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I have a question of the sign of a derivative. Let $y=\cos^{-1}x$ defined in some interval. Now we compute the derivative $\displaystyle\frac{\text{d}y}{\text{d}x}$. $$1=\frac{\text{d}x}{\text{d}x}=\frac{\text{d}(\cos y)}{\text{d}x}=-\sin y \frac{\text{d}y}{\text{d}x}.$$ Since $\,\,\,$ $\sin y=\sqrt{1-\cos^2 y}=\sqrt{1-x^2}$, $\displaystyle\frac{\text{d}y}{\text{d}x}=-\frac{1}{\sqrt{1-x^2}}$. It is possible that $\sin y=-\sqrt{1-\cos^2 y}=-\sqrt{1-x^2}$. Why we do not use "-" but use "+"?

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Related: math.stackexchange.com/questions/48752/… –  Aryabhata Jul 5 '11 at 18:10

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up vote 6 down vote accepted

Remember to check the definition of $\cos^{-1}$ carefully. A usual convention is that $\cos^{-1}:[-1,1]\to [0,\pi]$; that is, the range is chosen to be $[0,\pi]$. Then note that for all $y\in[0,\pi]$, $\sin(y)\geq 0$. That is why "$+$" is used, because it is a necessary consequence of the usual definition of $\cos^{-1}$ that $\sin(\cos^{-1}(x))\geq 0$ for all $x\in[-1,1]$.

It may also help to recall that $\cos^{-1}$ is a decreasing function, either by observing that $\cos^{-1}(-1)=\pi$ and $\cos^{-1}(1)=0$, or by looking at the graph. Correspondingly, its derivative is negative.

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