The derivative of an inverse function

Warm-up

Three quick questions from earlier lessons.

Discover

$f(2) = 5$ and the tangent to $f$ at $(2, 5)$ has slope 3. The graph of $f^{-1}$ is the reflection in $y = x$ , so it passes through $(5, 2)$ . Reflecting swaps rise and run. What is its slope there?

$\tfrac{1}{3}$
$3$

Key idea

Slopes of inverse functions are reciprocals at matching points.

If

f(a) = b

and

f'(a) \ne 0

, then

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

. In general

(f^{-1})'(x) = \dfrac{1}{f'\big(f^{-1}(x)\big)}

.

Worked example

$f(x) = x^3 + 2x$ , $g = f^{-1}$ . Find $g'(3)$ .

g(3) = 1

Find a with f(a) = 3: 1 + 2 = 3.

f'(1) = 5

f'(x) = 3x² + 2.

g'(3) = \tfrac{1}{5}

Reciprocal.

Your turn, guided

Your turn: same $f$ , find $g'(12)$ . Note $f(2) = 12$ .

f'(2) = 14

f'(2) = 3·4 + 2.

g'(12) = \tfrac{1}{14}

Reciprocal.

Practice ★

$f$ is invertible with $f(3) = 9$ and $f'(3) = 2$ . Find $(f^{-1})'(9)$ .
$f$ is invertible with $f(1) = 7$ and $f'(1) = 3$ . Find $(f^{-1})'(7)$ .

$g$ is the inverse of $f$ . Which expression equals $g'(7)$ ?
1. $\dfrac{1}{f'(g(7))}$
2. $\dfrac{1}{f'(7)}$
3. $f'(g(7))$
4. $\dfrac{1}{f(7)}$

Watch out!

Practice ★★

$f(x) = x^3 + 8x$ has an inverse $g$ . Note $f(1) = 9$ . Find $g'(9)$ .
$f(x) = x^3 + 7x$ has an inverse $g$ . Note $f(1) = 8$ . Find $g'(8)$ .

Exit ticket

Three questions to check you have got it.

$f$ is invertible with $f(3) = 9$ and $f'(3) = 2$ . Find $(f^{-1})'(9)$ .
$f$ is invertible with $f(1) = 7$ and $f'(1) = 3$ . Find $(f^{-1})'(7)$ .
$f$ is invertible with $f(4) = 7$ and $f'(4) = 6$ . Find $(f^{-1})'(7)$ .

Summary card

Derivative of an inverse

Find $a = f^{-1}(b)$ .
Compute $f'(a)$ .
$(f^{-1})'(b) = 1 / f'(a)$ .

Watch out: evaluate $f'$ at $a$ , the input of $f$ , not at $b$ .

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