Math Core

Many important functions are defined as inverses: $\ln x$ undoes $e^x$ , $\sqrt{x}$ undoes $x^2$ , and $\arcsin x$ undoes $\sin x$ . Often you can't even write a formula for an inverse, yet the AP exam still expects you to find its derivative. The trick is that the slope of an inverse is tied directly to the slope of the original function.

The picture: reciprocal slopes

The graph of $f^{-1}$ is the reflection of the graph of $f$ across the line $y = x$ . Reflecting swaps $x$ and $y$ , so every point $(a, b)$ on $f$ becomes $(b, a)$ on $f^{-1}$ .

Reflection also swaps rise and run. A tangent line to $f$ at $(a, b)$ with slope $\dfrac{\text{rise}}{\text{run}} = m$ becomes a tangent line to $f^{-1}$ at $(b, a)$ with slope $\dfrac{\text{run}}{\text{rise}} = \dfrac{1}{m}$ .

f(x) = x³ + 1 has slope 3 at (1, 2). Its inverse has slope 1/3 at the mirror point (2, 1).Open in grapher →

The formula

You can also get the rule from the chain rule. Let $g = f^{-1}$ , so $f(g(x)) = x$ for every $x$ in the domain of $g$ . Differentiate both sides:

f'(g(x)) \cdot g'(x) = 1 \quad\Longrightarrow\quad g'(x) = \frac{1}{f'(g(x))}.

Derivative of an inverse function

If $f$ is differentiable and one-to-one, $g = f^{-1}$ , and $f'(g(a)) \ne 0$ , then

g'(a) = \frac{1}{f'(g(a))}.

In words: to find the slope of the inverse at $x = a$ , find the matching input $b = g(a)$ (the number with $f(b) = a$ ), then take the reciprocal of $f'(b)$ .

The formula needs $f'(g(a)) \ne 0$ . Where $f$ has a horizontal tangent, the reflected tangent is vertical, and the inverse is not differentiable there.

Finding the matching input

The hardest part is usually finding $g(a)$ without a formula for $g$ . Remember what $g(a)$ means: it's the input $b$ that makes $f(b) = a$ . On the AP exam, $b$ is almost always a small integer you can find by inspection or read from a table.

Worked example: An inverse with no formula

Let $f(x) = x^3 + 2x - 1$ , and let $g$ be the inverse of $f$ . Find $g'(2)$ .

Solution. You can't easily solve $y = x^3 + 2x - 1$ for $x$ , so don't try. Instead, find the input that gives an output of 2. Try small integers: $f(1) = 1 + 2 - 1 = 2$ . ✓ So $g(2) = 1$ .

Next, $f'(x) = 3x^2 + 2$ , so $f'(1) = 5$ . Therefore

g'(2) = \frac{1}{f'(g(2))} = \frac{1}{f'(1)} = \frac{1}{5}.

Common mistake

Don't evaluate $f'$ at $a$ itself. In the example above, $f'(2) = 14$ , and $\dfrac{1}{14}$ is a very popular wrong answer. You need $f'$ at the matching input $g(2) = 1$ .

Worked example: From a table

The functions $f$ and $g$ are differentiable inverses. Selected values of $f$ and $f'$ are shown. Find $g'(7)$ .

$x$	2	4	7
$f(x)$	4	7	10
$f'(x)$	5	$\dfrac{2}{3}$	3

Solution. Find where $f$ outputs 7: the table shows $f(4) = 7$ , so $g(7) = 4$ . Then

g'(7) = \frac{1}{f'(4)} = \frac{1}{2/3} = \frac{3}{2}.

The entry $f'(7) = 3$ is a trap: it describes $f$ at the input 7, which is irrelevant here.

Worked example: Why the derivative of ln x is 1/x

You learned that $\dfrac{d}{dx}\ln x = \dfrac{1}{x}$ . Here's why. The function $g(x) = \ln x$ is the inverse of $f(x) = e^x$ , and $f'(x) = e^x$ . So

g'(x) = \frac{1}{f'(g(x))} = \frac{1}{e^{\ln x}} = \frac{1}{x}.

The same idea gives derivatives of inverse trig functions in the next lesson.

Worked example: Tangent line to an inverse

Let $f(x) = x + e^x$ , and let $g$ be the inverse of $f$ . Write an equation of the line tangent to the graph of $g$ at $x = 1$ .

Solution. Find the input with $f(b) = 1$ : $f(0) = 0 + e^0 = 1$ , so $g(1) = 0$ . The point of tangency on $g$ is $(1, 0)$ .

Next, $f'(x) = 1 + e^x$ , so $f'(0) = 2$ and $g'(1) = \dfrac{1}{2}$ . The tangent line is

y = 0 + \frac{1}{2}(x - 1) = \frac{1}{2}x - \frac{1}{2}.

Tip

Keep the coordinates straight by writing the point on $f$ first, then flipping it. For the last example: $f$ passes through $(0, 1)$ with slope 2, so $g$ passes through $(1, 0)$ with slope $\dfrac{1}{2}$ .

Practice

Practice 1

Let $f(x) = x^5 + x + 3$ , and let $g$ be the inverse of $f$ . Find $g'(5)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Let $f(x) = x^3 + 1$ . Find $\left(f^{-1}\right)'(9)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Let $f(x) = 2x + \cos x$ , and let $g$ be the inverse of $f$ . Find $g'(1)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

The functions $f$ and $g$ are differentiable inverses. The point $(3, -1)$ is on the graph of $g$ , and $g'(3) = 4$ . Find $f'(-1)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

(AP-style) The table gives values of a differentiable, one-to-one function $f$ and its derivative. If $g$ is the inverse of $f$ , what is $g'(5)$ ?

$x$	1	3	5
$f(x)$	3	5	8
$f'(x)$	2	4	6

Practice 6

(AP-style) Let $f(x) = e^{2x} + x$ , and let $g$ be the inverse of $f$ . What is $g'(1)$ ?

Practice 7

Let $f(x) = x^3 + x$ , and let $g$ be the inverse of $f$ . Write an equation of the line tangent to the graph of $g$ at $x = 2$ .

Enter an expression, e.g. 3x^2 - 2x + 1