Question

If $y={[x+\sqrt{x^2+a^2}]}^n,then\frac{dy}{dx}$ is equal to

• A. $\frac{ny}{\sqrt{x^2+a^2}}$
• B. $-\frac{ny}{\sqrt{x^2+a^2}}$
• C. $\frac{nx}{\sqrt{x^2+a^2}}$
• D. $-\frac{nx}{\sqrt{x^2+a^2}}$

Answer 1

The correct option is A

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

Using chain rule, we have
$\frac{dy}{dx}=\frac{d}{dx}\left[{\{x+\sqrt{x^2+a^2}\}}^n\right]$
$=n{\{x+\left\{\sqrt{x^2+a^2}\right\}\}}^{n-1}.\frac{d}{dx}\{x+\sqrt{x^2+a^2}\}$
$=n{\{x+\sqrt{x^2+a^2}\}}^{n-1}\left\{\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+a^2}}\right\}$
$=\frac{n{\{x+\sqrt{x^2+a^2}\}}^n}{\sqrt{x^2+a^2}}=\frac{ny}{\sqrt{x^2+a^2}}$