Derivatives of Inverse Functions and Inverse Trig

An inverse function undoes the original: the graph of $f^{-1}$ is the graph of $f$ reflected across $y = x$. Reflection swaps rise and run, so slopes turn into their reciprocals. Implicit Differentiation turns that picture into formulas, including those for the Inverse Trig Functions.

Key idea

If $f(a) = b$ and $f'(a) \ne 0$, then $$(f^{-1})'(b) = \frac{1}{f'(a)} = \frac{1}{f'\big(f^{-1}(b)\big)}.$$ In particular, $$\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}, \qquad \frac{d}{dx}\arctan x = \frac{1}{1 + x^2}, \qquad \frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1 - x^2}}.$$

$f(x) = x^2$ has slope 4 at $(2, 4)$ (tangent $T_1$). The inverse $\sqrt{x}$ has slope $\frac{1}{4}$ at the mirror point $(4, 2)$ (tangent $T_2$). The dashed line is $y = x$.

Why it works

Let $y = f^{-1}(x)$, so $f(y) = x$. Differentiate both sides implicitly: $f'(y)\frac{dy}{dx} = 1$, so $\frac{dy}{dx} = \frac{1}{f'(y)}$. For arcsine: $y = \arcsin x$ means $\sin y = x$ with $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$. Then $\cos y\,\frac{dy}{dx} = 1$, and on that range $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$ (positive). For arctangent: $\tan y = x$ gives $\sec^2 y\,\frac{dy}{dx} = 1$, and $\sec^2 y = 1 + \tan^2 y = 1 + x^2$.

Worked example: An inverse without a formula

$f(x) = x^3 + x$ is one-to-one. Find $(f^{-1})'(2)$.

Find the matching input: $f(1) = 1 + 1 = 2$, so $f^{-1}(2) = 1$.
$f'(x) = 3x^2 + 1$, so $f'(1) = 4$.
$(f^{-1})'(2) = \frac{1}{4}$.

Answer: $\frac{1}{4}$, even though we cannot write $f^{-1}$ with a simple formula.

Worked example: Inverse trig with the chain rule

Differentiate $\arcsin(2x)$ and $\arctan(x^2)$.

$\frac{1}{\sqrt{1 - (2x)^2}}\cdot 2 = \frac{2}{\sqrt{1 - 4x^2}}$.
$\frac{1}{1 + (x^2)^2}\cdot 2x = \frac{2x}{1 + x^4}$.

Answer: $\frac{2}{\sqrt{1 - 4x^2}}$ and $\frac{2x}{1 + x^4}$

Worked example: A derivative that reveals a constant

Differentiate $\arctan x + \arctan\left(\frac{1}{x}\right)$ for $x > 0$. What does the answer tell you?

$\frac{d}{dx}\arctan x = \frac{1}{1 + x^2}$.
Chain rule: $$\frac{d}{dx}\arctan\left(\frac{1}{x}\right) = \frac{1}{1 + \frac{1}{x^2}} \cdot \left(-\frac{1}{x^2}\right) = -\frac{1}{x^2 + 1}$$.
The sum of the derivatives is 0, so the function is constant for $x > 0$. At $x = 1$ it equals $\frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$.

Answer: The derivative is $0$, so $\arctan x + \arctan\frac{1}{x} = \frac{\pi}{2}$ for every $x > 0$: the two angles of a right triangle.

Derivatives of the inverse trig functions.
Function	Derivative	Domain of the derivative
$\arcsin x$	$\frac{1}{\sqrt{1 - x^2}}$	$-1 < x < 1$
$\arccos x$	$-\frac{1}{\sqrt{1 - x^2}}$	$-1 < x < 1$
$\arctan x$	$\frac{1}{1 + x^2}$	all real $x$

At $x = \pm 1$ the arcsine graph has vertical tangents, matching the horizontal tangents of $\sin y$ at $y = \pm\frac{\pi}{2}$: a flat slope reflects into a vertical one. The arctangent's slope is largest at 0 (slope 1) and fades toward 0 far out, as the graph levels off toward its asymptotes $y = \pm\frac{\pi}{2}$.

Common mistake

Using $\frac{1}{f'(b)}$ instead of $\frac{1}{f'(a)}$. In the first example, $\frac{1}{f'(2)} = \frac{1}{13}$, which is wrong. The slope of the inverse at $x = 2$ matches the slope of $f$ at the mirror point, whose $x$-coordinate is $f^{-1}(2) = 1$. Check: always find $a = f^{-1}(b)$ first.

Derivatives of Inverse Functions and Inverse Trig

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