Answer
$$\frac{d}{dx}(\cos^{-1}x) = \frac{-1}{ \sqrt{1-x^2}} $$
Work Step by Step
From the given figure, we see that
$$\sin \theta = \sqrt{1-x^2} $$
and
$$\cos \theta= x$$
Then, solving for the angle gives:
$$\theta =\cos^{-1}x$$
Now, take the derivative of
$$\cos \theta= x$$
with respect to $x$
$$\dfrac{d \cos \theta}{dx}=\dfrac{dx}{dx}$$
We get:
\begin{align*}
-\sin \theta \frac{d \theta}{d x}&=1 \\
\frac{d \theta}{d x}&=-\frac{1}{\sin \theta}\ \ \\
&= \frac{-1}{ \sqrt{1-x^2}}\\
\end{align*}
Thus
\begin{align*}
\frac{d}{dx}(\cos^{-1}x)&= \frac{-1}{ \sqrt{1-x^2}}
\end{align*}
