Derivatives of Inverse Functions

Sometimes you need the slope of an inverse function but can’t find a formula for the inverse, like for $f(x) = x^3 + 2x - 1$. Good news: you don’t need one. The slope of $f^{-1}$ at a point is just the reciprocal of the slope of $f$ at the matching point. This page shows why, and how to use it with equations, tables, and graphs.

Key ideas

Reflected points, reciprocal slopes

The graph of $y = f^{-1}(x)$ is the reflection of $y = f(x)$ in the line $y = x$. Reflecting swaps $x$ and $y$, so:

• If $(a, b)$ is on $f$, then $(b, a)$ is on $f^{-1}$.
• Reflecting a tangent line swaps its rise and run. A line with slope $\dfrac{\text{rise}}{\text{run}} = m$ becomes a line with slope $\dfrac{\text{run}}{\text{rise}} = \dfrac{1}{m}$.

So the slope of $f^{-1}$ at $(b, a)$ is the reciprocal of the slope of $f$ at $(a, b)$.

$f(x) = x^2 + 1$ ($x \ge 0$) has slope $2$ at $(1, 2)$, so $f^{-1}$ has slope $\tfrac{1}{2}$ at $(2, 1)$.

The formula

Here’s the algebra behind the picture. Since $f^{-1}$ undoes $f$,

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

Differentiate both sides, using the chain rule on the left:

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

Solve for the derivative of the inverse:

$$\big(f^{-1}\big)'(a) = \frac{1}{f'\big(f^{-1}(a)\big)}$$

This works whenever $f$ is differentiable and one-to-one near the point, and $f'\big(f^{-1}(a)\big) \ne 0$.

How to use it

To find $\big(f^{-1}\big)'(a)$:

1. Find $b = f^{-1}(a)$. That means solving $f(b) = a$. Usually you can spot $b$ by trying small numbers, or read it from a table or graph.
2. Find $f'(b)$.
3. Take the reciprocal: $\big(f^{-1}\big)'(a) = \dfrac{1}{f'(b)}$.

The key point: evaluate $f'$ at $b = f^{-1}(a)$, not at $a$.

When f’ is zero

If $f'(b) = 0$, then $f$ has a horizontal tangent at $(b, a)$. Its reflection is a vertical tangent, so $f^{-1}$ is not differentiable at $a$. For example, $f(x) = x^3$ has $f'(0) = 0$, and $f^{-1}(x) = \sqrt[3]{x}$ has a vertical tangent at the origin.

Where ln x comes from

This idea explains the derivative of $\ln x$. Since $\ln x$ is the inverse of $f(x) = e^x$, and $f'(x) = e^x$,

$$\frac{d}{dx}\ln x = \frac{1}{e^{\ln x}} = \frac{1}{x}$$

The same trick gives the derivatives of inverse trig functions.

Worked examples

Example 1: From an equation

Let $f(x) = x^3 + 2x - 1$. Find $\big(f^{-1}\big)'(2)$.

Solution. We can’t easily solve for $f^{-1}$, but we don’t need to.

Step 1: Find $b$ with $f(b) = 2$. Try $b = 1$: $1 + 2 - 1 = 2$. So $f^{-1}(2) = 1$.

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

Step 3:

$$\big(f^{-1}\big)'(2) = \frac{1}{f'(1)} = \frac{1}{5}$$

Example 2: From a table

The function $f$ is differentiable and increasing. Some values are given below.

$x$	$1$	$2$	$3$	$4$
$f(x)$	$3$	$5$	$8$	$12$
$f'(x)$	$1.5$	$2.5$	$4$	$6$

Find $\big(f^{-1}\big)'(5)$ and $\big(f^{-1}\big)'(8)$.

Solution. Look for $5$ in the $f(x)$ row, not the $x$ row. Since $f(2) = 5$, we have $f^{-1}(5) = 2$:

$$\big(f^{-1}\big)'(5) = \frac{1}{f'(2)} = \frac{1}{2.5} = 0.4$$

Similarly, $f(3) = 8$, so $f^{-1}(8) = 3$ and

$$\big(f^{-1}\big)'(8) = \frac{1}{f'(3)} = \frac{1}{4}$$

Example 3: Checking with a formula

Let $f(x) = x^2 + 1$ for $x \ge 0$ (shown in the figure). Find $\big(f^{-1}\big)'(2)$ two ways.

