15 worked examples24 exercises7 figuresbölüm 12 / 14
12Inverse trigonometric functions, hyperbolic functions, and indeterminate forms revisited
A student checks $e$ the slow way: take $1+\tfrac{1}{1000}$ and raise it to the power $1000$. Every extra factor moves the answer by a tenth of one percent, so it should stay near $1$. The calculator returns $2.7169$.
By the end you can name what kind of fight that expression is, turn it into a quotient a derivative can attack, and get the exact value.
In 60 seconds
Three new derivative families — general bases, inverse trigonometric, hyperbolic — come out of one idea, invert then differentiate; l'Hôpital's rule handles the limits they produce.
Substitution gives $0/0$ or $\infty/\infty$, and only then.
Three most common mistakes
Writing $(f^{-1})'(b)=1/f'(b)$. The denominator is $f'$ at the point that maps to $b$, not at $b$ itself.
Applying l'Hôpital's rule to a limit that is not indeterminate. The rule then hands back a wrong number instead of a warning.
Taking the logarithm of a power form and then reporting $\ln L$ as the answer. The last line of that method is always an exponential.
The weighting sheet gives two midterms at $28\%$ each, a final at $28\%$, quizzes $10\%$, homework $6\%$. It says nothing about how many questions come from any one week.
How much time do you have?
10 minutes
You leave able to differentiate the six standard inverses and to name the form of a limit before touching it.
In 60 seconds card, Reciprocal slopes: the derivative of an inverse, Naming the form and choosing the move, Formula card
45 minutes
Add the derivations, the sign family and the logarithm trick; you can now answer a full multi-part question instead of quoting a table.
everything in the 10 minute path, Restricting the sine until it can be inverted, Where the minus signs come from, l'Hôpital's rule and the two forms it accepts, The five forms that need a rewrite first, Scaffolding comes off, Full exam-style question
full reading
Everything above plus the general bases, the hyperbolic family, and practice graded from concept traps to problems that hide their type.
all blocks in order, Practice A, B, C and D, Mistake ledger, Check yourself
By the end of this section
Compute the derivative of an inverse function at a point from the derivative of the original function, without a formula for the inverse.
Differentiate $a^{x}$, $\log_a x$ and $x^{n}$ for a real exponent by rewriting each one through $e$ and $\ln$.
Derive the derivatives of $\arcsin$ and $\arctan$ by implicit differentiation on the , and apply them with the chain rule.
Explain where the minus signs and the absolute value in the inverse trigonometric derivative table come from, and state the interval on which each formula is valid.
Manipulate $\sinh$, $\cosh$ and $\tanh$ through their exponential definitions, and read an as a logarithm.
Decide whether l'Hôpital's rule may be applied to a given limit, and apply it repeatedly with a form check between applications.
Convert $0\cdot\infty$, $\infty-\infty$, $0^{0}$, $1^{\infty}$ and $\infty^{0}$ into a quotient the rule accepts, and undo the conversion at the end.
Syllabus coverage
6.4*
General logarithmic and exponential functions
Bases other than $e$: the derivatives of $a^{x}$ and $\log_a x$, the power rule for a real exponent, and the limit that produces $e$.
covered
6.8
and l'Hôpital's rule
The statement and its hypotheses, why it holds for $0/0$, the two ways it is misused, and the five forms that need a rewrite first.
covered
6.6
Principal branches, the derivations of the derivatives, the sign family from the complementary identity, and the interval each formula holds on.
covered
Hyperbolic and inverse hyperbolic functions
The week's syllabus line names no book section for these, so none of it is required material. It is here because the section title names them and because their inverses are the one place the reciprocal-slope argument can be checked against a closed form.
Every derivative in this section is a composite, and the inner factor is the one that gets dropped.
Implicit differentiation
If $y$ is a function of $x$, then $\dfrac{d}{dx}F(y)=F'(y)\cdot\dfrac{dy}{dx}$; differentiate both sides of an equation and solve for $\dfrac{dy}{dx}$.
It is the engine behind every inverse derivative here: we never solve for the inverse, we differentiate the equation that defines it.
Derivatives of the natural pair
$\dfrac{d}{dx}\ln x=\dfrac1x$ for $x>0$, and $\dfrac{d}{dx}e^{x}=e^{x}$.
Both were built in the previous section, and every general base is rewritten through them.
Pythagorean identities
$\sin^{2}\theta+\cos^{2}\theta=1$ and $1+\tan^{2}\theta=\sec^{2}\theta$.
They convert $\cos(\arcsin x)$ and $\sec^{2}(\arctan x)$ into algebra, which is where the roots in the table come from.
Monotone means invertible
If $f'>0$ on an interval, then $f$ is strictly increasing there, hence one to one and invertible on that interval.
Every inverse in this section exists only because a domain was cut down to a piece where the derivative keeps one sign.
Try it yourself first (2 questions)
1§12.0 — implicit differentiation, one line of it●●○○○
Everything in this section is built by differentiating an equation in which $y$ is not isolated. This is that step, on its own.
Given
$\sin y=x$, where $y$ is a function of $x$
Find
(a) Which line is the result of differentiating both sides with respect to $x$?
Hint 1/4
Ask what $y$ is here: a number or a function of $x$? That decides whether differentiating $\sin y$ needs the chain rule.
Hint 2/4
The chain rule on the left: $\dfrac{d}{dx}\sin y=\cos y\cdot\dfrac{dy}{dx}$, and the right side differentiates to $1$.
Hint 3/4
So the equation becomes $\cos y\cdot y'=1$, with $y'$ the unknown to be isolated.
Hint 4/4
The correct line is $\cos y\cdot y'=1$.
Show solutionDifferentiate both sides
$\frac{d}{dx}\sin y=\cos y\cdot\frac{dy}{dx}$
$y$ depends on $x$, so the chain rule applies on the left
$\frac{d}{dx}x=1$
the right side
$\cos y\cdot y'=1$
set the two equal
Answer $$\boxed{\cos y\cdot y'=1}$$
Check
Check on a case you can solve: $y=\arcsin x$ at $x=0$ gives $\cos 0\cdot y'=1$, so $y'=1$ — and $\arcsin$ does leave the origin with slope $1$.
2§12.0 — the natural logarithm and the chain rule●●○○○
The general bases here are all built on the natural logarithm, so this derivative has to be automatic before the new material starts.
Given
$y=\ln\left(x^{2}+1\right)$
Find
(a) Compute $y'$ and evaluate it at $x=2$.
Hint 1/4
Identify the inner function; the logarithm is applied to something that is not just $x$.
Hint 2/4
$\dfrac{d}{dx}\ln u=\dfrac{u'}{u}$.
Hint 3/4
Here $u=x^{2}+1$ and $u'=2x$, so $y'=\dfrac{2x}{x^{2}+1}$; the point asked for is $x=2$.
Hint 4/4
$y'(2)=\dfrac{4}{5}$.
Show solutionChain rule
$y'=\frac{2x}{x^{2}+1}$
outer derivative $1/u$, inner derivative $2x$
$y'(2)=\frac{4}{5}=0.8$
substitute
Answer $$\boxed{y'(2)=0.8}$$
Check
Numerical check with a symmetric difference across $x=2$: $\ln\left(2.01^{2}+1\right)-\ln\left(1.99^{2}+1\right)=1.617426-1.601426=0.016000$, and the two points are $0.02$ apart, so the measured slope is $0.800$.
Notation
symbol
reads as
means
watch out
$\arcsin x$
arc sine of x
the angle in $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ whose sine is $x$
also written $\sin^{-1}x$; the $-1$ means inverse function, never $1/\sin x$
$\arctan x$
arc tangent of x
the angle in $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ whose tangent is $x$
defined for every real $x$, unlike the arcsine
$\log_a x$
log base a of x
the exponent to which $a$ must be raised to give $x$
to differentiate it, first rewrite it as $\ln x/\ln a$
$\sinh x,\ \cosh x$
hyperbolic sine and hyperbolic cosine of x
the odd and even halves of $e^{x}$
neither is periodic and $\cosh x\ge 1$ always; the resemblance to the circular pair is in the algebra, not the graphs
$\operatorname{sech} x$
sech of x
$1/\cosh x$
appears squared in $(\tanh x)'$, and $\operatorname{sech}^{2}x=1-\tanh^{2}x$
$0\cdot\infty,\ \infty-\infty,\ 1^{\infty}$
the indeterminate shapes
shorthand for how the two parts of a limit behave, not arithmetic on numbers
a shape is a question; writing $1^{\infty}=1$ answers it without doing the work
Conventions used here
Angles are radians
Every derivative formula here assumes radians. In degrees the sine has derivative $\tfrac{\pi}{180}\cos x$ and the whole table changes.
The formulas descend from $\lim_{h\to 0}\tfrac{\sin h}{h}=1$, which is true only in radians.
Which branch an inverse returns
$\arcsin$ returns a value in $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$, $\arccos$ in $[0,\pi]$, $\arctan$ in $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$. When a question asks for an angle, that is the one meant.
