4.4 Derivatives of Inverse Functions

4.4 Derivatives of Inverse Functions

Quick Answer: The College Board outline lists "Derivative of the inverse of a function (including $\arcsin x$ and $\arctan x$)" as a single tested bullet. Section 4.3 handled the named inverse trigonometric cases; this section handles the general rule behind them. If $g = f^{-1}$, then $g'(x) = \frac{1}{f'(g(x))}$ — the reciprocal of $f'$ evaluated at the input of $f$, not at the input of $g$. That evaluation point is the entire difficulty of the topic, and it is where nearly every wrong answer comes from.

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1. The Inverse Function Theorem

Let $f$ be one-to-one and differentiable on an interval, and let $g = f^{-1}$ be its inverse. If $f'(g(a)) \neq 0$, then $g$ is differentiable at $a$ and

$$g'(a) = \frac{1}{f'\bigl(g(a)\bigr)}, \qquad \text{equivalently} \qquad \left(f^{-1}\right)'(a) = \frac{1}{f'\bigl(f^{-1}(a)\bigr)}$$

Derivation from the Chain Rule

The identity that defines an inverse is $f(g(x)) = x$ for every $x$ in the domain of $g$. Differentiate both sides with respect to $x$ and apply the Chain Rule from Section 4.1 to the left side:

$$\frac{d}{dx}\bigl[f(g(x))\bigr] = \frac{d}{dx}[x] \implies f'(g(x)) \cdot g'(x) = 1 \implies g'(x) = \frac{1}{f'(g(x))}$$

That single line is worth reproducing on scratch paper at the start of any inverse-derivative question — it regenerates the formula and simultaneously reminds you which function gets evaluated where.

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2. The Three-Step Protocol

Almost every CLEP-style inverse-derivative item gives you a value of $f$ and a value of $f'$ and asks for $g'$ at some number $a$. Work it in this fixed order:

Step 1 is the step candidates skip. You are handed the output $a$ and must run $f$ backwards to recover the input $c$.

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3. Worked Examples

Example 1: Recovering the input by inspection

Let $f(x) = x^3 + 2x + 1$, which is increasing on all of $\mathbb{R}$ (since $f'(x) = 3x^2 + 2 > 0$) and therefore invertible. Find $g'(4)$ where $g = f^{-1}$.

1. Find $g(4)$. Solve $f(c) = 4$: testing $c = 1$ gives $1 + 2 + 1 = 4$. So $g(4) = 1$.
2. Evaluate $f'$ at $c = 1$. $f'(x) = 3x^2 + 2$, so $f'(1) = 3 + 2 = 5$.
3. Reciprocate. $g'(4) = \dfrac{1}{5}$.

Note that $f'(4) = 50$ is irrelevant here. A distractor of $\frac{1}{50}$ is standard.

Example 2: A transcendental $f$ that cannot be inverted algebraically

Let $f(x) = x + e^{x}$. Find $\left(f^{-1}\right)'(1)$.

You cannot solve $x + e^x = 1$ with algebra, but you can guess and verify: $f(0) = 0 + e^0 = 1$, so $f^{-1}(1) = 0$. Then $f'(x) = 1 + e^x$ and $f'(0) = 1 + 1 = 2$, giving

$$\left(f^{-1}\right)'(1) = \frac{1}{f'(0)} = \frac{1}{2}$$

This is exactly why the theorem exists: it produces the inverse's derivative without ever writing down a formula for the inverse.

Example 3: The theorem reproduces the arcsine formula

Take $f(x) = \sin x$ on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so $g(x) = \arcsin x$. Then $f'(x) = \cos x$ and

$$g'(x) = \frac{1}{\cos(\arcsin x)} = \frac{1}{\sqrt{1 - x^2}}$$

using the reference triangle from Section 4.3 (adjacent $= \sqrt{1 - x^2}$, hypotenuse $= 1$). Every formula in the Section 4.3 master table is this one theorem applied to a restricted trigonometric function — memorize the table for speed, but derive from the theorem if your memory fails on exam day.

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4. Geometry: Reflection Across $y = x$

The graph of $f^{-1}$ is the mirror image of the graph of $f$ across the line $y = x$. Reflection swaps the roles of run and rise, so it inverts every slope:

• If $(c, a)$ lies on $f$ with tangent slope $m$, then $(a, c)$ lies on $f^{-1}$ with tangent slope $1/m$.
• A steep tangent on $f$ becomes a shallow tangent on $f^{-1}$, and vice versa.
• Increasing functions have increasing inverses; the reciprocal of a positive slope is positive.

When the derivative fails to exist

The theorem carries the hypothesis $f'(g(a)) \neq 0$ for a reason. If $f$ has a horizontal tangent at $c$, the reflected tangent on $f^{-1}$ is vertical, and the inverse is not differentiable at $a = f(c)$.

The standard illustration is $f(x) = x^3$ at $c = 0$: $f'(0) = 0$, so $g(x) = x^{1/3}$ has a vertical tangent at the origin and $g'(0)$ does not exist. Confirm it directly: $g'(x) = \frac{1}{3}x^{-2/3} \to \infty$ as $x \to 0$.

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5. CLEP Traps

• Trap 1 — Evaluating $f'$ at the wrong number. $g'(a) = 1/f'(g(a))$, never $1/f'(a)$. If a table lists both $f'(a)$ and $f'(g(a))$, one of them is a planted distractor.
• Trap 2 — Forgetting to reciprocate. Answering $f'(g(a))$ instead of its reciprocal is the second most common error.
• Trap 3 — Inverse function vs. reciprocal function. $f^{-1}(x)$ is the inverse; $[f(x)]^{-1} = 1/f(x)$ is the reciprocal. Their derivatives are unrelated: the reciprocal's derivative is $-f'(x)/[f(x)]^2$ by the Quotient Rule.
• Trap 4 — Assuming an inverse exists. The theorem needs $f$ to be one-to-one. On the CLEP this is usually established by showing $f' > 0$ (or $f' < 0$) everywhere, which is worth stating in your reasoning.