Derivatives of Inverse Functions: Examples

The derivative of an inverse function is the reciprocal of the original function’s derivative at the matching point. If f(a) = b, then $(f^{-1})'(b) = \frac{1}{f'(a)}$ , as long as $f'(a) \ne 0$ . You can find the slope of the inverse without finding a formula for the inverse.

Before this lesson: Inverse Functions, Chain Rule and Implicit Differentiation

Why the slopes are reciprocals

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

Slope is rise over run. Swapping x and y swaps the rise and the run, so the slope flips. If f has slope 4 at (1, 4), then $f^{-1}$ has slope $\frac{1}{4}$ at (4, 1).

Graph of f(x) = x³ + x + 2 (solid blue) and its inverse (solid orange), which are mirror images across the dashed gray line y = x. The point (1, 4) on f has a tangent line of slope 4, and the matching point (4, 1) on the inverse has a tangent line of slope 1/4. — Reflecting over y = x swaps rise and run, so the slope at (4, 1) is the reciprocal of the slope at (1, 4).

The inverse function derivative formula

\left(f^{-1}\right)'(b) = \frac{1}{f'(a)}, \quad \text{where } f(a) = b

Another way to write it, for every x in the domain of $f^{-1}$ :

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

To find the slope of the inverse at x = b, first find a: solve f(a) = b. This is the step students skip most often.
Find f′(x) and evaluate it at a, not at b.
Take the reciprocal: $\frac{1}{f'(a)}$ .

The rule needs $f'(a) \ne 0$ . Where f has a horizontal tangent, the inverse has a vertical tangent, and its derivative does not exist.

Derivatives of inverse functions examples

Example 1: Without a formula for the inverse

Let $f(x) = x^3 + x + 2$ . Find $(f^{-1})'(4)$ .

a^3 + a + 2 = 4

Find the input a with f(a) = 4.

a = 1

Try small whole numbers: 1 + 1 + 2 = 4, so a = 1.

f'(x) = 3x^2 + 1

Differentiate f with the power rule.

f'(1) = 3(1)^2 + 1 = 4

Evaluate f′ at a = 1, not at 4.

(f^{-1})'(4) = \frac{1}{f'(1)} = \frac{1}{4}

Take the reciprocal.

Answer: $(f^{-1})'(4) = \frac{1}{4}$ . Solving $y = x^3 + x + 2$ for x would be very hard, but you never had to.

Example 2: Checking with a known inverse

Let $f(x) = 2x + 5$ . Find $(f^{-1})'(x)$ with the formula, then check by finding the inverse.

f'(x) = 2

Differentiate f.

(f^{-1})'(x) = \frac{1}{2}

The reciprocal of 2. It is the same at every point, since f is a line.

y = 2x + 5 \Rightarrow x = 2y + 5

To check, swap x and y (see inverse functions).

f^{-1}(x) = \frac{x - 5}{2}

Subtract 5, then divide by 2, to solve for y.

\frac{d}{dx}\left[\frac{x - 5}{2}\right] = \frac{1}{2}

The slope of the inverse line is ½. It matches.

Answer: $(f^{-1})'(x) = \frac{1}{2}$ . A line with slope m has an inverse line with slope $\frac{1}{m}$ .

Example 3: The square root as an inverse

Let $f(x) = x^2$ for $x \ge 0$ , so $f^{-1}(x) = \sqrt{x}$ . Use the formula to find the derivative of $\sqrt{x}$ at x = 9.

a^2 = 9, \; a \ge 0 \Rightarrow a = 3

Solve f(a) = 9 in the domain x ≥ 0.

f'(x) = 2x \Rightarrow f'(3) = 6

Differentiate f and evaluate at a = 3.

(f^{-1})'(9) = \frac{1}{6}

Take the reciprocal.

\frac{1}{2\sqrt{9}} = \frac{1}{6}

Check with the power rule: the derivative of $x^{1/2}$ is $\frac{1}{2\sqrt{x}}$ .

Answer: the slope of $\sqrt{x}$ at x = 9 is $\frac{1}{6}$ . Both methods agree.

Example 4: Using a table of values

Let g be the inverse of f. You know f(2) = 7, f′(2) = 3 and f′(7) = 10. Find g′(7).

g(7) = 2

Since f(2) = 7, the inverse sends 7 back to 2.

g'(7) = \frac{1}{f'(g(7))}

Write the formula at x = 7.

