Derivatives of Inverse Functions

The formula comes from the chain rule. Since

f\left(f^{-1}(x)\right) = x

, differentiating both sides gives

f'\left(f^{-1}(x)\right) \cdot \left(f^{-1}\right)'(x) = 1.

Divide by

f'\left(f^{-1}(x)\right)

. The theorem is what guarantees that

f^{-1}

is differentiable, so the chain rule applies.

The graph of

f^{-1}

is the graph of

f

reflected across

y = x

. The reflection swaps rise and run, so a tangent of slope

m

(a, b)

becomes a tangent of slope

\frac{1}{m}

(b, a)

. Where

f'(a) = 0

, the reflection of a horizontal tangent is a vertical one, and

f^{-1}

isn't differentiable at

b

f(x) = x³ + x has slope 4 at (1, 2). Its inverse (blue) has slope 1/4 at the reflected point (2, 1).

From a formula

The value

b

is an output of

f

. To use the rule, first find the input

a

with

f(a) = b

, usually by inspection. If

f

is given by an equation in

x

and

y

, find the point

(a, b)

on its graph. Then

f'(a)

\frac{dy}{dx}

evaluated at

(a, b)

, and the inverse passes through

(b, a)

Example 2

Take as given that

f(x) = x + \sin x

is increasing, as its graph below suggests. So it has an inverse. Find

\left(f^{-1}\right)'

at (a)

0

, (b)

\frac{\pi}{2} + 1

, and (c)

\pi

. Here

f'(x) = 1 + \cos x

(a)

f(0) = 0

and

f'(0) = 2

, so

\left(f^{-1}\right)'(0) = \frac{1}{2}

(b)

f\left(\frac{\pi}{2}\right) = \frac{\pi}{2} + 1

and

f'\left(\frac{\pi}{2}\right) = 1

, so

\left(f^{-1}\right)'\left(\frac{\pi}{2} + 1\right) = 1

(c)

f(\pi) = \pi

, but

f'(\pi) = 1 + \cos\pi = 0

. The graph of

f

has a horizontal tangent at

(\pi, \pi)

, so the graph of

f^{-1}

has a vertical tangent there, and

\left(f^{-1}\right)'(\pi)

doesn't exist.

y = x + sin x and its inverse (blue). At (π, π) the graph of f is flat and the graph of the inverse is vertical. The marked pair of points, (π/2, π/2 + 1) and (π/2 + 1, π/2), both have slope 1.

From a table

Example 3

The function

f

is differentiable and increasing, with the values in the table, and

g = f^{-1}

is its inverse.

$x$	$0$	$1$	$2$	$3$
$f(x)$	$1$	$3$	$4$	$7$
$f'(x)$	$2$	$\tfrac{1}{2}$	$3$	$4$

(a) Find the tangent line to the graph of

g

x = 4

. (b) Let

h(x) = \big(g(x)\big)^2

. Find

h'(3)

(a) Find

4

in the

f(x)

row:

f(2) = 4

, so

g(4) = 2

and

g'(4) = \frac{1}{f'(2)} = \frac{1}{3}.

The tangent line is

y = 2 + \frac{1}{3}(x - 4)

(b) By the chain rule,

h'(3) = 2g(3)\,g'(3)

. Look for

3

in the

f(x)

row, not the

x

row:

f(1) = 3

, so

g(3) = 1

and

g'(3) = \frac{1}{f'(1)} = 2

. Reading the

x

row instead would give

\frac{1}{f'(3)}

, the slope of

g

f(3) = 7

. With the right values,

h'(3) = 2 \cdot 1 \cdot 2 = 4

Two familiar inverses

Example 4

Use the inverse rule to recover the derivatives of

\ln x

and

\sqrt{x}

$\ln x$ is the inverse of

f(x) = e^x

, with

f'(x) = e^x

. For

b > 0

\frac{d}{dx}\big[\ln x\big]\Big|_{x = b} = \frac{1}{e^{\ln b}} = \frac{1}{b}.

$\sqrt{x}$ is the inverse of

f(x) = x^2

for

x \ge 0

, with

f'(x) = 2x

. For

b > 0

, apply the rule on

(0, \infty)

with

a = \sqrt{b}

\frac{d}{dx}\big[\sqrt{x}\,\big]\Big|_{x = b} = \frac{1}{f'\left(\sqrt{b}\right)} = \frac{1}{2\sqrt{b}}.

b = 0

f'(0) = 0

, and the graph of

\sqrt{x}

has a vertical tangent from the right at the origin. That's the reflection of the parabola's horizontal tangent.