Solution.

Using the formula: $f(1) = 2$, so $f^{-1}(2) = 1$. Since $f'(x) = 2x$, $f'(1) = 2$, and $\big(f^{-1}\big)'(2) = \dfrac{1}{2}$.

Directly: Solve $y = x^2 + 1$ for $x$ and swap to get $f^{-1}(x) = \sqrt{x - 1}$. Then

$$\big(f^{-1}\big)'(x) = \frac{1}{2\sqrt{x - 1}}, \qquad \big(f^{-1}\big)'(2) = \frac{1}{2\sqrt{1}} = \frac{1}{2}$$

Both methods agree.

Example 4: Tangent line to an inverse

Let $f(x) = x + e^x$. Write the equation of the tangent line to $y = f^{-1}(x)$ at $x = 1$.

Solution. Find $f^{-1}(1)$: try $x = 0$, and $f(0) = 0 + 1 = 1$. So $f^{-1}(1) = 0$, and the point on the inverse is $(1, 0)$.

$f'(x) = 1 + e^x$, so $f'(0) = 2$, and the slope of the inverse is $\dfrac{1}{2}$.

$$y - 0 = \frac{1}{2}(x - 1) \quad\text{or}\quad y = \frac{1}{2}x - \frac{1}{2}$$

Notice you found a tangent line to $f^{-1}$ without ever having a formula for $f^{-1}$ (there isn’t a nice one).

Common mistakes

Evaluating f’ at the wrong number. $\big(f^{-1}\big)'(a)$ is $\dfrac{1}{f'(f^{-1}(a))}$, not $\dfrac{1}{f'(a)}$. In Example 1, $\dfrac{1}{f'(2)} = \dfrac{1}{14}$ is a very common wrong answer.

Reading the table the wrong way. To find $f^{-1}(a)$, look for $a$ among the outputs $f(x)$, then use the $x$ in that column.

Using the negative reciprocal. The slopes of $f$ and $f^{-1}$ at matching points are reciprocals, $m$ and $\dfrac{1}{m}$. The negative reciprocal is for perpendicular lines, which is a different idea.

Mixing up the two meanings of the -1. $f^{-1}(x)$ is the inverse function, not $\dfrac{1}{f(x)}$. The formula does contain a reciprocal, but of $f'$, at the right point.

Forgetting the point on the inverse is swapped. If $(0, 1)$ is on $f$, the point on $f^{-1}$ is $(1, 0)$. Use $(1, 0)$ when writing a tangent line to the inverse.

Practice

1. (Warm-up) Suppose $f(2) = 7$ and $f'(2) = 4$, and $f$ has an inverse. Find $\big(f^{-1}\big)'(7)$.

Solution

Since $f(2) = 7$, $f^{-1}(7) = 2$:

$$\big(f^{-1}\big)'(7) = \frac{1}{f'(2)} = \frac{1}{4}$$

2. (Warm-up) The tangent line to $y = f(x)$ at $(3, -1)$ has slope $5$. Name the matching point on $y = f^{-1}(x)$ and give the slope of the tangent there.

Solution

The matching point is $(-1, 3)$, and the slope there is $\dfrac{1}{5}$.

3. (Warm-up) Let $f(x) = 2x + 3$. Use the formula to find $\big(f^{-1}\big)'(x)$, then check by finding $f^{-1}(x)$.

Solution

$f'(x) = 2$ everywhere, so $\big(f^{-1}\big)'(x) = \dfrac{1}{2}$ everywhere.

Check: $f^{-1}(x) = \dfrac{x - 3}{2} = \dfrac{1}{2}x - \dfrac{3}{2}$, which has slope $\dfrac{1}{2}$.

4. (Core) Let $f(x) = x^3 + x$. Find $\big(f^{-1}\big)'(10)$.

Solution

Solve $f(b) = 10$: try $b = 2$, and $8 + 2 = 10$. So $f^{-1}(10) = 2$.

$f'(x) = 3x^2 + 1$, so $f'(2) = 13$, and

$$\big(f^{-1}\big)'(10) = \frac{1}{13}$$

5. (Core) The function $g$ is differentiable and increasing, with values shown below.

