Calculus 1

Derivatives of inverse functions

If we know how fast a function changes, how fast does its inverse change, even when the inverse has no formula we can write down?

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Implicit differentiationCalculus 1 Inverse trig functionsPrecalculus

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Antiderivatives of trig, exponential and log functionsCalculus 1 Derivatives of inverse trigonometric functionsCalculus 1

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Derivatives of inverse functions

Derivatives of inverse trigonometric functions

Why this exists

Many useful functions arrive as the inverse of a function we already understand: the natural logarithm undoes e^x, arcsine undoes a slice of sine. Others have inverses that exist but have no formula at all. The function f(x) = x^3 + x + 1 takes every real value exactly once, so it has an inverse, but finding f^{-1}(5) means solving the cubic x^3 + x - 4 = 0. We would still like to know how steep the graph of f^{-1} is. This lesson shows that the slope of an inverse is always one number divided by a slope of f, read at the right point.

Definition

Recall from One to one functions and the horizontal line test that a function has an inverse exactly when it is one-to-one: no output comes from two different inputs. A function that fails this, such as x^2, can be given an inverse by restricting its domain to a piece where it is one-to-one. For such an f,

f^{-1}(b) = a \iff f(a) = b.

So the point (a, b) lies on the graph of f exactly when (b, a) lies on the graph of f^{-1}, and the second graph is the first reflected in the line y = x (The graph of an inverse). We call (a, b) and (b, a) partner points.

The derivative of the inverse at b, written (f^{-1})'(b), is the slope of the graph of f^{-1} at the point (b, a). Two other objects look similar and are different functions: \dfrac{1}{f'(x)}, the reciprocal of f' at the same input, and the inverse of the function f'. Neither is (f^{-1})' in general.

Intuition

Figure 1Reflecting in y = x carries P = (1, 3) to Q = (3, 1) and the tangent at P to the tangent at Q. The step under the tangent at P has run \tfrac{1}{2} and rise 2, slope 4; its mirror image has run 2 and rise \tfrac{1}{2}, slope \tfrac{1}{4}.

Reflecting in y = x swaps every horizontal distance with a vertical one. A step along the tangent at P that runs \tfrac{1}{2} and rises 2 becomes, in the mirror, a step that runs 2 and rises \tfrac{1}{2}. Rise over run turns into run over rise, so the mirror tangent’s slope is the reciprocal: 4 becomes \tfrac{1}{4}.

The same fact in rates: if y = f(x) and y grows 4 times as fast as x near a point, then x grows \tfrac{1}{4} as fast as y there. Reading the relationship backwards inverts the rate.

Results

The derivative of an inverse. Let f be continuous and one-to-one on an open interval containing a, differentiable at a, with f'(a) \neq 0, and let b = f(a). Then f^{-1} is differentiable at b and

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

where a = f^{-1}(b). Writing a = f^{-1}(x) at every point gives the same rule as a formula in x:

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

In Leibniz notation, with y = f(x) and x = f^{-1}(y), it reads

\frac{dx}{dy} = \frac{1}{dy/dx}.

Where the rule gives nothing. If f is differentiable at a and f'(a) = 0, then f^{-1} is not differentiable at b = f(a). When f is smooth near a and one-to-one, the graph of f^{-1} has a vertical tangent at (b, a).

For example, with f(x) = e^x and f^{-1}(x) = \ln x, the formula gives (\ln x)' = \dfrac{1}{e^{\ln x}} = \dfrac{1}{x}, the result derived in Derivatives of logarithms and logarithmic differentiation.

Why it is true

The chain rule route. For every x in the domain of f^{-1},

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

Suppose f^{-1} is differentiable at b. Differentiate both sides at x = b; the left side is a composition, so the chain rule applies:

\begin{aligned} f'\big(f^{-1}(b)\big)\cdot(f^{-1})'(b) &= 1, \\ f'(a)\,(f^{-1})'(b) &= 1. \end{aligned}

If f'(a) \neq 0, dividing gives the formula. If f'(a) = 0, the left side is 0 for any value of (f^{-1})'(b), which contradicts 0 = 1; so f^{-1} cannot be differentiable at b. That proves the second result.