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

Replace g(7) with 2, then use f′(2) = 3.

Answer: $g'(7) = \frac{1}{3}$ . The value f′(7) = 10 is a trap: f′ is evaluated at the matching input 2, not at 7.

Example 5: Word problem with a water tank

The volume of water in a tank is $V = h^3 + 4h$ cubic feet when the water is h feet deep. When the tank holds 16 cubic feet, how fast does the depth change per cubic foot added (find dh/dV)?

h^3 + 4h = 16

Find the depth when V = 16.

h = 2

Try h = 2: 8 + 8 = 16.

\frac{dV}{dh} = 3h^2 + 4

Differentiate V with respect to h.

\frac{dV}{dh} = 3(2)^2 + 4 = 16

Evaluate at h = 2.

\frac{dh}{dV} = \frac{1}{16}

Depth as a function of volume is the inverse, so its derivative is the reciprocal.

Answer: $\frac{dh}{dV} = \frac{1}{16}$ foot per cubic foot. Near that level, each extra cubic foot of water raises the depth by about 0.0625 ft, or 0.75 inch.

Common mistakes

Mistake	Example of the error	Fix
Evaluating f′ at b instead of a	In Example 1, using f′(4) = 49 to get 1/49	First solve f(a) = 4, then use f′(1) = 4.
Confusing the inverse with the reciprocal	Writing $f^{-1}(x) = \frac{1}{f(x)}$	$f^{-1}$ undoes f. The reciprocal is $\frac{1}{f(x)}$ , which is different.
Forgetting to take the reciprocal	“ $(f^{-1})'(4) = f'(1) = 4$ “	The slopes are reciprocals: $\frac{1}{4}$ .
Ignoring the domain	For $x^2$ with $x \ge 0$ , using a = −3	Pick the a that lies in the domain of f.
Using the formula where f′(a) = 0	“ $(f^{-1})'(0) = \frac{1}{0}$ ” for $x^3$	The inverse has a vertical tangent there. Its derivative does not exist.

Related methods

Where the formula comes from

Since $f(f^{-1}(x)) = x$ for every x in the domain of $f^{-1}$ , differentiate both sides with the chain rule: $f'(f^{-1}(x)) \cdot (f^{-1})'(x) = 1$ . Divide by $f'(f^{-1}(x))$ to get the formula.

Implicit differentiation

You can also write $y = f^{-1}(x)$ as $f(y) = x$ and use implicit differentiation: $f'(y) \cdot y' = 1$ , so $y' = \frac{1}{f'(y)}$ . It is the same rule in a different form.

Coming next

This rule is how the derivatives of $\ln x$ and of $\arcsin x$ , $\arctan x$ and the other inverse trig functions are found. Those come later in this unit, after the derivatives of trig and exponential functions.

Practice problems

Let $f(x) = x^3 + 2$ . Find $(f^{-1})'(10)$ .
Let $f(x) = x^5 + x$ . Find $(f^{-1})'(2)$ .
Let $f(x) = 4x - 3$ . Find $(f^{-1})'(x)$ .
Let $f(x) = \sqrt{x + 1}$ . Find $(f^{-1})'(3)$ .
Let g be the inverse of f, with f(3) = −1 and f′(3) = 5. Find g′(−1).
Let $f(x) = x^3 + 3x - 1$ . Find $(f^{-1})'(3)$ .
Let $f(x) = x^3 + x$ . Find the tangent line to the graph of $f^{-1}$ at (2, 1).
Let $f(x) = x^2 + 2x$ for $x \ge -1$ . Find $(f^{-1})'(8)$ .
Explain why the inverse of $f(x) = x^3$ has no derivative at x = 0.
A sensor’s reading is $T = 2V^2 + 10$ degrees when the voltage is V volts (V ≥ 0). Find dV/dT when T = 28.

Show answers

a^3 + 2 = 10 \Rightarrow a^3 = 8

Solve f(a) = 10: subtract 2.

a = 2

Take the cube root.

f'(x) = 3x^2 \Rightarrow f'(2) = 12

Differentiate and evaluate at a = 2.

(f^{-1})'(10) = \frac{1}{12}

Take the reciprocal.

Answer: $\frac{1}{12}$

a^5 + a = 2 \Rightarrow a = 1

Solve f(a) = 2: 1 + 1 = 2.

f'(x) = 5x^4 + 1 \Rightarrow f'(1) = 6

Differentiate and evaluate at a = 1.

