Derivatives of Inverse Functions, calculus lesson

Reflect the graph of a function across the line $y = x$ and you get the graph of its inverse. A reflection swaps rise and run, so a tangent line of slope $m$ turns into a tangent line of slope $\frac{1}{m}$ . That one picture explains the whole lesson, including what goes wrong when $m = 0$ .

Key terms

Inverse function derivative

Definition Let $f$ be continuous and one-to-one on an open interval $I$ , and let $a$ be in $I$ with $b = f(a)$ . If $f$ is differentiable at $a$ and $f'(a) \neq 0$ , then $f^{-1}$ is differentiable at $b$ and

(f^{-1})'(b) = \frac{1}{f'(a)} = \frac{1}{f'(f^{-1}(b))}.

In plain words To find the slope of the inverse at $b$ , find the point $a$ that $f$ sends to $b$ , and take the reciprocal of the slope of $f$ there.

Example Let $f(x) = x^3 + x$ . Its derivative $3x^2 + 1$ is always positive, so $f$ is one-to-one on $\mathbb{R}$ . Since $f(1) = 2$ and $f'(1) = 4$ , the rule gives

(f^{-1})'(2) = \frac{1}{f'(1)} = \frac{1}{4},

without ever finding a formula for $f^{-1}$ .

Nonzero derivative condition

Definition Let $f$ be one-to-one on an open interval $I$ containing $a$ . If $f$ is differentiable at $a$ with $f'(a) = 0$ , then $f^{-1}$ is not differentiable at $b = f(a)$ . Indeed, if it were, differentiating $f^{-1}(f(x)) = x$ at $a$ by the chain rule would give $(f^{-1})'(b) \cdot f'(a) = 1$ , which is impossible when $f'(a) = 0$ .

In plain words A horizontal tangent on the graph of $f$ reflects into a vertical tangent on the graph of $f^{-1}$ , and a vertical line has no slope.

Example The function $f(x) = x^3$ is one-to-one with $f'(0) = 0$ . Its inverse $f^{-1}(x) = \sqrt[3]{x}$ has the difference quotient

\frac{\sqrt[3]{h} - \sqrt[3]{0}}{h} = h^{-2/3}, \qquad h \neq 0,

at $0$ , which grows without bound as $h \to 0$ . The graph has a vertical tangent at the origin.

Local invertibility

Definition Let $f$ be differentiable on an open interval containing $a$ , with $f'$ continuous at $a$ and $f'(a) \neq 0$ . Then there is an open interval $J$ containing $a$ on which $f'$ has the sign of $f'(a)$ . On $J$ the function $f$ is strictly monotonic, hence one-to-one, $f(J)$ is an open interval, and the inverse of $f$ on $J$ is differentiable on $f(J)$ with the formula above. The hypothesis that $f'$ is continuous at $a$ matters. Without it $f'(a) \neq 0$ alone does not make $f$ one-to-one near $a$ .

In plain words A function need not have an inverse on its whole domain. If its slope is not zero at a point, though, it is one-to-one close to that point, and you can invert that piece.

Example The function $f(x) = x^2$ is not one-to-one on $\mathbb{R}$ . At $a = 3$ the derivative is $f'(3) = 6 \neq 0$ , and on $(0, \infty)$ the function has the inverse $\sqrt{x}$ . Since $f(3) = 9$ , the rule predicts that the slope of $\sqrt{x}$ at $9$ is $\frac{1}{6}$ . The direct formula agrees.

\frac{d}{dx} \sqrt{x} \, \Big\vert_{x = 9} = \frac{1}{2\sqrt{9}} = \frac{1}{6}

Branch ranges

Definition The trigonometric functions are not one-to-one, so each inverse uses a restricted piece, called the principal branch.

Inverse	Inverse of	Domain	Range
$\arcsin x$	$\sin$ on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$	$[-1, 1]$	$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
$\arccos x$	$\cos$ on $[0, \pi]$	$[-1, 1]$	$[0, \pi]$
$\arctan x$	$\tan$ on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$	$\mathbb{R}$	$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
$\operatorname{arccot} x$	$\cot$ on $(0, \pi)$	$\mathbb{R}$	$(0, \pi)$
$\operatorname{arcsec} x$	$\sec$ on $\left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi\right]$	$\lvert x \rvert \ge 1$	$\left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi\right]$
$\operatorname{arccsc} x$	$\csc$ on $\left[-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right]$	$\lvert x \rvert \ge 1$	$\left[-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right]$

Some books choose other ranges for arcsec and arccsc, for example $\left[0, \frac{\pi}{2}\right) \cup \left[\pi, \frac{3\pi}{2}\right)$ for arcsec and $\left(0, \frac{\pi}{2}\right] \cup \left(\pi, \frac{3\pi}{2}\right]$ for arccsc. With those ranges the derivative formulas below have $x$ in place of $\lvert x \rvert$ , so their signs differ for $x < -1$ .