$x$	$0$	$1$	$2$	$3$
$g(x)$	$-2$	$1$	$4$	$9$
$g'(x)$	$2$	$3$	$5$	$7$

• (a) Find $\big(g^{-1}\big)'(4)$.
• (b) Find $\big(g^{-1}\big)'(1)$.
• (c) Write the equation of the tangent line to $y = g^{-1}(x)$ at $x = 9$.

Solution

(a) $g(2) = 4$, so $\big(g^{-1}\big)'(4) = \dfrac{1}{g'(2)} = \dfrac{1}{5}$.

(b) $g(1) = 1$, so $\big(g^{-1}\big)'(1) = \dfrac{1}{g'(1)} = \dfrac{1}{3}$.

(c) $g(3) = 9$, so the point on $g^{-1}$ is $(9, 3)$ and the slope is $\dfrac{1}{g'(3)} = \dfrac{1}{7}$:

$$y - 3 = \frac{1}{7}(x - 9)$$

6. (Core) Let $f(x) = 3x + \sin x$ (in radians). Explain why $f$ has an inverse, then find $\big(f^{-1}\big)'(0)$.

Solution

$f'(x) = 3 + \cos x \ge 3 - 1 = 2 \gt 0$, so $f$ is always increasing, which means it’s one-to-one and has an inverse.

$f(0) = 0 + 0 = 0$, so $f^{-1}(0) = 0$, and

$$\big(f^{-1}\big)'(0) = \frac{1}{f'(0)} = \frac{1}{3 + 1} = \frac{1}{4}$$

7. (Core) Let $f(x) = \sqrt{x^3 + 1}$ for $x \ge 0$. Find $\big(f^{-1}\big)'(3)$.

Solution

Solve $\sqrt{b^3 + 1} = 3$: $b^3 + 1 = 9$, so $b = 2$.

By the chain rule, $f'(x) = \dfrac{3x^2}{2\sqrt{x^3 + 1}}$, so

$$f'(2) = \frac{3(4)}{2\sqrt{9}} = \frac{12}{6} = 2$$

Therefore $\big(f^{-1}\big)'(3) = \dfrac{1}{2}$.

8. (Challenge) Let $f(x) = x^5 + 2x^3 + x - 4$.

• (a) Show that $f$ has an inverse.
• (b) Write the equation of the tangent line to $y = f^{-1}(x)$ at $x = 0$.

Solution

(a) $f'(x) = 5x^4 + 6x^2 + 1 \ge 1 \gt 0$ for all $x$, so $f$ is increasing and one-to-one.

(b) Find $f^{-1}(0)$: try $x = 1$, and $1 + 2 + 1 - 4 = 0$. So the point on $f^{-1}$ is $(0, 1)$.

$f'(1) = 5 + 6 + 1 = 12$, so the slope of $f^{-1}$ at $x = 0$ is $\dfrac{1}{12}$:

$$y - 1 = \frac{1}{12}x \quad\text{or}\quad y = \frac{1}{12}x + 1$$

9. (Challenge) Let $f(x) = x^3$, so $f^{-1}(x) = \sqrt[3]{x}$.

• (a) Use the formula to find $\big(f^{-1}\big)'(8)$. Check your answer by differentiating $x^{1/3}$.
• (b) What goes wrong if you try to find $\big(f^{-1}\big)'(0)$? What does the graph of $y = \sqrt[3]{x}$ look like there?

Solution

(a) $f^{-1}(8) = 2$ and $f'(x) = 3x^2$, so $\big(f^{-1}\big)'(8) = \dfrac{1}{f'(2)} = \dfrac{1}{12}$.

Check: $\dfrac{d}{dx}x^{1/3} = \dfrac{1}{3}x^{-2/3}$, and at $x = 8$, $\dfrac{1}{3} \cdot 8^{-2/3} = \dfrac{1}{3} \cdot \dfrac{1}{4} = \dfrac{1}{12}$.

(b) $f^{-1}(0) = 0$ and $f'(0) = 0$, so the formula would give $\dfrac{1}{0}$, which is undefined. The graph of $y = x^3$ has a horizontal tangent at the origin, so its reflection $y = \sqrt[3]{x}$ has a vertical tangent there, and $f^{-1}$ is not differentiable at $0$.