Derivatives of Inverse Functions — AP® Calculus AB Notes

An inverse function undoes a function: if $f(2) = 7$ , then $f^{-1}(7) = 2$ . Often we have no formula for $f^{-1}$ at all. The function $f(x) = x^3 + x$ is one-to-one, but solving $y = x^3 + x$ for $x$ is hard. Still, we can find the slope of its inverse at any point, using only $f$ and its derivative.

The picture: reflection swaps rise and run

The graph of $g = f^{-1}$ is the graph of $f$ reflected in the line $y = x$ . Every point $(a, b)$ of $f$ becomes the point $(b, a)$ of $g$ . Reflection also carries tangent lines along. Take a tangent line to $f$ at $(a, b)$ with slope $m$ : it rises $m$ units for every 1 unit it runs. After the reflection, the roles of rise and run swap, so the reflected line rises 1 unit for every $m$ units it runs. Its slope is $\frac{1}{m}$ .

The graph of f(x) = x³ + x (crimson) and its inverse g (blue), mirror images in the dashed line y = x. The tangent line to f at (1, 2) has slope 4; the tangent line to g at the reflected point (2, 1) has slope 1/4. — **Figure 1.** $f(x) = x^3 + x$ and its inverse $g$ . At $(1, 2)$ the slope of $f$ is 4; at the mirror point $(2, 1)$ the slope of $g$ is $\frac{1}{4}$ . LevlPrep original diagram.

So slopes at matching points are reciprocals. The only care needed is "matching": the slope of $g$ at the input $b$ goes with the slope of $f$ at the input $a = g(b)$ , not at $b$ .

The rule

If $g$ is the inverse of a differentiable function $f$ , and $f'(g(x)) \ne 0$ , then

$g'(x) = \frac{1}{f'(g(x))}$

The rule comes straight from the chain rule. Since $g$ undoes $f$ , $f(g(x)) = x$ for every $x$ in the domain of $g$ . Differentiate both sides: $f'(g(x)) \cdot g'(x) = 1$ . Divide by $f'(g(x))$ and the rule appears.

Using it: find the matching input first

Worked example 1. Let $f(x) = x^3 + x$ and $g = f^{-1}$ . Find $g'(2)$ .

Step 1, find $g(2)$ : which $x$ gives $f(x) = 2$ ? Try small numbers: $f(1) = 1 + 1 = 2$ . So $g(2) = 1$ .

Step 2, the slope of $f$ there: $f'(x) = 3x^2 + 1$ , so $f'(1) = 4$ .

Step 3, take the reciprocal: $g'(2) = \frac{1}{f'(1)} = \frac{1}{4}$ .

A common slip is to compute $\frac{1}{f'(2)} = \frac{1}{13}$ . That is the slope of $f$ at the wrong point.

On the exam the function often comes as a table. Then step 1 means finding the given number in the row of $f$ values and reading the $x$ above it.

Worked example 2. A one-to-one function has these values. Find $\left(f^{-1}\right)'(7)$ .

$x$	$1$	$2$	$3$
$f(x)$	$4$	$7$	$9$
$f'(x)$	$5$	$2$	$6$

The value 7 sits in the $f(x)$ row under $x = 2$ , so $f^{-1}(7) = 2$ .

Read $f'(2) = 2$ . Then $\left(f^{-1}\right)'(7) = \frac{1}{2}$ .

Worked example 3. Let $f(x) = e^{2x} + x$ and $g = f^{-1}$ . Write the line tangent to the graph of $g$ at $x = 1$ .

$f(0) = 1 + 0 = 1$ , so $g(1) = 0$ and the point is $(1, 0)$ .

$f'(x) = 2e^{2x} + 1$ , so $f'(0) = 3$ and $g'(1) = \frac{1}{3}$ .

The tangent line is $y = \frac{1}{3}(x - 1)$ .

When the rule fails

If $f'(a) = 0$ , the tangent line to $f$ at $(a, b)$ is flat. Its mirror image is a vertical line, which has no slope, so $g$ is not differentiable at $b$ . The formula agrees: $\frac{1}{f'(a)} = \frac{1}{0}$ has no value. For example, $f(x) = x^3$ has $f'(0) = 0$ , and its inverse $\sqrt[3]{x}$ has a vertical tangent line at the origin.

A check you can always do

When you do know a formula for the inverse, the rule must agree with differentiating it directly. Take $f(x) = x^3$ for $x > 0$ , so $g(x) = x^{1/3}$ . The rule gives $g'(x)$ $= \frac{1}{3\left(x^{1/3}\right)^2}$ $= \frac{1}{3x^{2/3}}$ . The power rule on $x^{1/3}$ gives $\frac{1}{3}x^{-2/3}$ , the same function.

Slips that cost points

Slip	Wrong	Right
Reading $f'$ at the given number	$\left(f^{-1}\right)'(7) = \frac{1}{f'(7)}$	$\frac{1}{f'(2)}$ , since $f(2) = 7$
Forgetting the reciprocal	$\left(f^{-1}\right)'(7) = f'(2)$	$\frac{1}{f'(2)}$
Reciprocal of a value	$\frac{1}{f(2)}$ or $\frac{1}{f^{-1}(7)}$	the reciprocal of a slope, $\frac{1}{f'(2)}$
Point of the wrong graph	tangent to $f^{-1}$ through $(2, 7)$	through $(7, 2)$