$\sin\theta=\tfrac12$ has infinitely many solutions; a function must pick one, and that choice is what makes the derivative formulas well defined.
The secant case is a convention, and we say so
For $x>1$ every book agrees that $\left(\operatorname{arcsec} x\right)'=\dfrac{1}{x\sqrt{x^{2}-1}}$. For $x<-1$ the sign depends on the branch the book chose, so here the secant inverse appears only where the question states its range.
An unstated branch turns a sign question into a coin toss, and a coin toss is not a solution.
How a limit is reported
Writing $\lim=+\infty$ says the limit does not exist and says how it fails. We keep it when true, and reserve does not exist for values that oscillate.
The infinite statement carries information that the bare denial throws away.
Indeterminate is about limits, not arithmetic
$0^{0}$, $1^{\infty}$, $\infty^{0}$, $0\cdot\infty$ and $\infty-\infty$ name the behaviour of a limit, not an arithmetic value; as limits they can come out to anything.
Two limits with the same shape were shown here to give $e$ and $e^{3}$, so the shape cannot be the answer.
Where a derivative formula stops
$(\arcsin x)'$ is quoted for $-11$ only: endpoints with a vertical tangent are excluded even though the function is defined there.
Substituting an endpoint divides by zero, and the division by zero is the graph reporting a vertical tangent.
Reciprocal slopes: the derivative of an inverse
The last section built one inverse pair by hand. This one needs six more, and none can be solved for with algebra, so the first job is to differentiate an inverse we cannot write down.
Invert $y=3x+5$ by hand and differentiate the result.
Differentiate $\ln x$, because the last section built it.
Not solvable yet
Solve $y=x^{3}+2x+1$ for $x$, so the inverse has no formula to differentiate.
Solve $x=\sin y$ for $y$ with algebra.
Write $\arcsin x$ with roots and quotients.
The graphs of $f$ and $f^{-1}$ are mirror images across $y=x$, so perhaps the slopes are mirror images too and $\left(f^{-1}\right)'(b)=\dfrac{1}{f'(b)}$.
Why it fails
Reflection does reciprocate the slope — it swaps rise and run — but it also moves the point, and the naive formula forgets that. With $f(x)=x^{3}+2x+1$ and $b=4$: since $f(1)=4$, the true value is $1/f'(1)=\tfrac{1}{5}$ and the naive $1/f'(4)=\tfrac{1}{50}$ is ten times too small.
TheoremTheorem 12.1: derivative of an inverse function
Conditions
$f$ is one to one and differentiable on an open interval $I$
The slope of the inverse at an output equals one over the slope of the original function at the input that produced that output.
Where it comes from
Differentiate $f\!\left(f^{-1}(x)\right)=x$ with the chain rule: $f'\!\left(f^{-1}(x)\right)\cdot\left(f^{-1}\right)'(x)=1$, then divide. What this does not prove is that $f^{-1}$ is differentiable at all — that is what the hypothesis $f'(a)\neq 0$ buys.
Reflection swaps the coordinates of a point and therefore swaps rise with run: a steep slope becomes a shallow one at the mirrored point, never at the same point.
Looks like this, but is not
$f(x)=x^{3}$ is one to one on the whole line, differentiable everywhere, and its inverse $x^{1/3}$ is written down explicitly. So the theorem should apply at $b=0$.
At $a=0$ we have $f'(0)=0$ and the conclusion fails: $x^{1/3}$ has a vertical tangent at the origin, so $\left(f^{-1}\right)'(0)$ does not exist. A zero slope reflects into a vertical one.
The slope of the inverse of x cubed plus 2x plus 1 at the output 4
$f(x)=x^{3}+2x+1$ has no usable inverse formula: solving $y=x^{3}+2x+1$ for $x$ needs the cubic formula. We want the slope of that inverse at $y=4$.
Given
$f(x)=x^{3}+2x+1$
the output $b=4$
Find
$\left(f^{-1}\right)'(4)$
SolutionCheck that an inverse exists
$f'(x)=3x^{2}+2\ge 2>0$
a sum of a square and a positive constant; positive derivative on an interval means strictly increasing, hence one to one
Find the input that produces 4
$f(1)=1+2+1=4$
small integers first: guessing is legitimate here because $f$ is increasing, so a hit is the only hit
$a=f^{-1}(4)=1$
that is the point where the rule wants the derivative
Independent numerical check: $f(1.02)=4.101208$, so the inverse takes $0.02$ of input for $0.101208$ of output, a slope of $0.1976$ near $y=4$ — that is $0.2$ to the accuracy this step size allows.
One guess, one derivative and one division; the cubic formula was never needed.
Read the two numbers in the right order and the rule is mechanical: the output says where to stand on the graph of $f$, and the slope found there gets flipped.
Checkpoint
§12.1 — slope of an inverse from a single value●●○○○
Everything you know about $f$ is three lines of data. Nobody has written a formula for the inverse, and you will not need one.
Given
$f$ is differentiable and increasing on all of $\mathbb{R}$
$f(3)=7$
$f'(3)=2$
Find
(a) Compute $g'(7)$, where $g=f^{-1}$.
Hint 1/4
You are asked for a slope of the inverse at the output $7$. First find which input of $f$ produced it; that is where the work happens.
Hint 2/4
The rule is $\left(f^{-1}\right)'(b)=\dfrac{1}{f'\!\left(f^{-1}(b)\right)}$.
Hint 3/4
Here $b=7$, and $f(3)=7$ means $f^{-1}(7)=3$. The data once more: $f(3)=7$ and $f'(3)=2$.
Hint 4/4
The answer is $g'(7)=\tfrac{1}{2}$.
Show solutionFind the input behind the output
$f(3)=7 \;\Longrightarrow\; f^{-1}(7)=3$
the inverse reads the same pair of numbers backwards
Apply the rule at that input
$g'(7)=\frac{1}{f'(3)}$
the rule evaluates $f'$ at $f^{-1}(7)=3$, not at $7$
$=\frac{1}{2}$
the given value $f'(3)=2$
Answer $$\boxed{g'(7)=\tfrac{1}{2}}$$
Check
$f$ is increasing, so $g$ is too and the answer had to be positive; $f$ climbs $2$ units of output per unit of input near $x=3$, so $g$ climbs half a unit the other way.
Nothing here needed a formula for $g$. That is the whole point of the rule: one value and one derivative of $f$ are enough.
⚠ Evaluating the derivative at the output instead of the input
The number $b$ is the one in the question, so the eye puts it in every slot it fits.
Before spending the reciprocal-slope theorem on trigonometry, spend it on what the last section left unfinished: powers and logarithms with a base that is not $e$.
DefinitionDefinition 12.2: general exponential and logarithm
Every base is the natural base in disguise: $a$ to a power is $e$ to that power times $\ln a$, and each derivative differs from the natural one by that constant — multiplying for the exponential, dividing for the logarithm.
Both derivatives in two lines
$\dfrac{d}{dx}e^{x\ln a}=e^{x\ln a}\ln a=a^{x}\ln a$ by the chain rule, the inner derivative being the constant $\ln a$. And $\log_a x$ is $\dfrac{1}{\ln a}$ times $\ln x$, so its derivative is $\dfrac{1}{\ln a}\cdot\dfrac{1}{x}$.
Every exponential passes through the same point, so what separates the bases is the steepness there — and that steepness is exactly $\ln a$. Base $e$ is the one with tangent slope $1$.
Looks like this, but is not
$\log_2$ is just $\ln$ with a different label on the base, so its derivative should also be $\dfrac{1}{x}$.
$\log_2 x=\dfrac{\ln x}{\ln 2}$ with $\ln 2\approx 0.693$, so the derivative is $\dfrac{1.443}{x}$ — about $44\%$ steeper than $1/x$ everywhere. A rescaled function has a rescaled slope.
shape
which rule
derivative
$x^{n}$, $n$ constant
power rule, valid for every real $n$ when $x>0$
$nx^{n-1}$
$a^{x}$, $a$ constant
general exponential
$a^{x}\ln a$
$u(x)^{v(x)}$
neither: take $\ln$ of both sides first
found case by case
Read the middle column before the third: the position of the variable, not the look of the expression, decides which line you are on.
Derivative of 3 to the power x squared
A constant base with a composite exponent: two rules meet in one line.
Given
$y=3^{x^{2}}$
Find
$y'$
SolutionName the outer and the inner function
$y=3^{u},\quad u=x^{2}$
the outer function is a constant to a power, the inner one is the exponent
Differentiate outer times inner
$\frac{dy}{du}=3^{u}\ln 3$
the general base rule
$\frac{du}{dx}=2x$
power rule on the exponent
$y'=3^{x^{2}}\ln 3\cdot 2x$
chain rule, then put $u=x^{2}$ back
Answer $$\boxed{y'=2x\,3^{x^{2}}\ln 3}$$
Check
Independent check by symmetry: $3^{x^{2}}$ is even, and an even function differentiable at $0$ has derivative $0$ there. The formula gives $2\cdot 0\cdot 1\cdot\ln 3=0$.
