Question

If $y={(x+\sqrt{1+x^2})}^n$, then $(1+x^2)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}$ is equal to

• A. $n^{2}y$
• B. $-n^{2}y$
• C. $-y$
• D. $2x^{2}y$

Answer 1

The correct option is A $n^{2}y$

Given, $y={(x+\sqrt{1+x^2})}^n$
On differentiating with respect to $x$, we get
$\frac{dy}{dx}=n{(x+\sqrt{1+x^2})}^{n-1}(1+\frac{x}{\sqrt{1+x^2}})$

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

$\Rightarrow \left(\sqrt{1+x^2}\right)\frac{dy}{dx}=n{(x+\sqrt{1+x^2})}^n$

Again differentiating with respect to $x$, we get
$\frac{d^{2}y}{d{x}^2}\cdot \sqrt{1+x^2}+\frac{dy}{dx}\left(\frac{x}{\sqrt{1+x^2}}\right)$ $=n^2{(x+\sqrt{1+x^2})}^{n-1}(1+\frac{x}{\sqrt{1+x^2}})$

$\Rightarrow (1+x^2)\frac{d^{2}y}{d{x}^2}+x\cdot \frac{dy}{dx}=n^2{(x+\sqrt{1+x^2})}^n$

$\Rightarrow (1+x^2)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}=n^{2}y$