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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?

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    $\begingroup$ 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 $0$. $\endgroup$
    – balddraz
    May 4, 2023 at 1:09
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    $\begingroup$ 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 $|x|\le 1$ $\endgroup$
    – jjagmath
    May 4, 2023 at 1:10
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    $\begingroup$ 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. $\endgroup$
    – David
    May 4, 2023 at 1:16
  • $\begingroup$ Also in case you want visuals, here is a wolfram plot: wolframalpha.com/… $\endgroup$
    – balddraz
    May 4, 2023 at 1:17
  • $\begingroup$ Another plot of the function: desmos.com/calculator/9vkuq2kt2y $\endgroup$
    – David K
    May 4, 2023 at 1:26

1 Answer 1

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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:

$$f'(x) = \left\{\begin{array}{lr} -1, & \text{for } x \in \displaystyle\bigcup_{n\in\mathbb{Z}}(2n\pi,(2n+1)\pi)\\ 1, & \text{for } x \in \displaystyle\bigcup_{n\in\mathbb{Z}} ((2n+1)\pi,(2n+2)\pi). \end{array} \right.$$

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  • $\begingroup$ I'm not sure if that is a simplification over $-\frac{|\sin x|}{\sin x}$ 😆 $\endgroup$
    – jjagmath
    May 4, 2023 at 1:19

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