In plain words The branch decides which angle the inverse returns. It also decides the sign of square roots when you derive the derivative formulas.

Example Both $\frac{\pi}{6}$ and $\frac{5\pi}{6}$ have sine $\frac{1}{2}$ , but $\arcsin \frac{1}{2} = \frac{\pi}{6}$ , the one in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ . Likewise $\arccos\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$ and $\arctan(-1) = -\frac{\pi}{4}$ .

Inverse trigonometric derivatives

Definition With the branches above,

\frac{d}{dx} \arcsin x = \frac{1}{\sqrt{1 - x^2}}, \qquad \frac{d}{dx} \arccos x = -\frac{1}{\sqrt{1 - x^2}}, \qquad -1 < x < 1,

\frac{d}{dx} \arctan x = \frac{1}{1 + x^2}, \qquad \frac{d}{dx} \operatorname{arccot} x = -\frac{1}{1 + x^2}, \qquad x \in \mathbb{R},

\frac{d}{dx} \operatorname{arcsec} x = \frac{1}{\lvert x \rvert \sqrt{x^2 - 1}}, \qquad \frac{d}{dx} \operatorname{arccsc} x = -\frac{1}{\lvert x \rvert \sqrt{x^2 - 1}}, \qquad \lvert x \rvert > 1.

For arcsin, put $y = \arcsin x$ with $-1 < x < 1$ , so $\sin y = x$ and $-\frac{\pi}{2} < y < \frac{\pi}{2}$ . On that open interval $\cos y > 0$ , so the derivative of sine is not zero and the inverse rule gives

\frac{dy}{dx} = \frac{1}{\cos y} = \frac{1}{\sqrt{1 - \sin^2 y}} = \frac{1}{\sqrt{1 - x^2}}.

The positive square root is the right one because $\cos y > 0$ on this branch.

In plain words The inverse trigonometric functions have algebraic derivatives, with no trigonometry left in them. That is why they show up so often as answers to integrals later.

Example The slope of $y = \arcsin x$ at $x = \frac{1}{2}$ is

\frac{1}{\sqrt{1 - 1/4}} = \frac{2}{\sqrt{3}} \approx 1.155.

The slope of $y = \arctan x$ at $x = 1$ is $\frac{1}{1 + 1} = \frac{1}{2}$ .

Endpoint restrictions

Definition The functions $\arcsin$ and $\arccos$ are continuous on $[-1, 1]$ but not differentiable at $x = \pm 1$ . There $\sin$ and $\cos$ have derivative zero at the matching angles, so the nonzero derivative condition fails, and the one-sided difference quotients grow without bound in absolute value. The same happens for $\operatorname{arcsec}$ and $\operatorname{arccsc}$ at $x = \pm 1$ .

In plain words At the ends of its domain the graph of arcsin stands vertical. The formula $\frac{1}{\sqrt{1 - x^2}}$ has a zero denominator there, and that matches what the graph does.

Example For $\arcsin$ at the right endpoint $1$ , take the left-hand difference quotient

\frac{\arcsin 1 - \arcsin(1 - h)}{h}, \qquad h > 0.

It is about $14.15$ for $h = 0.01$ and about $141.4$ for $h = 0.0001$ . It keeps growing as $h \to 0^{+}$ .

Check yourself

Let $f(x) = e^x + x$ . Find $(f^{-1})'(1)$ .

Show the answer

Here $f(0) = 1$ and $f'(x) = e^x + 1$ , so $f'(0) = 2$ and $(f^{-1})'(1) = \frac{1}{2}$ .

Differentiate $y = \arctan(2x)$ .

Show the answer

By the chain rule

y' = \frac{1}{1 + (2x)^2} \cdot 2 = \frac{2}{1 + 4x^2},

valid for all real $x$ .

Why is $\arccos x$ not differentiable at $x = 1$ ?

Show the answer

The matching angle is $\arccos 1 = 0$ , and there $\cos'(0) = -\sin 0 = 0$ , so the inverse rule would need to divide by zero. The graph has a vertical tangent there, and the formula $-\frac{1}{\sqrt{1 - x^2}}$ is undefined at $x = 1$ .

Derivatives of Inverse Functions

Theory, part 59

Practice, part 60

Key terms

Inverse function derivative

Nonzero derivative condition

Local invertibility

Branch ranges

Inverse trigonometric derivatives

Endpoint restrictions

Check yourself

How this lesson connects

Builds on

Used later in