The same computation is implicit differentiation: y = f^{-1}(x) means f(y) = x, and differentiating gives f'(y)\,\dfrac{dy}{dx} = 1.

Why f^{-1} is differentiable. The argument above assumed it. Here is the reason, with one fact taken without proof: a continuous one-to-one function on an interval has a continuous inverse. Write y near b as y = f(x) with x = f^{-1}(y). Then

\begin{aligned} \frac{f^{-1}(y) - f^{-1}(b)}{y - b} &= \frac{x - a}{f(x) - f(a)} \\ &= 1 \Big/ \frac{f(x) - f(a)}{x - a}. \end{aligned}

As y \to b, continuity of f^{-1} gives x \to a (and x \neq a when y \neq b, since f is one-to-one), so the denominator tends to f'(a) \neq 0 and the quotient tends to \dfrac{1}{f'(a)}.

Worked examples

Example 1

A restricted parabola, two ways

Let f(x) = x^2 + 4x + 1 for x \geq -2. Find (f^{-1})'(13) with the formula, then check it by finding f^{-1}.

Solution

On x \geq -2 the parabola is to the right of its vertex (-2, -3), so f is one-to-one there and has an inverse.

The partner point: solve f(a) = 13, that is a^2 + 4a - 12 = 0, or (a + 6)(a - 2) = 0. The root a = -6 lies outside the domain, so a = 2. Since f'(x) = 2x + 4, f'(2) = 8, and

(f^{-1})'(13) = \frac{1}{f'(2)} = \frac{1}{8}.

Check: completing the square, y = (x + 2)^2 - 3, so \displaystyle x = -2 + \sqrt{y + 3} (the positive root, because x \geq -2). Thus \displaystyle f^{-1}(x) = -2 + \sqrt{x + 3} and \displaystyle (f^{-1})'(x) = \dfrac{1}{2\sqrt{x + 3}}, which at x = 13 is \dfrac{1}{2 \cdot 4} = \dfrac{1}{8}. The two routes agree.

At the vertex, f'(-2) = 0, so the graph of f^{-1} starts at (-3, -2) with a vertical tangent.

Example 2

An inverse with no formula

Yerbol is asked for the slope of y = f^{-1}(x) at x = 3, where f(x) = x^3 + x + 1. Find it, and the tangent line there, without a formula for f^{-1}.

Solution

First, f has an inverse: f'(x) = 3x^2 + 1 \geq 1, so the graph rises everywhere and never takes a value twice (see Increasing, decreasing, constant, positive and negative intervals).

Yerbol needs the partner a with f(a) = 3, which is a^3 + a + 1 = 3. He cannot solve cubics in general, but he can try small numbers: a = 1 gives 1 + 1 + 1 = 3. Since f is one-to-one, a = 1 is the only partner. Then f'(1) = 4 and

(f^{-1})'(3) = \frac{1}{f'(1)} = \frac{1}{4}.

The tangent line to f^{-1} passes through the partner point (3, 1) with slope \tfrac{1}{4}:

\begin{aligned} y &= 1 + \tfrac{1}{4}(x - 3) \\ y &= \frac{1}{4}x + \frac{1}{4}. \end{aligned}

This is the tangent at Q in Figure 1.

Example 3

A formula in terms of the inverse itself

Let f(x) = x + e^x and g = f^{-1}. Find g'(1) and g'(1 + e), then write g'(x) using only x and g(x).

