Derivatives of Inverse Functions

Derivatives of Inverse Functions and Inverse Trig

By the end of this lesson, you will be able to find the derivative of an inverse function using the inverse function derivative theorem, and differentiate inverse trigonometric functions.

The Inverse Function Derivative Theorem

If $f$ is one-to-one and differentiable with inverse $f^{-1}$, then:

$$\left(f^{-1}\right)'(b) = \frac{1}{f'(a)}$$

where $b = f(a)$ (i.e., $a = f^{-1}(b)$).

Intuition: the graph of $f^{-1}$ is the reflection of the graph of $f$ over $y = x$. If $f$ has slope $m$ at $(a, b)$, then $f^{-1}$ has slope $1/m$ at $(b, a)$.

Example: Finding $(f^{-1})'$ at a Point

Given $f(x) = x^3 + 2x$, find $(f^{-1})'(3)$.

Step 1: Find $a$ where $f(a) = 3$: try $a = 1$: $f(1) = 1 + 2 = 3$ ✓

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

Step 3: $(f^{-1})'(3) = \frac{1}{f'(1)} = \frac{1}{5}$

Derivatives of Inverse Trig Functions

These are derived by applying the inverse function theorem to the trig functions:

Function	Derivative
$\arcsin x$	$\dfrac{1}{\sqrt{1-x^2}}$
$\arccos x$	$-\dfrac{1}{\sqrt{1-x^2}}$
$\arctan x$	$\dfrac{1}{1+x^2}$
$\text{arcsec}, x$	$\dfrac{1}{

Note: $\arcsin$ and $\arccos$ are negatives of each other; together they give $\frac{d}{dx}[\arcsin x + \arccos x] = 0$ (since $\arcsin x + \arccos x = \pi/2$ for $|x| \leq 1$).

Deriving $\frac{d}{dx}[\arctan x]$

Let $y = \arctan x$, so $\tan y = x$.

Differentiate implicitly: $\sec^2 y \cdot \frac{dy}{dx} = 1$

$\frac{dy}{dx} = \frac{1}{\sec^2 y} = \frac{1}{1 + \tan^2 y} = \frac{1}{1 + x^2}$

(Using $\sec^2 y = 1 + \tan^2 y = 1 + x^2$)

Examples with Chain Rule

$\frac{d}{dx}[\arctan(3x)] = \frac{1}{1+(3x)^2} \cdot 3 = \frac{3}{1+9x^2}$

$\frac{d}{dx}[\arcsin(x^2)] = \frac{1}{\sqrt{1-x^4}} \cdot 2x = \frac{2x}{\sqrt{1-x^4}}$

$\frac{d}{dx}[\arctan(e^x)] = \frac{e^x}{1+e^{2x}}$

Summary

Inverse function derivative: $(f^{-1})'(b) = \frac{1}{f'(f^{-1}(b))}$. Find the input $a$, evaluate $f'(a)$, take the reciprocal.