A constant base never leaves; it only picks up the factor $\ln a$.
Where the number e comes from as a limit
The definition $a^{x}=e^{x\ln a}$ takes $e$ for granted. Here is $e$ produced by a derivative — the exact statement the opening calculator experiment was groping at.
Given
$f(x)=\ln x$
$f'(1)=1$
Find
$\lim_{h\to 0}(1+h)^{1/h}$
SolutionWrite the known derivative as its difference quotient
$1=f'(1)=\lim_{h\to 0}\frac{\ln(1+h)-\ln 1}{h}$
the derivative of $\ln$ at $1$ is $1/1$, and it is also a limit by definition
$=\lim_{h\to 0}\frac{1}{h}\ln(1+h)$
$\ln 1=0$
Move the factor inside the logarithm
$=\lim_{h\to 0}\ln\left[(1+h)^{1/h}\right]$
a coefficient in front of a logarithm is an exponent inside it
Undo the logarithm
$\lim_{h\to 0}(1+h)^{1/h}=e^{1}=e$
$\exp$ is continuous, so it may be moved through the limit
Answer $$\boxed{\lim_{h\to 0}(1+h)^{1/h}=e\quad\text{and, with }h=\tfrac{1}{n},\quad \lim_{n\to\infty}\left(1+\tfrac{1}{n}\right)^{n}=e}$$
Check
Numerical check at $n=1000$: $(1.001)^{1000}=2.71692$ against $e=2.71828$. The gap shrinks like $1/n$, which is why the calculator needed a large exponent.
The base creeping to $1$ and the exponent running to infinity fight, and neither wins outright. That fight has a name and a method, both later in this section.
Checkpoint
§12.2 — differentiating a constant to a variable power●●○○○
Thirty seconds. The base is a constant and the exponent is the variable, which is the opposite arrangement from the power rule.
Given
$y=5^{x}$
Find
(a) Which expression is $\dfrac{dy}{dx}$?
Hint 1/4
Decide first which of the two positions the variable occupies: base or exponent. That decision picks the rule; nothing else does.
Hint 2/4
For a constant base, $\dfrac{d}{dx}a^{x}=a^{x}\ln a$, because $a^{x}=e^{x\ln a}$.
Hint 3/4
Here $a=5$, so the constant that the chain rule pulls out is $\ln 5$, and the function $5^{x}$ itself survives untouched.
Hint 4/4
The derivative is $5^{x}\ln 5$.
Show solutionRewrite through the exponential we can differentiate
$5^{x}=e^{x\ln 5}$
$\ln 5$ is a constant, so this is $e$ to a constant times $x$
$\frac{d}{dx}e^{x\ln 5}=e^{x\ln 5}\cdot\ln 5$
chain rule; the inner derivative is the constant $\ln 5$
$=5^{x}\ln 5$
translate back
Answer $$\boxed{5^{x}\ln 5}$$
Check
Sanity check on size: $\ln 5\approx 1.609>1$, so $5^{x}$ grows faster than $e^{x}$ does at the same height, which is what a base larger than $e$ should do.
⚠ Using the power rule on a constant base
The shape $\text{something}^{\text{something}}$ triggers the most practised rule.
wrong$\frac{d}{dx}2^{x}=x\,2^{x-1}$
right$\frac{d}{dx}2^{x}=2^{x}\ln 2$
⚠ Putting the constant on the wrong side of the fraction
Both formulas contain $\ln a$, and only the position distinguishes them.
wrong$\frac{d}{dx}\log_a x=\frac{\ln a}{x}$
right$\frac{d}{dx}\log_a x=\frac{1}{x\ln a}$
Restricting the sine until it can be inverted
The theorem needs a one-to-one function, and no trigonometric function is one to one. So the first move is not calculus but surgery on the domain.
TheoremTheorem 12.3: the two derivatives you will use most
Conditions
$\arcsin:[-1,1]\to\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$, the branch on which the sine increases
The arcsine climbs at one over the cosine of the angle it returns, and on its branch that cosine is $\sqrt{1-x^{2}}$; the arctangent climbs at $1/(1+x^{2})$, never zero, so it is defined for every real number.
Both, by implicit differentiation
Put $y=\arcsin x$, so $\sin y=x$ with $y\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$. Differentiating both sides with respect to $x$ gives $\cos y\cdot y'=1$, hence $y'=1/\cos y$. On that branch $\cos y\ge 0$, so $\cos y=+\sqrt{1-\sin^{2}y}=\sqrt{1-x^{2}}$ — the sign is decided by the branch, not by taste. For the arctangent, $\tan y=x$ gives $\sec^{2}y\cdot y'=1$ and $\sec^{2}y=1+\tan^{2}y=1+x^{2}$.
Why an algebraic root appears in the derivative of a trigonometric inverse: write $\theta=\arcsin \textcolor{#1f6feb}{x}$ into a right triangle with hypotenuse $1$, and $\cos\theta$ is read off as $\textcolor{#d1690a}{\sqrt{1-x^{2}}}$.
Looks like this, but is not
The notation $\sin^{-1}x$ has an exponent $-1$ in it, and $u^{-1}$ means $1/u$ everywhere else in algebra, so $\sin^{-1}\left(\tfrac12\right)$ should be $2$.
$\sin^{-1}\left(\tfrac12\right)=\tfrac{\pi}{6}\approx 0.524$, an angle, while $\dfrac{1}{\sin\left(\tfrac12\right)}\approx 2.086$, a ratio. In this notation $-1$ means inverse function; the reciprocal is $\csc$.
Derivative of arctan of e to the x
A composite whose inner function is itself an exponential; two known rules meet.
Given
$y=\arctan\left(e^{x}\right)$
Find
$y'$
SolutionSplit into outer and inner
$y=\arctan u,\quad u=e^{x}$
the arctangent is applied to something, so it is the outer function
Two independent checks. Sign: $\arctan$ and $e^{x}$ both increase, so the derivative must be positive, and it is a quotient of positive numbers. End behaviour: as $x\to\infty$ the formula tends to $0$, matching the flattening curve.
Inverse trigonometric functions are ordinary outer functions; nothing about them changes the chain rule.
The combination whose derivative collapses to arcsine
This one looks heavier than it is, and the payoff is a result you will meet again when integrating.
the two awkward terms are exact negatives, which is the reason this combination is worth remembering
Answer $$\boxed{F'(x)=\arcsin x}$$
Check
Numerical check at $x=\tfrac12$, where the answer should be $\arcsin 0.5=0.523599$: a symmetric difference quotient with step $0.01$ gives $\dfrac{F(0.51)-F(0.49)}{0.02}=0.523612$, agreeing to four decimals.
Product rule once, chain rule once, and the whole middle of the answer cancels.
Read backwards this says that $x\arcsin x+\sqrt{1-x^{2}}$ is an antiderivative of $\arcsin$, which is how the next chapter will use it.
Checkpoint
§12.3 — the arcsine derivative at a point●●○○○
Thirty seconds, no chain rule involved. The formula is quoted at a single point well inside the domain.
Given
$y=\arcsin x$
the point $x=\tfrac{1}{2}$
Find
(a) Evaluate $y'\left(\tfrac{1}{2}\right)$ and leave the answer exact.
Hint 1/4
You are asked for a number, not a function: quote the derivative formula and then substitute the one value of $x$.
Hint 2/4
On the principal branch, $(\arcsin x)'=\dfrac{1}{\sqrt{1-x^{2}}}$ for $|x|<1$.
Hint 3/4
With $x=\tfrac{1}{2}$ the square inside is $\tfrac14$, so the root is $\sqrt{1-\tfrac14}=\sqrt{\tfrac34}$.
Hint 4/4
The value is $\dfrac{2}{\sqrt3}=\dfrac{2\sqrt3}{3}$.
Each pair of co-functions adds up to a right angle, so their derivatives add up to zero: whatever one of them does, its partner does with the opposite sign.
Why the sum is a right angle, and what that costs
$\cos\left(\tfrac{\pi}{2}-\theta\right)=\sin\theta$, and if $\theta=\arcsin x$ lies in $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ then $\tfrac{\pi}{2}-\theta$ lies in $[0,\pi]$ — the branch $\arccos$ returns. So $\arccos x=\tfrac{\pi}{2}-\arcsin x$; differentiate and the minus sign appears by itself. The secant case differs: books disagree on the branch for negative $x$, so that sign is a convention. For $x>1$, $\sec y=x$ gives $\sec y\tan y\cdot y'=1$ with $\tan y=\sqrt{x^{2}-1}$, hence $y'=1/\left(x\sqrt{x^{2}-1}\right)$.
The identity as a picture: reflecting the arcsine graph in the horizontal line $y=\pi/4$ produces the arccosine graph, and a reflection reverses every slope.
Looks like this, but is not
The two graphs differ by the constant $\tfrac{\pi}{2}$, and adding a constant changes no slope. So $\arcsin$ and $\arccos$ should have equal derivatives.
