Differentiation — Maths STD 12 Science — Question

Differentiating w.r.t. x, we get

$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[\sin ^{-1}\left(\sqrt{\frac{1+x^2}{2}}\right)\right] \\ & =\frac{1}{\sqrt{1-\left(\sqrt{\frac{1+x^2}{2}}\right)^2}} \cdot \frac{d}{d x}\left(\sqrt{\frac{1+x^2}{2}}\right) \\ & =\frac{1}{\left.\sqrt{1-\left(\frac{1+x^2}{2}\right.}\right)} \times \frac{1}{\sqrt{2}} \frac{d}{d x}\left(\sqrt{1+x^2}\right) \\ & =\frac{\sqrt{2}}{\sqrt{2-1-x^2}} \times \frac{1}{\sqrt{2}} \times \frac{1}{2 \sqrt{1+x^2}} \cdot \frac{d}{d x}\left(1+x^2\right) \\ & =\frac{1}{\sqrt{1-x^2}} \times \frac{1}{2 \sqrt{1+x^2}} \cdot(0+2 x) \\ & =\frac{x}{\sqrt{\left(1-x^2\right)\left(1+x^2\right)}}=\frac{x}{\sqrt{1-x^4}} .\end{aligned}$