what it reminds you of.
Hint 2/4
Either factor $x^{5}-32$ as $(x-2)\left(x^{4}+2x^{3}+4x^{2}+8x+16\right)$, or notice the pattern $\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=f'(a)$ with $f(x)=x^{5}$ and $a=2$, and use $\frac{d}{dx}x^{n}=nx^{n-1}$.
Hint 3/4
By the second route: $f'(x)=5x^{4}$, so the limit is $f'(2)=5\cdot2^{4}$. By the first, substituting $x=2$ into the long factor gives $16+16+16+16+16$.
five equal terms at $x=2$ — a nice check that the factorisation was done correctly
Answer $$\boxed{80}$$
Check
The two routes are genuinely independent — one uses the power rule, the other uses polynomial division — and they agree. Numerical check: at $x=2.001$ the quotient is $80.08$, closing in on $80$.
A $\frac{0}{0}$ limit whose numerator is $(\text{expression})-(\text{that expression's value at }a)$ is a derivative in disguise. From now on that recognition is usually the fastest route.
2§03.3 — repairing a hole, then testing the repair●●●○○
A function with a gap in its domain, a constant to choose, and then a question about the repaired function. The two halves come from different weeks of the course.
Given
$f(x)=\dfrac{x^{2}-9}{x-3}$ for $x\neq3$, and $f(3)=k$
Find
(a) Find the value of $k$ that makes $f$ continuous at $3$.
(b) With that $k$, decide whether $f'(3)$ exists, and find it if it does.
Hint 1/4
Simplify the formula for $x\neq3$ before doing anything else; the whole problem changes appearance once you do.
Hint 2/4
Continuity at $3$ needs $\lim_{x\to3}f(x)=f(3)=k$. Then differentiability is decided by $\lim_{x\to3}\frac{f(x)-f(3)}{x-3}$.
Hint 3/4
$\frac{x^{2}-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3$ for $x\neq3$, so $\lim_{x\to3}f(x)=6$; with $k=6$ the rule $f(x)=x+3$ holds at $x=3$ as well.
Hint 4/4
So $k=6$, and $f$ is the line $y=x+3$, whose derivative is $1$ everywhere: $f'(3)=1$.
the difference quotient is constant, which is what a straight line's always is
$f'(3)=\lim_{x\to3}1=1$
so the derivative exists and equals the slope of the line
Answer $$\boxed{k=6,\qquad f'(3)=1}$$
Check
Check the alternative: any other $k$ leaves a hole with a point floating above or below it, and Theorem 3.3 then rules out a derivative immediately — so $k=6$ is not just convenient, it is the only value for which part (b) can even be asked.
Filling a removable discontinuity with the limit value does more than restore continuity: it can hand back a function that is differentiable there too. That is exactly why the discontinuity is called removable.
3§03.1 — a difference quotient that oscillates●●●●●
The hardest problem in the set, and a genuine bonus part on past papers. Note that no derivative of the sine function is needed anywhere — only the fact that sine stays between $-1$ and $1$.
Given
$f(x)=x^{2}\sin\!\left(\dfrac{1}{x}\right)$ for $x\neq0$, and $f(0)=0$
$-1\le\sin\theta\le1$ for every $\theta$
Find
(a) Compute the difference quotient of $f$ at $0$ and explain why substitution cannot finish the job.
(b) Show that $f'(0)=0$.
Hint 1/4
Write the quotient at $a=0$ using $f(0)=0$, and see what is left. Then ask which theorem from the limits chapter deals with a factor that refuses to settle.
Hint 2/4
The Squeeze Theorem: if $g\le F\le h$ near $0$ and $g,h$ both tend to $L$, then $F\to L$. Combine it with the bound $\left\vert\sin\left(\frac1h\right)\right\vert\le1$.
Hint 3/4
$\frac{f(h)-f(0)}{h}=\frac{h^{2}\sin\left(\frac1h\right)}{h}=h\sin\left(\frac1h\right)$, and $-\vert h\vert\le h\sin\left(\frac1h\right)\le\vert h\vert$ because the sine factor never leaves $[-1,1]$.
Hint 4/4
Both bounds tend to $0$, so the middle does too, and $f'(0)=0$.
Show solution(a) Write the quotient and see what survives
one factor of $h$ cancels; the remaining $h$ is what will save the calculation
$\sin\!\left(\tfrac1h\right)\ \text{has no limit as}\ h\to0$
as $h$ shrinks, $\frac1h$ runs through ever more full turns, so the sine keeps sweeping between $-1$ and $1$ — the product law needs both factors to have limits, so it does not apply
the quotient is trapped between two things that go to zero, so it has no room to do anything else
Answer $$\boxed{f'(0)=0}$$
Check
Numerical spot check: at $h=0.01$ the quotient is $0.01\sin(100)=-0.0051$, at $h=0.001$ it is $0.001\sin(1000)=0.00083$ — the values jump around in sign, exactly as the oscillation predicts, but their size is bounded by $\vert h\vert$ and collapses. Bound check: every value listed is inside $[-\vert h\vert,\vert h\vert]$.
A derivative can exist at a point where the function wobbles infinitely often nearby. The lesson that transfers: when a factor refuses to settle but is bounded, do not evaluate it — trap it.