The formula is $\arccos x=\tfrac{\pi}{2}-\arcsin x$, not $\tfrac{\pi}{2}+\arcsin x$: the minus flips the graph before the constant slides it. At $x=\tfrac12$ the slopes are $+1.155$ and $-1.155$.
function
domain
principal values
derivative
$\arcsin x$
$[-1,1]$
$\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$
$\dfrac{1}{\sqrt{1-x^{2}}}$
$\arctan x$
$\mathbb{R}$
$\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$
$\dfrac{1}{1+x^{2}}$
$\operatorname{arcsec} x$
$\vert x\vert\ge 1$
stated by the question
$\dfrac{1}{x\sqrt{x^{2}-1}}$ for $x>1$
$\arccos x$
$[-1,1]$
$[0,\pi]$
$-\dfrac{1}{\sqrt{1-x^{2}}}$
$\operatorname{arccot} x$
$\mathbb{R}$
$(0,\pi)$
$-\dfrac{1}{1+x^{2}}$
$\operatorname{arccsc} x$
$\vert x\vert\ge 1$
stated by the question
$-\dfrac{1}{x\sqrt{x^{2}-1}}$ for $x>1$
The derivative column holds on the open part of the domain: the arcsine, arccosine and both secant rows lose it at $x=\pm 1$, where the tangent turns vertical, while $1+x^{2}$ never vanishes.
Proving the identity instead of quoting it
Suppose you cannot remember whether the sum is $\tfrac{\pi}{2}$ or $\pi$. Calculus settles it in two lines, for any suspected identity.
Independent check at a different point: $x=1$ gives $\tfrac{\pi}{2}+0=\tfrac{\pi}{2}$, and $x=-1$ gives $-\tfrac{\pi}{2}+\pi=\tfrac{\pi}{2}$. Two more agreements, neither used in the proof.
Zero derivative plus one evaluated point is a complete proof of an identity. Remember the technique rather than the six formulas.
Checkpoint
§12.4 — what a constant sum forces on two derivatives●●○○○
A single sentence to judge. It is the kind of claim that sounds harmless because the identity quoted inside it is true.
Given
$\arcsin x+\arccos x=\dfrac{\pi}{2}$ for every $x$ in $[-1,1]$
Find
(a) True or false: because the two functions differ only by the constant $\tfrac{\pi}{2}$, they have the same derivative.
Hint 1/4
Rewrite the claim as an equation between the two functions and look at what kind of operation turns one into the other.
Hint 2/4
If $u+v$ is constant then $u'+v'=0$, so $v'=-u'$: the derivatives are opposite, not equal.
Hint 3/4
Here $\arccos x=\tfrac{\pi}{2}-\arcsin x$; the arcsine is not shifted, it is shifted and flipped.
Hint 4/4
The statement is false: the derivatives are negatives of one another.
an identity may be differentiated on both sides; a constant differentiates to zero
$(\arcsin x)'+(\arccos x)'=0$
the derivative of a sum
$(\arccos x)'=-\frac{1}{\sqrt{1-x^{2}}}$
move one term across
Answer $$\boxed{(\arccos x)'=-(\arcsin x)'}$$
Check
Check against the graph: $\arccos$ falls from $\pi$ to $0$ as $x$ runs from $-1$ to $1$, so its derivative must be negative everywhere; the formula is.
⚠ Copying the arcsine root into the arcsecant formula
Both formulas contain a square root and a difference of squares, and the order of the two terms is easy to swap.
right$x>1:\ \frac{1}{x\sqrt{x^{2}-1}};\qquad x<-1:\ \text{sign fixed by the branch the question states}$
The exponential pair that behaves like sine and cosine
The section title names one more family. It is built from $e^{x}$ with no new machinery, and its inverses are the one place a closed form can be checked by hand.
DefinitionDefinition 12.5: the hyperbolic functions
Split $e^{x}$ into its even and odd halves: the even half is $\cosh$, the odd half $\sinh$, each is the derivative of the other with no minus sign, and the difference of their squares is one.
The identity in one line
$\left(e^{x}+e^{-x}\right)^{2}-\left(e^{x}-e^{-x}\right)^{2}=4e^{x}e^{-x}=4$; divide both sides by $4$. The same computation with a plus sign gives $\cosh(2x)$, which is certainly not constant.
Where the name comes from: $(\cos t,\sin t)$ runs around a circle, and $(\cosh t,\sinh t)$ runs along one branch of a hyperbola, because the identity it satisfies has a minus sign.
Looks like this, but is not
Near the origin $\cosh x$ matches the parabola $1+\tfrac{x^{2}}{2}$ — same value, slope and curvature at $x=0$ — so it is presumably a parabola in disguise.
At $x=5$ the parabola gives $13.5$ while $\cosh 5=74.21$. A parabola grows like a power, $\cosh$ like $e^{x}/2$; agreement near one point says nothing about the far field.
function
definition or closed form
derivative
$\sinh x$
$\dfrac{e^{x}-e^{-x}}{2}$
$\cosh x$
$\cosh x$
$\dfrac{e^{x}+e^{-x}}{2}$
$\sinh x$
$\tanh x$
$\dfrac{\sinh x}{\cosh x}$
$\operatorname{sech}^{2}x=\dfrac{1}{\cosh^{2}x}$
$\sinh^{-1}x$
$\ln\left(x+\sqrt{x^{2}+1}\right)$, all $x$
$\dfrac{1}{\sqrt{1+x^{2}}}$
$\cosh^{-1}x$
$\ln\left(x+\sqrt{x^{2}-1}\right)$, $x\ge 1$
$\dfrac{1}{\sqrt{x^{2}-1}}$, $x>1$
$\tanh^{-1}x$
$\tfrac12\ln\dfrac{1+x}{1-x}$, $\vert x\vert<1$
$\dfrac{1}{1-x^{2}}$
The last three rows are logarithms, so column three could be obtained by differentiating column two; the reciprocal-slope theorem is faster.
Solving x equals sinh y to get the logarithmic form of the inverse
Every inverse so far had to stay unwritten. This one can be solved for, because the equation is secretly a quadratic.
Given
$x=\sinh y=\dfrac{e^{y}-e^{-y}}{2}$
Find
$y=\sinh^{-1}x$ in closed form
SolutionTurn it into a quadratic
$2x=e^{y}-e^{-y}$
clear the denominator
$2xe^{y}=e^{2y}-1$
multiply through by $e^{y}$, which is never zero so nothing is lost
$t^{2}-2xt-1=0,\quad t=e^{y}$
naming $t=e^{y}$ makes the quadratic visible
Solve and discard the impossible root
$t=x\pm\sqrt{x^{2}+1}$
quadratic formula with $a=1$, $b=-2x$, $c=-1$
$t=x+\sqrt{x^{2}+1}$
$t=e^{y}>0$, and $\sqrt{x^{2}+1}>\vert x\vert$ makes the minus root negative
Exact check at $x=\tfrac34$: the formula gives $\ln\left(0.75+\sqrt{1.5625}\right)=\ln(0.75+1.25)=\ln 2$, and running it forward, $\sinh(\ln 2)=\tfrac{2-\tfrac12}{2}=\tfrac34$. The two directions agree.
Differentiating this closed form gives $\left(\sinh^{-1}x\right)'=\dfrac{1}{\sqrt{1+x^{2}}}$, which is what the reciprocal-slope theorem gives through $1/\cosh y$. With an inner function the chain rule applies as usual: $\left(\sinh^{-1}(3x)\right)'=\dfrac{3}{\sqrt{1+9x^{2}}}$.
Checkpoint
§12.5 — the hyperbolic identity, sign and all●●○○○
The trigonometric identity next door is $\sin^{2}+\cos^{2}=1$, and the habit of writing a plus sign there is strong.
Given
$\cosh x=\dfrac{e^{x}+e^{-x}}{2}$
$\sinh x=\dfrac{e^{x}-e^{-x}}{2}$
Find
(a) True or false: $\cosh^{2}x+\sinh^{2}x=1$ for every real $x$.
Hint 1/4
Do not test the claim at $x=0$ only; that point makes several different claims look true. Pick a second value as well.
Hint 2/4
Expanding the two definitions gives $\cosh^{2}x-\sinh^{2}x=1$: the identity has a minus sign, and $\cosh^{2}x+\sinh^{2}x$ is $\cosh(2x)$ instead.
Hint 3/4
At $x=1$: $\cosh 1\approx 1.5431$ and $\sinh 1\approx 1.1752$, so the sum of squares is about $2.381+1.381=3.762$.
Hint 4/4
False: the sum of squares is $\cosh(2x)$, and only the difference is $1$.
one counterexample is enough to kill a claim made for every $x$
$\cosh^{2}1-\sinh^{2}1\approx 2.381-1.381=1$
the same two numbers satisfy the identity with the minus sign
Answer $$\boxed{\cosh^{2}x-\sinh^{2}x=1}$$
Check
Algebraic confirmation, independent of the decimals: $\left(e^{x}+e^{-x}\right)^{2}-\left(e^{x}-e^{-x}\right)^{2}=4$, and dividing by $4$ gives exactly $1$.