Solution

f'(x) = 1 + e^x > 0, so f is one-to-one. The partners are visible: f(0) = 0 + 1 = 1 and f(1) = 1 + e. So

\begin{aligned} g'(1) &= \frac{1}{f'(0)} = \frac{1}{2}, \\ g'(1 + e) &= \frac{1}{f'(1)} = \frac{1}{1 + e}. \end{aligned}

For a general x, the rule gives g'(x) = \dfrac{1}{1 + e^{g(x)}}. The identity f(g(x)) = x says g(x) + e^{g(x)} = x, so e^{g(x)} = x - g(x) and

g'(x) = \frac{1}{1 + x - g(x)}.

At x = 1, where g(1) = 0, this gives \tfrac{1}{2} again. The derivative of g is known everywhere once g itself is, even though g has no formula built from familiar functions.

Example 4

Reading a table

Sholpan has this table for a one-to-one differentiable function f.

x	1	2	5	7
f(x)	2	5	7	9
f'(x)	4	3	\tfrac{2}{3}	\tfrac{1}{2}

Find (f^{-1})'(5) and (f^{-1})'(7). Then let k(x) = x\,f^{-1}(x) and find k'(5).

Solution

The input 5 of f^{-1} is an output of f, so Sholpan looks for 5 in the f(x) row: f(2) = 5, so f^{-1}(5) = 2 and

(f^{-1})'(5) = \frac{1}{f'(2)} = \frac{1}{3}.

The entry f'(5) = \tfrac{2}{3} belongs to a different point of the graph and plays no part here. For 7: f(5) = 7, so (f^{-1})'(7) = \dfrac{1}{f'(5)} = \dfrac{3}{2}.

For k, the product rule gives k'(x) = f^{-1}(x) + x\,(f^{-1})'(x), so

k'(5) = f^{-1}(5) + 5\,(f^{-1})'(5) = 2 + \frac{5}{3} = \frac{11}{3}.

Example 5

A flat point and a vertical tangent

Let f(x) = x^3 - 3x^2 + 3x. Find (f^{-1})'(2), and describe f^{-1} at x = 1.

Solution

Factoring by the binomial pattern, f(x) = (x - 1)^3 + 1. Cubing takes every real value exactly once, so f is one-to-one, with \displaystyle f^{-1}(x) = 1 + \sqrt[3]{x - 1}. Also f'(x) = 3x^2 - 6x + 3 = 3(x - 1)^2.

For b = 2: (a - 1)^3 + 1 = 2 gives a = 2, and f'(2) = 3, so (f^{-1})'(2) = \tfrac{1}{3}. Check with the formula: (f^{-1})'(x) = \dfrac{1}{3}(x - 1)^{-2/3}, which is \tfrac{1}{3} at x = 2.

For b = 1: f(1) = 1 and f'(1) = 0, so the rule would divide by zero, and by the second result f^{-1} is not differentiable at 1. The formula shows what happens: \dfrac{1}{3}(x - 1)^{-2/3} grows without bound as x \to 1 from either side, so the graph of f^{-1} has a vertical tangent at (1, 1), the mirror image of the horizontal tangent of f (Figure 2). Vertical tangents are discussed in Differentiability: corners, cusps and piecewise functions.

Figure 2Example 5. Where f has a horizontal tangent, at (1, 1), its mirror image has a vertical one, and f^{-1} has no derivative there.

Example 6

The second derivative of an inverse

Show that (f^{-1})''(b) = -\dfrac{f''(a)}{f'(a)^3}, where b = f(a), for a twice-differentiable one-to-one f with f'(a) \neq 0. Use it to find (f^{-1})''(3) for f(x) = x^3 + x + 1.

Solution

Write g = f^{-1}, so g'(x) = \dfrac{1}{f'(g(x))} = \big(f'(g(x))\big)^{-1}. Differentiate with the chain rule twice, once for the power -1 and once for f'(g(x)), whose derivative is f''(g(x))\,g'(x):

g''(x) = -\frac{f''(g(x))\,g'(x)}{f'(g(x))^2} = -\frac{f''(g(x))}{f'(g(x))^3},

using g'(x) = \dfrac{1}{f'(g(x))} in the last step. At x = b, g(b) = a, which is the claim.

