Answer
$\frac{d}{dx}(cos^{-1}~x) = -\frac{1}{\sqrt{1-x^2}}$
Work Step by Step
$y = cos^{-1}~x$
$cos~y = x$ and $0 \leq y \leq \pi$
We can use implicit differentiation:
$-sin~y~\frac{dy}{dx} = 1$
$\frac{dy}{dx} = -\frac{1}{sin~y}$
Since $0 \leq y \leq \pi$, then $sin~y \geq 0$
$sin~y = \sqrt{1-cos^2~y} = \sqrt{1-x^2}$
Then:
$\frac{dy}{dx} = -\frac{1}{sin~y} = -\frac{1}{\sqrt{1-x^2}}$
Therefore:
$\frac{d}{dx}(cos^{-1}~x) = -\frac{1}{\sqrt{1-x^2}}$
