# Derivative of $\arcsin(\cos x)$

In my calculus course, my teacher asked me to differentiate the function $$f(x) = \arcsin(\cos x).$$ My work looked like so, $$f'(x)=\frac{1}{\sqrt{1-\cos^2x}}\cdot-\sin x = -\frac{\sin x}{\sqrt{\sin^2 x}} = -\frac{\sin x}{|\sin x|} = -\frac{|\sin x|}{\sin x}.$$ However, as I went to confirm my answer with my instructor, they claimed that my answer was incorrect. Instead, they claimed my work should have gone like so: $$f'(x)=\frac{1}{\sqrt{1-\cos^2x}}\cdot-\sin x = -\frac{\sin x}{\sqrt{\sin^2 x}} = -\frac{\sin x}{\sin x} = -1.$$ I attempted to explain how I thought $$\sqrt{x^2}$$ simplified to $$|x|$$, but they keep asserting I'm incorrect, claiming that the function $$\arcsin x$$ does not exist for $$|x|>\frac{\pi}{2}$$, and that therefore $$-1$$ should indeed be the correct answer within the domain of the function, which they said was $$|x|<\frac{\pi}{2}$$.

This has just left my confused about the whole matter. Can anyone explain which derivative is right here, and why?

• Did your instructor mean $0 < x < \pi/2$ instead of $|x| < \pi/2$? Because otherwise for $-\pi/2 < x < 0$ $\arcsin(\cos(x))$ is definitely increasing with derivative $+1$ not $-1$ as your instructor claimed. Also the function is not differentiable at $0$. May 4, 2023 at 1:09
• Your instructor is wrong. It doesn't matter that $\arcsin x$ is undefined for $|x|>\tfrac{\pi}{2}$ because you're not applying $\arcsin$ to $x$, but to $\cos x$ which is always between $-1$ and $1$, where the $\arcsin$ IS defined. By the way, the domain of $\arcsin$ is NOT $|x|< \tfrac{\pi}{2}$ , is $|x|\le 1$ May 4, 2023 at 1:10
• If your instructor is saying that $f'(x)=-1$ for all $x$ in $[-\frac\pi2,\frac\pi2]$, then this is clearly wrong because it means $f$ is always strictly decreasing, which is impossible since $f$ is even. May 4, 2023 at 1:16
• Also in case you want visuals, here is a wolfram plot: wolframalpha.com/… May 4, 2023 at 1:17
• Another plot of the function: desmos.com/calculator/9vkuq2kt2y May 4, 2023 at 1:26

You are correct. The domain of the function is $$\mathbb{R}$$ and not $$[-\pi/2,\pi/2]$$ because the domain of the cosine function is $$\mathbb{R}$$ and the range is $$[-1,1]$$, and so the $$\arcsin$$ function doesn't limit the domain of $$f$$. Your derivative simplifies to:
$$f'(x) = \left\{\begin{array}{lr} -1, & \text{for } x \in \displaystyle\bigcup_{n\in\mathbb{Z}}(2n\pi,(2n+1)\pi)\\ 1, & \text{for } x \in \displaystyle\bigcup_{n\in\mathbb{Z}} ((2n+1)\pi,(2n+2)\pi). \end{array} \right.$$
• I'm not sure if that is a simplification over $-\frac{|\sin x|}{\sin x}$ 😆 May 4, 2023 at 1:19