⚠ Importing the minus sign from the circular derivative
$(\cos x)'=-\sin x$ is drilled far harder than its hyperbolic neighbour.
wrong$(\cosh x)'=-\sinh x$
right$(\cosh x)'=+\sinh x$
⚠ Writing the identity with a plus sign
The Pythagorean identity has a plus, and the two families look alike on paper.
wrong$\cosh^{2}x+\sinh^{2}x=1$
right$\cosh^{2}x-\sinh^{2}x=1$
l'Hôpital's rule and the two shapes it accepts
Every family here produces limits substitution refuses to answer: the opening calculator experiment, and any quotient in which both parts vanish. One theorem covers them, with hypotheses worth obeying.
TheoremTheorem 12.6: l'Hôpital's rule
Conditions
$f$ and $g$ are differentiable near $a$, except possibly at $a$ itself
$g'(x)\neq 0$ near $a$
either $f(x)\to 0$ and $g(x)\to 0$, or $|f(x)|\to\infty$ and $|g(x)|\to\infty$
the limit on the right exists or is $\pm\infty$
$a$ may be a number, a one-sided approach, or $\pm\infty$
When both parts of a quotient collapse together, compare how fast each one is collapsing instead: the ratio of the speeds answers the question that the ratio of the sizes could not.
Why it is true in the simplest case
Suppose $f(a)=g(a)=0$, both differentiable at $a$ with $g'(a)\neq 0$. Divide top and bottom by $x-a$: $\dfrac{f(x)}{g(x)}=\dfrac{\left(f(x)-f(a)\right)/(x-a)}{\left(g(x)-g(a)\right)/(x-a)}\longrightarrow\dfrac{f'(a)}{g'(a)}$ — nothing but the definition of the derivative. The general statement, with $a$ infinite or the parts blowing up, needs the two-function Mean Value Theorem and is proved in the book.
The reason the rule works, drawn: near a common zero each curve is almost its own tangent line, and two lines through one point have heights in the ratio of their slopes.
Looks like this, but is not
$\displaystyle\lim_{x\to 0}\frac{x+1}{x^{2}+2}$ is a quotient of differentiable functions, so differentiating top and bottom should give $\dfrac{1}{2x}$, which runs off to infinity.
Substitution gives $\tfrac12$, so there was nothing to resolve. The rule is not a technique for quotients but for two specific collapses. Used outside them it returns a confident wrong answer instead of an error.
The limit that needs the rule twice
A quotient where one application is not enough — and where the form has to be checked again before the second.
each part differentiated on its own; this is not the quotient rule
$\frac{e^{0}-1}{0}=\frac{0}{0}$
check again: still indeterminate, so a second application is legal
Apply it once more and read off the answer
$L=\lim_{x\to 0}\frac{e^{x}}{2}=\frac{1}{2}$
now substitution works, which is the signal to stop
Answer $$\boxed{L=\tfrac12}$$
Check
Numerical check at $x=0.1$: $\dfrac{e^{0.1}-1-0.1}{0.01}=\dfrac{0.00517092}{0.01}=0.517$, and the values keep falling towards $0.5$ as $x$ shrinks.
Two applications, and one form check between them; the check is the step people skip.
Every application needs its own permission. Had the second check come out $0/2$, the answer was already $0$ and differentiating again would have been a mistake.
How an exponential beats a polynomial
The other accepted shape, infinity over infinity, in the comparison that decides most end-behaviour questions.
both parts grow without bound, which is the second accepted shape
Differentiate until the top runs out
$L=\lim_{x\to\infty}\frac{2x}{e^{x}}$
first application; still $\infty/\infty$
$L=\lim_{x\to\infty}\frac{2}{e^{x}}=0$
second application; now the top is a constant and the bottom still grows, so the shape is $2/\infty$ and the limit is $0$
Answer $$\boxed{L=0}$$
Check
Numerical check at $x=20$: $\dfrac{400}{e^{20}}\approx 8.2\times 10^{-7}$, already a millionth, and each further unit of $x$ divides it by roughly $e$.
Two differentiations flattened the polynomial and left the exponential untouched. With $x^{n}$ it takes $n$ steps and ends the same way — the precise sense in which exponentials outrun polynomials.
Checkpoint
§12.6 — deciding whether the rule is allowed●●●○○
Before differentiating anything, substitute. Only two shapes give the rule permission to start, and one of these four limits has one of them.
Given
l'Hôpital's rule applies to the shapes $\dfrac{0}{0}$ and $\dfrac{\infty}{\infty}$ only
Find
(a) To which limit may the rule be applied directly, with no rewriting first?
Hint 1/4
Substitute the limit point into each numerator and each denominator separately and write down the pair of numbers.
Hint 2/4
The rule needs both parts to go to zero, or both to blow up. A number over a number, a number over zero, zero over a number: none is its business.
Hint 3/4
The four substitutions give, in order: $0/0$; $1/0$; $0/1$; $0/4$.
Hint 4/4
Only $\dfrac{\cos x-1}{x^{2}}$ has an indeterminate shape at $x\to 0$.
Show solutionSubstitute in each
$\frac{\cos 0-1}{0}=\frac{0}{0}$
indeterminate: the rule may start here
$\frac{\cos 0}{0}=\frac{1}{0}$
a non-zero number over zero is not indeterminate; the size runs to infinity
$\frac{0+0}{0+1}=\frac{0}{1}=0$
already an answer
$\frac{4-4}{2+2}=\frac{0}{4}=0$
already an answer; the zero on top is not matched below
right$\lim_{x\to 0}\frac{e^{x}}{2}=\frac12\quad\text{stop: the shape is no longer indeterminate}$
The five forms that need a rewrite before the rule is allowed
The rule accepts exactly two shapes. Everything else — the calculator experiment included — has to be pushed into one of them first, and there are only three pushes.
MethodMethod 12.7: the three rewrites
Conditions
$0\cdot\infty$: send one factor to the denominator
$\infty-\infty$: combine into a single fraction, or factor the dominant term out
$0^{0}$, $1^{\infty}$, $\infty^{0}$: take the logarithm, then handle the $0\cdot\infty$ that appears, and exponentiate at the end
A product becomes a quotient by moving one factor downstairs; a difference becomes a quotient over a common denominator; a power becomes a product by taking logarithms — which is why every power problem ends in an exponential.
Why the logarithm step is legal
$u^{v}=e^{v\ln u}$ for $u>0$ and $\exp$ is continuous, so the limit passes through it: if $v\ln u\to c$ then $u^{v}\to e^{c}$, and if $v\ln u\to-\infty$ then $u^{v}\to 0$. That continuity is also what makes the final exponentiation compulsory.
The whole decision in one chart: the left column is what substitution reported, the right column the single move that shape needs. Every move but the last sends you back to the top to substitute again.
Looks like this, but is not
$1$ raised to any power is $1$, so a limit of shape $1^{\infty}$ must be $1$ — settling the opening experiment without any work.
$\left(1+\tfrac{1}{n}\right)^{n}\to e=2.71828$ and $\left(1+\tfrac{3}{n}\right)^{n}\to e^{3}=20.0855$: same shape, different answers, neither $1$. The base is only heading to $1$, and how fast decides the outcome.
The calculator experiment from the first paragraph, finished exactly
The opening scene had base $1.001$ and exponent $1000$. Here is the general version, with a constant in the numerator so the mechanism is visible.
chain rule; the inner derivative of $3/x$ is $-3/x^{2}$
$\frac{d}{dx}\frac{1}{x}=-\frac{1}{x^{2}}$
the denominator
$\ln L=\lim_{x\to\infty}\frac{3}{1+3/x}=3$
the two factors $-1/x^{2}$ cancel, which is the reason $1/x$ was the right thing to divide by
Undo the logarithm
$L=e^{3}\approx 20.0855$
the step that turns $\ln L$ into $L$; leaving it out is the classic lost mark
Answer $$\boxed{L=e^{3}}$$
Check
Numerical check at $x=1000$: $(1.003)^{1000}=19.995$, within half a percent of $e^{3}=20.0855$. With $3$ replaced by $1$ it gives $(1.001)^{1000}=2.7169$ against $e=2.71828$ — the number in the first paragraph.
One logarithm, one rewrite, one application of the rule, one exponential.
The general result is $\lim_{x\to\infty}\left(1+\tfrac{k}{x}\right)^{x}=e^{k}$, worth recognising on sight: it runs every continuous-growth model.
A difference of two things that both blow up
Neither term has a limit, so the difference has to be assembled into one object first.
second application; the bottom is now $2$ at $x=0$, so the shape is settled
Answer $$\boxed{L=0}$$
Check
Numerical check at $x=0.1$: $\dfrac{1}{0.1}-\dfrac{1}{\sin 0.1}=-0.016686$, shrinking towards $0$. The limit is $0$ but the approach is from below.
Two infinite quantities can differ by anything. Combining them into one fraction is not cosmetic; it is what creates something the rule can act on.
Checkpoint
§12.7 — choosing the first move for a power form●●●○○
The exponent contains the variable, so this is not a quotient and the rule cannot touch it as written.
