Differentiating Inverse Trigonometric Functions

When an angle depends on a length, the angle is an inverse trig function of that length: a camera tracking a rising rocket points at the angle $\arctan\left(\frac{\text{height}}{\text{distance}}\right)$. To differentiate $\arcsin x$, $\arccos x$ or $\arctan x$ you don't need a new rule. You turn the statement back into an ordinary trig equation, differentiate it with the chain rule, and solve for the derivative.

What arcsin, arccos and arctan return

Sine takes every value between $-1$ and $1$ infinitely often, so "the angle whose sine is $x$" only makes sense once you agree on which angle. Each inverse trig function returns the one angle in a fixed interval, chosen so that the original function is one-to-one there and covers all its values.

So $\arcsin\frac{1}{2} = \frac{\pi}{6}$, not $\frac{5\pi}{6}$, even though both have sine $\frac{1}{2}$, because only $\frac{\pi}{6}$ is in the output interval. Likewise $\arccos\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$ and $\arctan(-1) = -\frac{\pi}{4}$. The outputs are angles in radians, and every derivative formula below assumes radians.

The exam also writes these as $\sin^{-1}x$, $\cos^{-1}x$ and $\tan^{-1}x$. That $-1$ means inverse function, not reciprocal: $\sin^{-1}x$ is an angle, while $\frac{1}{\sin x}$ is $\csc x$.

Deriving the derivative of arcsine

Let $y = \arcsin x$. By definition, this says $\sin y = x$ with $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$. Both sides are functions of $x$, so differentiate both with respect to $x$. On the left, $y$ depends on $x$, so the chain rule puts a factor $\frac{dy}{dx}$ next to the derivative of sine:

That answer is in terms of $y$, and you want it in terms of $x$. The Pythagorean identity gives $\cos^2 y = 1 - \sin^2 y = 1 - x^2$, so $\cos y = \pm\sqrt{1 - x^2}$. The sign is not a guess: $y$ lies between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, where cosine is never negative, so $\cos y = \sqrt{1-x^2}$ and

At $x = \pm 1$ the denominator is 0, because there $y = \pm\frac{\pi}{2}$ and $\cos y = 0$. The graph of arcsine has vertical tangent lines at its two endpoints, so its derivative doesn't exist there.

The same result comes out of the inverse function formula from topic 3.3, $(f^{-1})'(x) = \dfrac{1}{f'\left(f^{-1}(x)\right)}$, with $f(x) = \sin x$ on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Then $f'\left(f^{-1}(x)\right) = \cos(\arcsin x) = \sqrt{1-x^2}$. Check it at $x = \frac{1}{2}$: sine's slope at $\frac{\pi}{6}$ is $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$, and arcsine's slope at $\frac{1}{2}$ is $\frac{1}{\sqrt{1 - 1/4}} = \frac{2}{\sqrt{3}}$, the reciprocal. Reflecting a graph across $y = x$ swaps rise and run, so slopes at matching points are reciprocals.

Arctangent and arccosine

Arctangent. Let $y = \arctan x$, so $\tan y = x$ with $-\frac{\pi}{2} < y < \frac{\pi}{2}$. Differentiating, $\sec^2 y \cdot \frac{dy}{dx} = 1$. The identity $\sec^2 y = 1 + \tan^2 y = 1 + x^2$ turns this into

No sign choice is needed, since $\sec^2 y$ is a square. The formula matches the graph: the slope is 1 at $x = 0$ and shrinks toward 0 as $|x|$ grows, because the graph of arctangent flattens toward its horizontal asymptotes $y = \pm\frac{\pi}{2}$.

Arccosine. Let $y = \arccos x$, so $\cos y = x$ with $0 \le y \le \pi$. Differentiating, $-\sin y \cdot \frac{dy}{dx} = 1$. On $[0, \pi]$ sine is never negative, so $\sin y = \sqrt{1 - x^2}$ and

The minus sign says arccosine is decreasing: as $x$ runs from $-1$ to $1$, the angle runs from $\pi$ down to $0$. There is also a shortcut to this result. For every $x$ in $[-1, 1]$, $\arcsin x + \arccos x = \frac{\pi}{2}$ (for $0 < x < 1$, they are the two acute angles of a right triangle with hypotenuse 1 and one leg $x$). A constant sum means the two derivatives add to 0, so they are negatives of each other.

With an inside function

On the exam the input is rarely plain $x$. With $u$ a differentiable function of $x$, the chain rule (topic 3.1) multiplies each formula by $u'$, and the $x^2$ in each formula becomes $u^2$, the square of the whole inside.

For example, $\frac{d}{dx}\arctan(5x) = \frac{5}{1+25x^2}$, and $\frac{d}{dx}\arcsin\left(x^2\right) = \frac{2x}{\sqrt{1-x^4}}$ for $-1 < x < 1$.

Worked examples

Where students lose points

On the exam

You need the exact values of these functions at the standard inputs without a calculator: $\arcsin$ and $\arccos$ at $0, \pm\frac{1}{2}, \pm\frac{\sqrt{2}}{2}, \pm\frac{\sqrt{3}}{2}, \pm 1$, and $\arctan$ at $0, \pm\frac{\sqrt{3}}{3}, \pm 1, \pm\sqrt{3}$. Inverse trig derivatives also come up in contexts like Example 3, where an angle is defined through a ratio of lengths, and in table problems such as finding the derivative of $\arctan(f(x))$ from given values of $f$ and $f'$.