For f(x) = x^3 + x + 1, Example 2 found a = 1 and f'(1) = 4; also f''(x) = 6x, so f''(1) = 6 and

(f^{-1})''(3) = -\frac{6}{4^3} = -\frac{3}{32}.

f is concave up at (1, 3) and rising, and its mirror image is concave down at (3, 1): reflection in y = x turns a curve bending upward into one bending the other way (see Second and higher derivatives).

Boundaries

f must be one-to-one near a. Otherwise there is no inverse to differentiate. For f(x) = x^2 the question “what is (f^{-1})'(4)” has no meaning until a piece is chosen, and the two pieces give \tfrac{1}{4} and -\tfrac{1}{4}.
f'(a) must not be 0. At such a point f^{-1} has no derivative; for smooth f its graph has a vertical tangent there.
f must be differentiable at a. If f has a corner at a, so does its mirror image at b.
The derivative of f is read at the partner a = f^{-1}(b), not at b. The input of f^{-1} is an output of f.
The full statement that f^{-1} is differentiable rests on the continuity of f^{-1}, which this lesson uses without proof.

Where this goes next

The next lesson, Derivatives of inverse trigonometric functions, applies the same rule to arcsine, arccosine and arctangent on the slices chosen in Inverse trig functions, and turns \dfrac{1}{f'(f^{-1}(x))} into expressions with no trigonometry left in them. The rule also gives the derivative of every logarithm from the derivative of the matching exponential, and it explains why an inverse found only numerically, as by Newton's method, still has a known slope at every point.

Exercises

Partner points with a formula

For each function, find the partner a and (f^{-1})' at the given number with the rule; then check by finding f^{-1}.

f(x) = 4x - 7; find (f^{-1})'(5).
f(x) = 2x^3 + 3; find (f^{-1})'(19).
\displaystyle f(x) = \sqrt{3x + 1} for x \geq -\tfrac{1}{3}; find (f^{-1})'(4).
f(x) = x^2 + 2x for x \geq -1; find (f^{-1})'(8).
f(x) = \dfrac{x}{x + 1} for x > -1; find (f^{-1})'\big(\tfrac{1}{2}\big).

Inverses with no formula

f(x) = x^5 + x^3 + x. Explain why f has an inverse and find (f^{-1})'(3).
f(x) = 2x + \sin x. Find (f^{-1})'(\pi + 1).
f(x) = e^{2x} + x. Find (f^{-1})'(1).
f(x) = x + \ln x for x > 0. Find (f^{-1})'(e + 1).
f(x) = x^3 + x + 1. Find the tangent line to y = f^{-1}(x) at x = 1.
f(x) = x^3 + x + 1. Find every point on the graph of f^{-1} where the slope is \tfrac{1}{13}.

Tables and rates

Exercises 12 to 15 use this table for a one-to-one differentiable f.

x	-2	0	1	3
f(x)	7	4	1	-2
f'(x)	-1	-2	-5	-\tfrac{1}{2}

Find (f^{-1})'(4).
Find (f^{-1})'(1).
Let h(x) = f^{-1}(x^2). Find h'(2).
Find the tangent line to y = f^{-1}(x) at x = -2.
A one-to-one function has f(4) = 10 and f'(4) = 5. Find the tangent line to f at x = 4 and the tangent line to f^{-1} at x = 10, and show that each is the reflection of the other in y = x.
Arman rides a bike along a straight path, always forward, and is f(t) meters from the start after t seconds. At t = 8 he is at 100 meters, moving at 5 meters per second. Find (f^{-1})'(100) with its units and say what it means.