Given
$\displaystyle L=\lim_{x\to 0^{+}}(\sin x)^{x}$
as $x\to 0^{+}$ the base tends to $0$ and the exponent tends to $0$
Find
(a) What is the correct first move?
Hint 1/4
Ask what shape this is first: base and exponent both head to zero, so it is a power form, not a quotient.
Hint 2/4
Power forms are handled by taking the logarithm: if $L=\lim u^{v}$ then $\ln L=\lim v\ln u$, provided the limit on the right exists.
Hint 3/4
Here $u=\sin x$ and $v=x$, so the logarithm gives $\ln L=\lim_{x\to 0^{+}}x\ln(\sin x)$, which is $0\cdot(-\infty)$ and needs one more rewrite before the rule is allowed.
Hint 4/4
Take logarithms first; the answer, after the dust settles, is $L=1$.
Show solutionTake the logarithm
$\ln L=\lim_{x\to 0^{+}}x\ln(\sin x)$
logarithm of a power brings the exponent down
$=\lim_{x\to 0^{+}}\frac{\ln(\sin x)}{1/x}$
the product $0\cdot(-\infty)$ becomes $-\infty/\infty$
Numerical check at $x=0.01$: $\sin(0.01)=0.0099998$, and $(0.0099998)^{0.01}=e^{0.01\ln 0.0099998}=e^{-0.04605}=0.955$, moving towards $1$ as slowly as a logarithm does.
⚠ Reporting the logarithm as the answer
The hard work ends when $\ln L$ is found, and the last line feels like bookkeeping.
wrong$\ln L=3\ \Longrightarrow\ L=3$
right$\ln L=3\ \Longrightarrow\ L=e^{3}$
⚠ Splitting an infinity minus infinity into two limits
The limit laws are usually applied without checking that each piece has a limit.
A question gives you $f$, one value of $f$ and one value of $f'$, and asks about $f^{-1}$.
Check it has an inverse
Show $f'$ keeps one sign on the interval, so $f$ is strictly monotone and one to one.
Walk the output back to its input
Find the $a$ with $f(a)=b$; try small integers. A strictly monotone $f$ has at most one, so a lucky guess is a proof.
Differentiate the original, not the inverse
Compute $f'(a)$ and confirm it is not zero.
Reciprocate
Report $\left(f^{-1}\right)'(b)=1/f'(a)$, and sanity-check the sign against the direction $f$ runs.
Where it goes wrong
Evaluating $f'$ at $b$ instead of at $a$.
Skipping the monotonicity check, so the inverse does not exist and the answer is about nothing.
Meeting $f'(a)=0$ and dividing anyway; that case has a vertical tangent and no derivative.
Naming the form and choosing the move
Substitution produced a shape rather than a number.
Substitute and name
Write down what each part tends to. The name of the shape is the whole decision.
If it is $0/0$ or $\infty/\infty$, differentiate top and bottom
Separately, never with the quotient rule.
Otherwise rewrite first
Product: one factor downstairs. Difference: one fraction. Power: logarithm.
Re-check after every application
The shape can stop being indeterminate at any point, and that is the signal to stop and substitute.
Undo what you did
If a logarithm was taken, the answer is $e$ to the limit you just computed.
Where it goes wrong
Applying the rule to a shape that was never indeterminate; it returns a wrong number with no warning.
Differentiating a third time out of momentum, after the shape has already resolved.
Reporting $\ln L$ as $L$.
Derivative of x to the power pi
The variable is in the base and the constant is upstairs.
Given
$y=x^{\pi}$, $x>0$
Find
$y'$
SolutionRecognise the position of the variable
$y=x^{\pi}$
constant exponent, so this is the power rule, valid for every real exponent when $x>0$
Apply the power rule
$y'=\pi x^{\pi-1}$
bring the exponent down and reduce it by one
Answer $$\boxed{y'=\pi x^{\pi-1}}$$
Check
Check through the definition: $x^{\pi}=e^{\pi\ln x}$, so $y'=e^{\pi\ln x}\cdot\dfrac{\pi}{x}=\pi x^{\pi-1}$. The power rule for irrational exponents follows; it is not an extra assumption.
Derivative of pi to the power x
The same two symbols, swapped.
Given
$y=\pi^{x}$
Find
$y'$
SolutionRecognise the position of the variable
$y=\pi^{x}$
constant base, variable exponent, so this is the general exponential rule
Apply the exponential rule
$y'=\pi^{x}\ln\pi$
the function survives and picks up the factor $\ln$ of the base
Answer $$\boxed{y'=\pi^{x}\ln\pi}$$
Check
Check at $x=0$: the tangent to any $a^{x}$ at $(0,1)$ has slope $\ln a$, and the formula gives $\ln\pi\approx 1.145$ — steeper than $e^{x}$, as a bigger base should be.
Same two symbols, two different rules, and the answers do not even have the same shape: a power of $x$ against an exponential.
How to tell them apart
Point at the $x$. In the base means power rule; in the exponent means multiply by $\ln$ of the base; in both means take logarithms first.
Scaffolding comes off
The common skeleton
Substitute the limit point and write down the shape you get.
If the shape is $0/0$ or $\infty/\infty$, differentiate top and bottom separately.
If it is a product, a difference or a power, rewrite it into one of those two shapes first.
After every application, substitute again and re-name the shape.
Undo any transformation you made, and check the size of the answer against a value near the limit point.
1 · fully worked
A product where one factor vanishes and the other blows up
The full solution, every reason written out.
Given
$\displaystyle L=\lim_{x\to 0^{+}}x^{2}\ln x$
Find
$L$
SolutionName the shape
$x^{2}\to 0,\qquad \ln x\to-\infty$
the shape is $0\cdot(-\infty)$, which the rule does not accept
Move one factor downstairs
$x^{2}\ln x=\frac{\ln x}{x^{-2}}$
the factor that blows up stays on top; putting $x^{2}$ downstairs as $x^{-2}$ is what turns a product into a quotient
simplify first, substitute second; the shape is no longer indeterminate
Answer $$\boxed{L=0}$$
Check
Numerical check at $x=0.001$: $x^{2}\ln x=10^{-6}\times(-6.908)=-6.9\times 10^{-6}$, already within seven millionths of zero. The power beats the logarithm, which is the general lesson.
Which factor goes downstairs is a choice; sending $\ln x$ down instead produces a mess that never resolves.
2 · you write the reasoning
The same skeleton on a lighter problem: $\displaystyle\lim_{x\to 0^{+}}x\ln x$. Steps given, reasons not. Say why each line is legal before opening the model answer.
Substituting gives $0\cdot(-\infty)$, so no rule applies yet.
reasoning
Substitution is always the first move, and here it produces a shape rather than a number: $0\cdot(-\infty)$ is not something the rule accepts.
Rewrite as a quotient: $\displaystyle\lim_{x\to 0^{+}}\frac{\ln x}{1/x}$.
reasoning
Dividing by $1/x$ is multiplying by $x$, so nothing changed except the arrangement — and the arrangement is now a quotient.
The shape is now $-\infty/\infty$, so differentiate top and bottom.
reasoning
Both parts now run to infinity in size, the second shape the rule accepts, so differentiating top and bottom separately is legal.
Simplify and substitute: the answer is $0$.
reasoning
The quotient $\dfrac{1/x}{-1/x^{2}}$ simplifies to $-x$ before any limit is taken; simplifying first is what keeps the last step trivial.
3 · find the buried error
A worked solution to $\displaystyle L=\lim_{x\to 0^{+}}(1+2x)^{1/x}$ is written below. It reaches an answer, and the answer is wrong. Two separate errors are buried in it.
Step 1. As $x\to 0^{+}$ the base tends to $1$ and the exponent $1/x$ tends to $\infty$, so the shape is $1^{\infty}$ and we take logarithms.
Step 2. Written as a quotient this is $\dfrac{\ln(1+2x)}{x}$, and substituting gives $\dfrac{0}{0}$, so the rule applies.
Step 3. Differentiating, $\dfrac{d}{dx}\ln(1+2x)=\dfrac{1}{1+2x}$ and $\dfrac{d}{dx}x=1$, so $\ln L=\lim_{x\to 0^{+}}\dfrac{1}{1+2x}=1$.
Step 4. Therefore $L=1$.
the two buried errors (2)
⚠ step 3
The chain rule was dropped inside the logarithm: $\dfrac{d}{dx}\ln(1+2x)=\dfrac{2}{1+2x}$, not $\dfrac{1}{1+2x}$.
The outer derivative of $\ln$ is the visible one and the inner factor is a single digit, so it disappears unnoticed — the same slip costs the same mark in every logarithmic differentiation question.
right
With the factor restored, $\ln L=\lim_{x\to 0^{+}}\dfrac{2}{1+2x}=2$.
⚠ step 4
The logarithm was never undone: even with the value found in step 3, the answer would be $L=e^{1}$, not $L=1$.