(f^{-1})'(2) = \frac{1}{6}

Take the reciprocal.

Answer: $\frac{1}{6}$

f'(x) = 4

Differentiate f.

(f^{-1})'(x) = \frac{1}{4}

Take the reciprocal.

Answer: $\frac{1}{4}$ for every x

\sqrt{a + 1} = 3 \Rightarrow a + 1 = 9

Solve f(a) = 3: square both sides.

a = 8

Subtract 1.

f'(x) = \frac{1}{2\sqrt{x + 1}}

Chain rule on $(x + 1)^{1/2}$ .

f'(8) = \frac{1}{2 \cdot 3} = \frac{1}{6}

Evaluate at a = 8.

(f^{-1})'(3) = 6

The reciprocal of ⅙ is 6.

Answer: 6

g(-1) = 3

Since f(3) = −1.

g'(-1) = \frac{1}{f'(3)} = \frac{1}{5}

Take the reciprocal of f′ at the matching input.

Answer: $\frac{1}{5}$

a^3 + 3a - 1 = 3 \Rightarrow a = 1

Solve f(a) = 3: 1 + 3 − 1 = 3.

f'(x) = 3x^2 + 3 \Rightarrow f'(1) = 6

Differentiate and evaluate at a = 1.

(f^{-1})'(3) = \frac{1}{6}

Take the reciprocal.

Answer: $\frac{1}{6}$

f(1) = 1 + 1 = 2

The point (2, 1) on the inverse matches (1, 2) on f.

f'(x) = 3x^2 + 1 \Rightarrow f'(1) = 4

Differentiate and evaluate at 1.

\text{slope} = \frac{1}{4}

The inverse’s slope at (2, 1) is the reciprocal.

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

Point-slope form (see point-slope form).

y = \frac{1}{4}x + \frac{1}{2}

Distribute, then add 1.

Answer: $y = \frac{1}{4}x + \frac{1}{2}$

a^2 + 2a - 8 = 0

Solve f(a) = 8: subtract 8.

(a + 4)(a - 2) = 0

Factor the trinomial (see factoring trinomials).

a = 2

Reject a = −4, which is not in the domain x ≥ −1.

f'(x) = 2x + 2 \Rightarrow f'(2) = 6

Differentiate and evaluate at a = 2.

(f^{-1})'(8) = \frac{1}{6}

Take the reciprocal.

Answer: $\frac{1}{6}$

f(0) = 0

The point (0, 0) on the inverse matches a = 0.

f'(x) = 3x^2 \Rightarrow f'(0) = 0

Differentiate and evaluate at 0.

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

The reciprocal of 0 is undefined.

Answer: f has a horizontal tangent at the origin, so its inverse, the cube root, has a vertical tangent there and no derivative.

10.

2V^2 + 10 = 28 \Rightarrow V^2 = 9

Subtract 10, then divide by 2.

V = 3

Take the positive root, since V ≥ 0.

\frac{dT}{dV} = 4V = 12

Differentiate T and evaluate at V = 3.

\frac{dV}{dT} = \frac{1}{12}

Take the reciprocal.

Answer: $\frac{1}{12}$ volt per degree

Derivatives of inverse functions FAQ

What is the formula for the derivative of an inverse function?

$(f^{-1})'(b) = \frac{1}{f'(a)}$ , where f(a) = b. The slope of the inverse at (b, a) is the reciprocal of the slope of f at (a, b).

Do I need to find the inverse function first?

No. You only need the input a that gives f(a) = b and the derivative f′. That is the point of the formula.

Why are the slopes reciprocals?

The inverse’s graph is the reflection of f over y = x. Reflecting swaps rise and run, so the slope m becomes 1/m.

When does the inverse not have a derivative?

Where f′(a) = 0. The graph of f is flat there, so the inverse’s graph is vertical, and a vertical line has no slope.

Summary

$(f^{-1})'(b) = \frac{1}{f'(a)}$ , where f(a) = b.
Solve f(a) = b first, then evaluate f′ at a.
The graphs of f and its inverse are reflections over y = x, so their slopes at matching points are reciprocals.
If f′(a) = 0, the inverse has a vertical tangent and no derivative there.
This rule leads to the derivatives of ln x and the inverse trig functions.

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