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

• 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, $y={(x+\sqrt{1+x^2})}^n$
$\frac{dy}{dx}=n{(x+\sqrt{1+x^2})}^{n-1}(1+\frac{1}{2}{(1+x^2)}^{-1/2}.2x)$
$\frac{dy}{dx}=n{(x+\sqrt{1+x^2})}^{n-1}\frac{(\sqrt{1+x^2}+x)}{\sqrt{1+x^2}}=\frac{n{(x+\sqrt{1+x^2})}^n}{\sqrt{1+x^2}}$

$\sqrt{1+x^2}\frac{dy}{dx}=ny$

or

$\sqrt{1+x^2}{y}_1=ny(y_1=\frac{dy}{dx})$

On squaring, we get

$(1+x^2)y_1^2=n^2{y}^2$

Differentiating $(1+x^2)y_2+x{y}_1=n^{2}y$