After the hard part is over, the final exponential feels like bookkeeping rather than mathematics, and $\ln L$ is quietly read as $L$.
right
The correct last line is $L=e^{2}\approx 7.389$.
4 · the bare problem
§12.7 — a bare power form, no scaffolding●●●○○
Nothing is set up for you here. Name the shape, choose the move, and finish the job including the last line.
Given
$\displaystyle L=\lim_{x\to\infty}x^{1/x}$
Find
(a) Evaluate $L$, showing the shape at each stage.
Hint 1/4
Look at base and exponent separately as $x$ grows: one runs to infinity, the other to zero. That combination is a power form.
Hint 2/4
For a power form, take the logarithm: $\ln L=\lim v\ln u$ with $u=x$ and $v=1/x$. Undo it at the end.
Hint 3/4
Here $\ln L=\lim_{x\to\infty}\dfrac{\ln x}{x}$, which is $\infty/\infty$ and ready for the rule; the data again: $u=x$, $v=1/x$.
Hint 4/4
$\ln L=0$, so $L=e^{0}=1$.
Show solutionName the shape
$x\to\infty,\quad \frac{1}{x}\to 0$
the shape is $\infty^{0}$: a power form, not a quotient
the exponent comes down; the result is already a quotient of shape $\infty/\infty$
Apply the rule and undo the logarithm
$\ln L=\lim_{x\to\infty}\frac{1/x}{1}=0$
differentiate top and bottom
$L=e^{0}=1$
exponentiate
Answer $$\boxed{L=1}$$
Check
Numerical check at $x=1000$: $1000^{0.001}=e^{0.0069078}=1.00693$, and at $x=10^{6}$ it is $1.0000138$. The approach is slow, as anything driven by a logarithm is.
$\infty^{0}$ came out $1$ here, but the shape did not decide that — the rate did. Another pair with the same shape can give any positive answer.
Full exam-style question
Three parts in the shape a final examination asks themexam format
One question touching all three families. Nothing is harder than the worked examples; the difficulty is that the three parts do not announce which method they want.
Given
$h(x)=\arctan\left(x^{2}\right)+5^{x}$
$f(x)=x+\arcsin x$ on $(-1,1)$, which is strictly increasing there
Part (a) at $x=0$: the formula gives $\ln 5\approx 1.609$, and $h$ near zero is $5^{x}$ plus something quadratically small. Part (c) at $x=0.05$: $(\cos 0.05)^{400}=0.60640$ against $e^{-1/2}=0.60653$.
Three parts, three different first moves; the recognition is the examined skill.
Notice what part (b) did not need: a formula for $f^{-1}$. A strictly increasing function plus one convenient value is always asking for the reciprocal-slope theorem.
Practice
A · concept 3 questions
1§12.1 — which slot the number goes into●●○○○
The data below is everything a question of this kind gives you. The claim underneath uses all of it and still gets the answer wrong.
Given
$f$ is one to one and differentiable on $\mathbb{R}$
$f(2)=5$
$f'(2)=4$
Find
(a) True or false: it follows that $\left(f^{-1}\right)'(2)=\tfrac14$.
(b) Whatever you answered, write down the one statement about $f^{-1}$ that the data does support.
Hint 1/4
Draw the arrow: $f$ sends $2$ to $5$. Now ask which of those two numbers the inverse takes as its input.
Hint 2/4
The rule is $\left(f^{-1}\right)'(b)=1/f'\!\left(f^{-1}(b)\right)$, so the number written inside $\left(f^{-1}\right)'$ is an output of $f$.
Hint 3/4
The data again: $f(2)=5$ and $f'(2)=4$, so the reciprocal $\tfrac14$ belongs at the output $5$.
Hint 4/4
False; the supported statement is $\left(f^{-1}\right)'(5)=\tfrac14$.
Consistency check with slopes: $f$ climbs $4$ units of output per unit of input at $x=2$, so the mirrored graph climbs a quarter of a unit near the mirrored point.
2§12.7 — what a shape does and does not decide●●○○○
Two limits with the same shape are placed side by side so that the claim can be tested rather than argued about.
Check the first term for evenness: $\arctan\left(3x^{2}\right)$ is even, so its derivative must be odd and must vanish at $0$. The expression $6x/(1+9x^{4})$ does both.
2§12.3 — arcsine of a square root●●●○○
A composite whose inner function has its own domain restriction, so the answer comes with an interval attached.
Given
$y=\arcsin\left(\sqrt{x}\right)$
Find
(a) Differentiate and simplify.
(b) State the interval on which your answer is valid.
Hint 1/4
Identify the inner function and note where it is allowed to live before differentiating.
Hint 2/4
$(\arcsin u)'=\dfrac{u'}{\sqrt{1-u^{2}}}$ with $u=\sqrt{x}$, whose derivative is $\dfrac{1}{2\sqrt{x}}$.
Hint 3/4
Substituting: $u^{2}=x$, so the root becomes $\sqrt{1-x}$, and the product is $\dfrac{1}{\sqrt{1-x}}\cdot\dfrac{1}{2\sqrt{x}}$.
Hint 4/4
$y'=\dfrac{1}{2\sqrt{x(1-x)}}$ on $0
Show solutionChain rule with the inner square root
Symmetry check that does not repeat the computation: $\arcsin\sqrt{x}+\arcsin\sqrt{1-x}=\tfrac{\pi}{2}$, so differentiating forces $y'(x)=y'(1-x)$ — and the answer is unchanged when $x$ and $1-x$ swap.
Whenever an inner function has a restricted domain, the interval is part of the answer, not decoration.
3§12.2 — variable in the base and in the exponent●●●○○
Neither the power rule nor the exponential rule applies, because the variable is in both positions at once. There is exactly one move.
Given
$y=x^{\tan x}$ for $x>0$ and $\cos x\neq 0$
Find
(a) Find $y'$.
(b) Evaluate $y'$ at $x=1$, to three decimals.
Hint 1/4
Check both positions of the variable before choosing a rule; when it sits in both, no memorised derivative applies.
Hint 2/4
Take logarithms of both sides and differentiate implicitly: $\ln y=v\ln u$ gives $\dfrac{y'}{y}=(v\ln u)'$.
Hint 3/4
Here $\ln y=\tan x\ln x$, whose derivative by the product rule is $\sec^{2}x\ln x+\dfrac{\tan x}{x}$, and $y$ itself is $x^{\tan x}$.
Hint 4/4
$y'=x^{\tan x}\left(\sec^{2}x\ln x+\dfrac{\tan x}{x}\right)$, which at $x=1$ is $\tan 1\approx 1.557$.
Show solutionTake logarithms to separate the two positions
$\ln y=\tan x\ln x$
the logarithm turns a variable exponent into a factor, which is the only reason this problem is doable
Numerical check at $x=1$ with a symmetric difference of step $0.01$: $\dfrac{1.01^{\tan 1.01}-0.99^{\tan 0.99}}{0.02}=1.5583$, against $\tan 1=1.5574$.
This is the third row of the table in the general-base block, and it is always this move.
4§12.6 — two applications with a check between them●●●○○
A quotient that stays indeterminate after the first differentiation, so the shape has to be read twice.
(a) Evaluate $L$, naming the shape before each application.
Hint 1/4
Substitute first and write down the shape; that decides whether you may start at all.
Hint 2/4
The rule differentiates top and bottom separately, and may be repeated while the shape stays indeterminate.
Hint 3/4
Here $\tfrac{1-\cos 0}{0}=\tfrac{0}{0}$, and one application leaves $\dfrac{\sin x}{2x}$, which is $\tfrac00$ again.
Hint 4/4
The second application gives $\dfrac{\cos x}{2}\to\tfrac12$.
Show solutionShape check, then first application
$\frac{1-\cos 0}{0^{2}}=\frac{0}{0}$
indeterminate, so the rule may start
$L=\lim_{x\to 0}\frac{\sin x}{2x}$
top and bottom differentiated separately, not as a quotient
Shape check, then second application
$\frac{\sin 0}{0}=\frac{0}{0}$
still indeterminate, so a second round is legal
$L=\lim_{x\to 0}\frac{\cos x}{2}=\frac12$
substitution works now, so stop
Answer $$\boxed{L=\tfrac12}$$
Check
Numerical check at $x=0.05$: $\dfrac{1-\cos 0.05}{0.0025}=\dfrac{0.00124974}{0.0025}=0.49990$.
Stopping is a decision too. Once the shape is a number the rule is no longer allowed, and a third differentiation would replace a correct answer with a meaningless one.
5§12.7 — a difference of two quantities that both blow up●●●●○
Neither term has a limit on its own, so the limit laws do not apply until the two are made into one object.
Numerical check: $f(1.02)=1.061208+3.06+2=6.121208$, so near the output $6$ the inverse gains $0.02$ of input for $0.121208$ of output, a rate of $0.165$ against the exact $0.1667$.
The coincidence $f(1)=6$ and $f'(1)=6$ is bait: one is an output, the other a slope, and only the second gets reciprocated.