Vertical tangents, restricted domains, second derivatives

f(x) = x^3 + 3x^2 + 3x + 2. Find the point where the graph of f^{-1} has a vertical tangent, and find (f^{-1})'(28).
f(x) = x^2 restricted to x \leq 0. Find (f^{-1})'(4).
f(x) = x^3 - 12x restricted to [-2, 2]. Find (f^{-1})'(-11), and the points of the graph of f^{-1} where its tangent is vertical.
f(x) = x^4 + 2 for x \geq 0. Find (f^{-1})'(18) and the point where the graph of f^{-1} has a vertical tangent.
f(x) = x + e^x. Use Example 6 to find (f^{-1})''(1).

Reasoning

Explain why (f^{-1})'(b) always has the same sign as f'(a), and what this says about the graph of the inverse of a rising function and of a falling one.
A function is its own inverse, and f(c) = c with f'(c) \neq 0. Prove that f'(c) = 1 or f'(c) = -1. Check both cases with f(x) = \dfrac{1}{x} at c = 1 and f(x) = x.
Yerbol says that because f(x) = x^3 + x + 1 has f'(x) \geq 1 everywhere, the graph of f^{-1} has no vertical tangent, and every slope of f^{-1} is greater than 0 and at most 1. Explain why both claims are true, and find where the slope equals 1.

