Answer
\[\frac{dy}{dx}=\frac{-1}{2\sqrt{x-x^2}}\]
Work Step by Step
\[y=g(x)=arc\cos\sqrt x\]
\[\Rightarrow \cos y=\sqrt x\;\;\;...(1)\]
Differentiate (1) implicitly with respect to $x$
\[-\sin y\frac{dy}{dx}=\frac{1}{2\sqrt x}\]
\[\frac{dy}{dx}=\frac{-1}{2\sin y\sqrt x}\]
\[\frac{dy}{dx}=\frac{-1}{2\sqrt{1-\cos^2 y}\sqrt x}\]
Using (1)
\[\frac{dy}{dx}=\frac{-1}{2\sqrt{1-x}\sqrt x}\]
\[\frac{dy}{dx}=\frac{-1}{2\sqrt{x-x^2}}\]
Hence \[\frac{dy}{dx}=\frac{-1}{2\sqrt{x-x^2}}\]