Mistake ledger (17 entries)
⚠ Setting h = 0 before cancelling
the instruction let h go to zero gets read as substitute zero, which worked for every limit met before this one
$f$ is defined on an open interval containing $a$; the limit below exists and is a finite number
The tangent line at a point
${\ y-f(a)=f'(a)\,(x-a)\ }$
$f'(a)$ exists; the height $f(a)$ comes from $f$; the slope $f'(a)$ comes from $f'$
The derivative function
${\ f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\ }$
the same limit as Definition 3.1, with the fixed number $a$ replaced by a variable $x$; the domain of $f'$ is the set of $x$ at which the limit exists, and it can be smaller than the domain of $f$
$f$ and $g$ are differentiable at $x$, and $c$ is a constant; $n$ is any real number — the derivation below covers positive integers, and the roots and reciprocals of this course obey the same formula
Close the page. On one sheet, write from memory: the definition of $f'(a)$ as a limit, the four rules with the condition each one carries, and the three names for the ways a derivative can fail to exist. Then open the formula card and mark what was missing. What you missed is your revision list — the rest of the section is not.
Compute $f'(2)$ from the definition for $f(x)=x^{2}-4x$, showing the cancellation of $h$?
c-derivative-at-a-point
Say what $f'(3)=-2$ tells you about the graph, and what it does not tell you?
c-derivative-as-function
State the implication between differentiable and continuous in the right direction, and give the counterexample for the wrong one?
c-differentiability
Differentiate $5\sqrt{x}-\frac{3}{x^{2}}$ without hesitating over the rewriting step?
c-power-sum-rules
Write the quotient rule with its numerator in the correct order, and say when you would avoid it altogether?
c-product-quotient
Find the acceleration from a position formula, and explain why $f''$ is not $(f')^{2}$?
c-higher-derivatives
Glossary (18 terms)
derivativetürev
The limit $\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$ when it exists as a finite number; the slope of the tangent line at $a$, and the instantaneous rate of change of $f$ there.
derivative functiontürev fonksiyonu
The function $f'$ that assigns to each $x$ the derivative of $f$ at $x$. Its domain is the set of points where that limit exists, which can be smaller than the domain of $f$.
differentiabletürevlenebilir
Having a derivative. A function differentiable at a point is automatically continuous there; the converse fails.
average rate of changeortalama değişim oranı
$\frac{f(b)-f(a)}{b-a}$: the slope of the line joining two points of the graph, computed over an interval rather than at a point.
Leibniz notationLeibniz gösterimi
Writing the derivative as $\frac{dy}{dx}$ rather than $f'(x)$. It names the variables and suggests a ratio of changes, but it is a single symbol, not a fraction.
cornerköşe
A point where the graph is unbroken but the one sided difference quotients approach two different finite numbers, so no derivative exists. Example: $\vert x\vert$ at $0$.
cuspsivri uç
A point where the one sided difference quotients run to infinity with opposite signs. Example: $x^{2/3}$ at $0$.
vertical tangentdüşey teğet
A point where the one sided difference quotients run to infinity with the same sign; the tangent line is vertical and the derivative does not exist. Example: $x^{1/3}$ at $0$.
power rulekuvvet kuralı
$\frac{d}{dx}x^{n}=n\,x^{n-1}$. Applies to every real exponent, which is why roots and reciprocals must first be rewritten as powers.
constant multiple rule
$(cf)'=c\,f'$: a constant factor passes through differentiation untouched.
sum ruletoplam kuralı
$(f\pm g)'=f'\pm g'$: differentiation distributes over addition and subtraction, so polynomials can be handled term by term.
product ruleçarpım kuralı
$(fg)'=f'g+fg'$. Two terms, not one: growing both factors at once adds two strips of area, which is what the rectangle picture shows.
quotient rulebölüm kuralı
$\left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^{2}}$, valid where $g\neq0$. The order in the numerator is fixed, since subtraction is not symmetric.
second derivativeikinci türev
$f''=(f')'$, written $\frac{d^{2}y}{dx^{2}}$ in Leibniz notation. For motion it is acceleration; it is not the square of the first derivative.
yüksek mertebeden türev
$f^{(n)}$, obtained by differentiating $n$ times. For a polynomial of degree $n$ every derivative past the $n$th is zero.
one sided derivativetek yanlı türev
The limit of the difference quotient taken from one side only, $\lim_{x\to a^{\pm}}\frac{f(x)-f(a)}{x-a}$. The two sided derivative exists exactly when both are finite and equal.
velocityhız
$s'(t)$ for a position function $s$: the rate at which position changes, signed, so a negative value means motion in the negative direction.
accelerationivme
$s''(t)=v'(t)$: the rate at which velocity changes. Constant acceleration produces a straight line velocity graph and a bending position graph.
What comes next
§04 · Chain rule, implicit differentiation, and derivatives of trigonometric functions
Everything here was built for functions written as sums, products and quotients of powers. Next week the functions start arriving nested inside one another — $\sin\left(x^{2}\right)$, $\sqrt{x^{2}+9}$ — and none of this week's rules reaches them. One more rule does, and it is the one that makes every later application possible.
Sources
James Stewart, Calculus, Metric Version, Ninth Edition — sections 2.1, 2.2 and 2.3 The topic split follows the week 3 line of the course syllabus; section numbers are quoted only as they appear there.
Course syllabus: assessment weights and the minimum requirement Midterm 1 28%, Midterm 2 28%, Final 28%, quizzes 10%, homework 6%; a combined midterm score under 40 out of 200 gives an FZ.
Standard conventions used here Continuity and differentiability are asked only at points of the domain; an infinite difference quotient is reported as *does not exist* plus a geometric name.