Answers

\tfrac{1}{4}. a = 3 since f(3) = 5; f'(3) = 4. Check: f^{-1}(x) = \dfrac{x + 7}{4} has slope \tfrac{1}{4} everywhere.
\tfrac{1}{24}. 2a^3 + 3 = 19 gives a = 2; f'(x) = 6x^2, f'(2) = 24. Check: f^{-1}(x) = \big(\tfrac{x - 3}{2}\big)^{1/3} has derivative \tfrac{1}{6}\big(\tfrac{x - 3}{2}\big)^{-2/3}, which is \tfrac{1}{6} \cdot \tfrac{1}{4} = \tfrac{1}{24} at 19.
\tfrac{8}{3}. \displaystyle \sqrt{3a + 1} = 4 gives a = 5; \displaystyle f'(x) = \dfrac{3}{2\sqrt{3x + 1}}, f'(5) = \tfrac{3}{8}. Check: f^{-1}(x) = \dfrac{x^2 - 1}{3} for x \geq 0, derivative \tfrac{2x}{3} = \tfrac{8}{3} at 4.
\tfrac{1}{6}. a^2 + 2a = 8 gives a = 2 or a = -4, and only a = 2 is in the domain; f'(2) = 6. Check: \displaystyle f^{-1}(x) = -1 + \sqrt{x + 1}, derivative \displaystyle \dfrac{1}{2\sqrt{x + 1}} = \tfrac{1}{6} at 8.
4. \dfrac{a}{a + 1} = \tfrac{1}{2} gives a = 1; f'(x) = \dfrac{1}{(x + 1)^2}, f'(1) = \tfrac{1}{4}. Check: f^{-1}(x) = \dfrac{x}{1 - x}, derivative \dfrac{1}{(1 - x)^2} = 4 at \tfrac{1}{2}.
\tfrac{1}{9}. f'(x) = 5x^4 + 3x^2 + 1 \geq 1, so the graph rises everywhere and f is one-to-one. f(1) = 3, f'(1) = 9.
\tfrac{1}{2}. f'(x) = 2 + \cos x \geq 1, so f is one-to-one. f\big(\tfrac{\pi}{2}\big) = \pi + 1 and f'\big(\tfrac{\pi}{2}\big) = 2 + 0 = 2.
\tfrac{1}{3}. f'(x) = 2e^{2x} + 1 > 0. f(0) = 1, f'(0) = 3.
\dfrac{e}{e + 1}. f'(x) = 1 + \dfrac{1}{x} > 0 for x > 0. f(e) = e + 1, f'(e) = 1 + \dfrac{1}{e} = \dfrac{e + 1}{e}.
y = x - 1. f(0) = 1, so the partner point is (1, 0); f'(0) = 1, so the slope is 1.
(11, 2) and (-9, -2). The slope of f^{-1} at (f(a), a) is \dfrac{1}{3a^2 + 1}, which is \tfrac{1}{13} when a^2 = 4; f(2) = 11 and f(-2) = -9.
-\tfrac{1}{2}. f(0) = 4, so (f^{-1})'(4) = \dfrac{1}{f'(0)} = -\tfrac{1}{2}.
-\tfrac{1}{5}. f(1) = 1, so (f^{-1})'(1) = \dfrac{1}{f'(1)} = -\tfrac{1}{5}.
-2. Chain rule: h'(x) = (f^{-1})'(x^2) \cdot 2x, so h'(2) = (f^{-1})'(4) \cdot 4 = -\tfrac{1}{2} \cdot 4.
y = -2x - 1. f(3) = -2, so the point is (-2, 3) and the slope is \dfrac{1}{-1/2} = -2; y = 3 - 2(x + 2).
y = 5x - 10 and y = \tfrac{1}{5}x + 2. The second passes through (10, 4) with slope \tfrac{1}{5}. Swapping x and y in the first gives x = 5y - 10, that is y = \tfrac{1}{5}x + 2, the second line.
\tfrac{1}{5} second per meter. f^{-1}(s) is the time at which Arman is s meters out, so (f^{-1})'(100) = \dfrac{1}{f'(8)} = \tfrac{1}{5}. Near the 100-meter mark each further meter takes him about 0.2 seconds.
(1, -1), and \tfrac{1}{27}. f(x) = (x + 1)^3 + 1 and f'(x) = 3(x + 1)^2 is 0 only at x = -1, where f(-1) = 1. For 28: (a + 1)^3 = 27 gives a = 2, f'(2) = 27.
-\tfrac{1}{4}. On x \leq 0 the partner of 4 is a = -2, and f'(-2) = -4. Check: \displaystyle f^{-1}(x) = -\sqrt{x}, derivative \displaystyle -\dfrac{1}{2\sqrt{x}} = -\tfrac{1}{4} at 4.
-\tfrac{1}{9}; vertical tangents at (16, -2) and (-16, 2). f'(x) = 3x^2 - 12 < 0 inside [-2, 2], so f falls and is one-to-one. f(1) = -11 and 1 is the only partner in [-2, 2], f'(1) = -9. f'(\pm 2) = 0 with f(-2) = 16, f(2) = -16: the graph of f^{-1} is vertical at both of its ends.
\tfrac{1}{32}; (2, 0). a^4 + 2 = 18 gives a = 2, f'(2) = 4 \cdot 8 = 32. f'(0) = 0 and f(0) = 2, so the graph of f^{-1} starts at (2, 0) with a vertical tangent.
-\tfrac{1}{8}. a = 0, f'(0) = 2, f''(x) = e^x so f''(0) = 1, and -\dfrac{1}{2^3} = -\tfrac{1}{8}.
A number and its reciprocal have the same sign, and (f^{-1})'(b) = \dfrac{1}{f'(a)}. So where f rises its inverse rises, and where f falls its inverse falls; the table for Exercises 12 to 15 shows a falling f with all slopes of f^{-1} negative.
If f = f^{-1} then f'(c) = (f^{-1})'(c) = \dfrac{1}{f'(f^{-1}(c))} = \dfrac{1}{f'(c)}, since f^{-1}(c) = c. So f'(c)^2 = 1. For \tfrac{1}{x}: f'(1) = -1. For x: f'(1) = 1.
The slope of f^{-1} at (f(a), a) is \dfrac{1}{f'(a)} = \dfrac{1}{3a^2 + 1}. The denominator is never 0, so no slope is undefined and no tangent is vertical. Since 3a^2 + 1 \geq 1, the slope satisfies 0 < \dfrac{1}{3a^2 + 1} \leq 1, with equality only at a = 0, that is at the point (1, 0).