Question

If $y=\{x+\sqrt{x^2+a^2}{\}}^n$, Show that $\frac{dy}{dx}=\frac{ny}{\sqrt{x^2+a^2}}$.

Answer 1

$y=\{x+\sqrt{x^2+a^2}{\}}^n$

$\frac{dy}{dx}=n\{x+\sqrt{x^2+a^2}{\}}^{n-1}\frac{d}{dx}(x+\sqrt{x^2+a^2})$

$=n\frac{\{x+\sqrt{x^2+a^2}{\}}^n}{x+\sqrt{x^2+a^2}}(1+\frac{1}{2\sqrt{x^2+a^2}})\frac{d}{dx}(x^2+a^2)$

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

$=\frac{ny}{x+\sqrt{x^2+a^2}}\left(\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+a^2}}\right)$

$=\frac{ny}{\sqrt{x^2+a^2}}$