Question:
Differentiate the following functions with respect to $x$ :
$\cos ^{-1}\left\{\sqrt{\frac{1+x}{2}}\right\},-1
Solution:
$y=\cos ^{-1}\left\{\sqrt{\frac{1+x}{2}}\right\}$
let $x=\cos 2 \theta$
Now
$y=\cos ^{-1}\left\{\sqrt{\frac{1+\cos 2 \theta}{2}}\right\}$
$y=\cos ^{-1}\left\{\sqrt{\frac{2 \cos ^{2} \theta}{2}}\right\}$
Using $\cos 2 \theta=2 \cos ^{2} \theta-1$
$y=\cos ^{-1}(\cos \theta)$
Considering the limits,
$-1 $-1<\cos 2 \theta<1$ $0<2 \theta<\pi$ $0<\theta<\frac{\pi}{2}$ Now, $y=\cos ^{-1}(\cos \theta)$ $y=\theta$ $y=\frac{1}{2} \cos ^{-1} x$ Differentiating w.r.t $\mathrm{x}$, we get $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}\left(-\frac{1}{\sqrt{1-\mathrm{x}^{2}}}\right)$