Differentiation — Maths STD 12 Science — Question

Differentiating w.r.t. x, we get

$\frac{d y}{d x}=\frac{d}{d x}\left(x^2+\sqrt{x^2+1}\right)^{\frac{1}{2}}$

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