Differentiating Inverse Functions

The function $f(x) = x^3 + x$ has an inverse $g$, but you cannot easily write a formula for it: finding $g(2)$ means solving $x^3 + x = 2$. Yet you can find the slope of $g$ at $x = 2$ in one line. Since $f(1) = 2$, the inverse sends 2 back to 1, and since $f'(1) = 3 + 1 = 4$, the slope of $g$ at 2 is $\frac{1}{4}$. This works whenever the slope of $f$ at the matching input is not 0, and the reason is the chain rule from topic 3.1 applied to the equation that defines an inverse.

What an inverse undoes

If $f$ sends $a$ to $b$, its inverse $f^{-1}$ sends $b$ back to $a$:

For that to make sense, $f$ must be one-to-one: no two inputs share an output, so each $b$ comes from a single $a$. A function whose derivative is positive everywhere on an interval is increasing there, so it never repeats an output and has an inverse on that interval. The same is true when the derivative is negative everywhere. Topic 5.3 proves the link between the sign of $f'$ and increasing or decreasing.

Two facts follow directly from the definition.

• Undoing. $f\left(f^{-1}(x)\right) = x$ for every $x$ in the domain of $f^{-1}$, and $f^{-1}\left(f(x)\right) = x$ for every $x$ in the domain of $f$.
• Swapped points. The point $(a, b)$ is on the graph of $f$ exactly when $(b, a)$ is on the graph of $f^{-1}$. Swapping coordinates is reflection across the line $y = x$, so the two graphs are mirror images in that line.

The notation $f^{-1}(x)$ means the inverse function, not the reciprocal $\frac{1}{f(x)}$. The two are unrelated: for $f(x) = x^3$, the inverse is $\sqrt[3]{x}$, while the reciprocal is $\frac{1}{x^3}$.

Why the slopes are reciprocals

Reflecting across $y = x$ swaps horizontal and vertical. A segment that rises 4 units over a run of 1 becomes, after reflection, a segment that rises 1 over a run of 4. So if the tangent to $f$ at $(a, b)$ has slope $m$, the reflected tangent, which touches $f^{-1}$ at $(b, a)$, has slope $\frac{1}{m}$.

The chain rule turns that picture into a proof. Write $g = f^{-1}$ and start from the undoing equation,

Differentiate both sides. The left side is a composition with outside $f$ and inside $g$, so by the chain rule

Dividing is allowed only when $f'(g(x)) \neq 0$. The same result comes from implicit differentiation (topic 3.2): $y = g(x)$ means $f(y) = x$, and differentiating with respect to $x$ gives $f'(y)\frac{dy}{dx} = 1$. In Leibniz notation, with $x = f(y)$, this reads $\frac{dy}{dx} = \frac{1}{dx/dy}$: the rate of the inverse is the reciprocal of the original rate, measured at the same point of the curve.

The step that takes thought is finding $a$. You are given $b$, an output of $f$, and you need the input that produced it, so you solve $f(a) = b$. With a formula, try small integers first, as with $x^3 + x = 2$ in the opening; on a calculator question, use the calculator's solver. With a table, look for $b$ in the $f(x)$ column, not the $x$ column. With a graph, find the point of the graph whose height is $b$. For instance, if the graph of $f$ passes through $(2, 5)$ with a tangent slope of 3 there, then the graph of $f^{-1}$ passes through $(5, 2)$ with slope $\frac{1}{3}$.

When the inverse has no derivative

If $f'(a) = 0$, the formula would divide by zero, and the inverse really does fail to have a derivative at $b$. The picture explains it: the tangent to $f$ at $(a, b)$ is horizontal, and reflecting a horizontal line across $y = x$ gives a vertical one. So the graph of $f^{-1}$ has a vertical tangent at $(b, a)$.

The simplest case is $f(x) = x^3$, which is one-to-one with inverse $g(x) = \sqrt[3]{x}$. Since $f'(0) = 0$, the cube root has a vertical tangent at the origin. Directly, $g'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}}$, which grows without bound as $x \to 0$.

The chain-rule derivation assumed $g'(b)$ exists and then found its value. Slopes of secant lines show why it exists when $f'(a) \neq 0$. Take a point $x$ near $b$ and let $y = g(x)$, so $f(y) = x$. The secant slope of $g$ between $b$ and $x$ is

the reciprocal of a secant slope of $f$ between $a$ and $y$. The graph of $g$ is the unbroken graph of $f$ reflected across $y = x$, so as $x \to b$, $y \to a$, and the bottom fraction approaches $f'(a)$. When $f'(a) \neq 0$, the whole expression approaches $\frac{1}{f'(a)}$, so $g'(b)$ exists and has that value. When $f'(a) = 0$, the secant slopes of $g$ grow without bound in size, which is the vertical tangent. So whenever you use the formula, the nonzero denominator is the condition that makes the answer valid.

Derivatives you already use, explained

Two derivatives from Unit 2 come straight from this rule.

The natural log. $\ln x$ is the inverse of $f(x) = e^x$, and $f'(x) = e^x$. With $g(x) = \ln x$,

The denominator $e^{\ln x}$ is never 0, which is why $\ln x$ is differentiable on its entire domain, $x > 0$.

The square root. On $x \geq 0$, $\sqrt{x}$ is the inverse of $f(x) = x^2$, and $f'(x) = 2x$. So

which matches the power rule with exponent $\frac{1}{2}$. At $x = 0$ the denominator is $f'(0) = 0$, and the square root has a vertical tangent there.

The same method gives the derivatives of $\arcsin x$, $\arctan x$ and the other inverse trigonometric functions, which topic 3.4 covers.

Worked examples

Where students lose points

On the exam

Inverse-derivative questions give $f$ by a formula, a table, or a graph and ask for $(f^{-1})'(b)$ or a tangent line to $f^{-1}$. They appear in multiple choice and as parts of free-response questions, with or without a calculator. The work a grader looks for is the matching point, written as $f(a) = b$ or $g(b) = a$, then the reciprocal of $f'(a)$. If finding $a$ from $f(a) = b$ needs a solver, the question will be calculator-active, and the answer is expected to three decimal places.