2§12.5 — the hyperbolic tangent, end to end●●●●○
Three parts on one function, in the order an examination builds them: a structural fact, an inverse derivative by two routes, then a limit needing the rule twice.
Given
$\tanh x=\dfrac{\sinh x}{\cosh x}$
$\tanh^{-1}x=\tfrac12\ln\dfrac{1+x}{1-x}$ for $\vert x\vert<1$
$\cosh^{2}x-\sinh^{2}x=1$
Find
(a) Show that $\tanh$ is strictly increasing on $\mathbb{R}$.
(b) Find $\left(\tanh^{-1}\right)'(x)$ twice: once from the reciprocal-slope theorem and once by differentiating the logarithmic form.
Part (a) is about a sign, part (b) about doing one job two ways, part (c) about naming a shape before touching it.
Hint 2/4
Use $(\tanh x)'=\operatorname{sech}^{2}x$ for (a); $\left(f^{-1}\right)'=1/f'\!\left(f^{-1}\right)$ for (b); and l'Hôpital's rule, repeatedly and with a check each time, for (c).
Hint 3/4
In (b) the theorem needs $\operatorname{sech}^{2}y$ where $\tanh y=x$, and $\operatorname{sech}^{2}y=1-\tanh^{2}y=1-x^{2}$; the logarithmic form is $\tfrac12\left[\ln(1+x)-\ln(1-x)\right]$.
Hint 4/4
The answers are: $\operatorname{sech}^{2}x>0$; $\dfrac{1}{1-x^{2}}$ by both routes; and $-\tfrac13$.
Two independent checks. Part (b): the two routes were computed by different methods and agree. Part (c) numerically at $x=0.1$: $\dfrac{\tanh 0.1-0.1}{0.001}=-0.3320$, against $-\tfrac13=-0.3333$.
The answer $-\tfrac13$ says that $\tanh x\approx x-\tfrac{x^{3}}{3}$ near zero, which is why $\tanh$ falls away from the line $y=x$ so quickly.
3§12.7 — a power form under examination conditions●●●○○
Exponent and base both move, and the answer is one of four numbers, each matching a different slip in the same method.
Given
$\displaystyle L=\lim_{x\to 0^{+}}(1+3x)^{2/x}$
Find
(a) Which value is $L$?
Hint 1/4
Name the shape first: as $x\to 0^{+}$ the base tends to $1$ and the exponent to $\infty$.
Hint 2/4
For a power form, $\ln L=\lim v\ln u$, and the answer is $e$ to whatever that limit is.
Hint 3/4
Here $\ln L=\lim_{x\to 0^{+}}\dfrac{2\ln(1+3x)}{x}$, a $\tfrac{0}{0}$ quotient; the data again: base $1+3x$, exponent $2/x$.
Independent check of the slope: the curve climbs from $0.78540$ at $x=1$ to $0.80520$ at $x=1.02$, a measured average of $0.99$, consistent with a slope of exactly $1$ at the left end.
Linear approximation and the new derivative table are one question here, which is how the material actually gets examined.
2§12.3 — an angle changing in time●●●○○
A camera on a straight road tracks a car. Distances in metres, time in seconds, angle measured from the perpendicular to the road.
Given
the camera stands $50$ m from the road, at the closest point
$x$ is the car's distance along the road from that closest point
$\tan\theta=\dfrac{x}{50}$
$\dfrac{dx}{dt}=20$ m/s
at the instant asked, $x=50$ m
Find
(a) How fast is $\theta$ changing at that instant, in radians per second?
Hint 1/4
Two quantities change in time and one equation ties them together, so the first step is to write $\theta$ as a function of $x$.
Hint 2/4
$\theta=\arctan\left(\dfrac{x}{50}\right)$, and the chain rule in time gives $\dfrac{d\theta}{dt}=\dfrac{d\theta}{dx}\cdot\dfrac{dx}{dt}$.
Hint 3/4
With $\dfrac{d\theta}{dx}=\dfrac{1}{1+\left(x/50\right)^{2}}\cdot\dfrac{1}{50}$ and the data $x=50$, $\dfrac{dx}{dt}=20$: the bracket is $1+1=2$.
Size check without the formula: at $x=50$ the car is at $45$ degrees, $50\sqrt2\approx 70.7$ m away, and its velocity across the line of sight is $20\cos 45^{\circ}\approx 14.1$ m/s; $14.1/70.7=0.2$ rad/s.
Inverse trigonometric derivatives are where related rates and this section meet, and the meeting is common in examinations.
3§12.3 — a derivative of an integral, unannounced●●●●○
The variable appears as a limit of integration rather than inside the integrand, which changes which theorem opens the problem.
Size check: on $[0,\pi/4]$ the integrand $e^{t^{2}}$ lies between $1$ and $1.85$, so a derivative just under $1$ at $x=1$ is the right order, and positive as an increasing $G$ requires.
No antiderivative of $e^{t^{2}}$ exists in elementary terms, which is the reason the question is phrased this way.
Mistake ledger (16 entries)
⚠ Evaluating the derivative at the output instead of the input
The number $b$ is the one in the question, so the eye puts it in every slot it fits.
⚠ Buried error found in the scaffolding ladder, step 3
The chain rule was dropped inside the logarithm: $\dfrac{d}{dx}\ln(1+2x)=\dfrac{2}{1+2x}$. The outer derivative is the visible one and the inner factor is a single digit, so it disappears unnoticed.
⚠ Buried error found in the scaffolding ladder, step 4
The logarithm was never undone: with the step-3 value the answer would be $L=e^{1}$, not $L=1$. Once the hard part is over, $\ln L$ is quietly read as $L$.
Close the page and write from memory: the derivative of an inverse at a point; the derivative of $a^{x}$; $\arcsin$ and $\arctan$ with their intervals; the hyperbolic identity with its sign; the two shapes the rule accepts; the three rewrites. Then check against the formula card and mark what you invented rather than recalled.
Given $f(2)=9$ and $f'(2)=3$, produce $\left(f^{-1}\right)'(9)$ and say why the derivative is evaluated at $2$?
c-inverse-slope
Differentiate $4^{x}$, $\log_3 x$ and $x^{\sqrt2}$ without confusing the three rules?
c-general-base
Derive $(\arcsin x)'$ from $\sin y=x$, including the sentence that fixes the sign of the root?
c-arcsin-arctan
Explain the minus sign in $(\arccos x)'$ without quoting a table, and state where each formula stops being valid?
c-arcsec-sign-family
Write $\cosh$ and $\sinh$ from memory, prove $\cosh^{2}-\sinh^{2}=1$ in one line, and recover $\sinh^{-1}x$ as a logarithm?
c-hyperbolic
Look at a quotient and say in one line whether the rule is allowed, including the case where it is allowed but useless?
c-lhopital
Take $\lim_{x\to 0^{+}}x^{x}$ from its shape to its value without help, and remember the last line?
c-recast-forms
Glossary (10 terms)
principal branchesas dal
The piece of a periodic function's domain on which it is one to one, chosen once so an inverse exists; for the sine, $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$.
A function returning the angle on the principal branch whose sine, cosine or tangent is the given number; written $\arcsin$, $\arccos$, $\arctan$.
genel üstel fonksiyon
The function $a^{x}=e^{x\ln a}$ for a fixed base $a>0$; its derivative is itself multiplied by the constant $\ln a$.
general logarithmgenel logaritma
The function $\log_a x=\dfrac{\ln x}{\ln a}$, the inverse of $a^{x}$; its derivative is $\dfrac{1}{x\ln a}$.
ters fonksiyon teoremi
The statement that a one-to-one differentiable function with non-zero derivative has a differentiable inverse whose slope at an output is the reciprocal of the slope at the matching input.
hyperbolic functionhiperbolik fonksiyon
A combination of $e^{x}$ and $e^{-x}$ written $\sinh$, $\cosh$ or $\tanh$, satisfying $\cosh^{2}-\sinh^{2}=1$ and parametrising a hyperbola.
The inverse of a hyperbolic function; unusually it has an elementary closed form, since solving for it means solving a quadratic in $e^{y}$.
indeterminate formbelirsizlik
The symbol a substitution produces when it decides nothing, such as zero over zero; it reports that the expression must be rewritten, never what the answer is.
l'Hôpital's ruleL'Hôpital kuralı
The theorem that a limit of shape $\tfrac{0}{0}$ or $\tfrac{\infty}{\infty}$ equals the limit of the quotient of the derivatives, provided that second limit exists.
The standard move for a power form: take logarithms, evaluate the resulting product or quotient, and exponentiate at the end.
What comes next
§13 · Integration techniques: parts and trigonometric integrals
Every derivative built here gets read backwards next: the arcsine and arctangent formulas become the antiderivatives behind trigonometric substitution, and the product rule reversed becomes integration by parts.
Sources
James Stewart, Calculus, Ninth Edition Sections 6.4*, 6.6 and 6.8, which are the three named on this week's syllabus line.
Standard results assumed from earlier in this course The chain rule, implicit differentiation, the consequence that a zero derivative on an interval means a constant, and the derivatives of $\ln x$ and $e^{x}$.