Answer
Proof given below.
Work Step by Step
Let $y=\cos^{-1}x$
Then, $y\in[0,\pi]$ and $\cos y=x$
Differentiating (implicitly)
$ \displaystyle \frac{d}{dx}[\cos y]= \frac{d}{dx}[x]$
$-\displaystyle \sin y\cdot\frac{dy}{dx}=1$
$\displaystyle \frac{dy}{dx}=\frac{1}{-\sin y}$
Since y is from either the 1st or 2nd quadrant, where sine is positive,
we take the + sign when solving $\quad\sin^{2}y+\cos^{2}=1$
$\sin y=+\sqrt{1-\cos^{2}y}$
Because of the way we defined y, this equals $\sqrt{1-x^{2}}.$
Thus,
$\displaystyle \frac{dy}{dx}=\frac{1}{-\sqrt{1-x^{2}}}$
That is,
$ \displaystyle \frac{d}{dx}[\cos^{-1}x]=\frac{1}{-\sqrt{1-x^{2}}}$
