# If $$y= [ x + \sqrt{x^2+a^2}]^n.$$ then prove that : $\large\frac{dy}{dx}=\frac{ny}{\sqrt{x^2+a^2}}$

Toolbox:
• $\large\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}$
Step 1:
$y=[x+\sqrt{x^2+a^2}]^n$
Let $u=x+\sqrt{x^2+a^2}$
Differentiating w.r.t $x$ we get,
$\large\frac{du}{dx}$$=1+\large\frac{1}{2}$$(x^2+a^2)^{\Large\frac{-1}{2}}(2x)$
$\quad\;\;=1+\large\frac{x}{\sqrt{x^2+a^2}}$
Step 2:
$y=u^n$
Differentiating w.r.t $x$ we get,
$\large\frac{dy}{du}$$=nu^{n-1} \large\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx} \large\frac{dy}{dx}$$=nu^{n-1}\times (1+\large\frac{x}{\sqrt{x^2+a^2}})$
Substituting for $u$ we get,