Question

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

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

Answer 1

Answer: (A)

Since, $y=(x+\sqrt{1+x^2}{)}^n$
On differentiating $w.r.t.x$, we get
$\frac{dy}{dx}=n{\left(x+\sqrt{1+x^2}\right)}^{n-1}\cdot \left(1+\frac{2x}{2\sqrt{1+x^2}}\right)$
$\Rightarrow \frac{dy}{dx}=\frac{n{\left(x+\sqrt{1+x^2}\right)}^n}{\sqrt{1+x^2}}$
$\Rightarrow \left(1+x^2\right){\left(\frac{dy}{dx}\right)}^2=n^2{y}^2$
Again differentiating w.r.t. $x$, we get
$\left(1+x^2\right)\cdot 2\frac{dy}{dx}\cdot \frac{d^{2}y}{d{x}^2}+2x{\left(\frac{dy}{dx}\right)}^2$
$=n^{2}2y\frac{dy}{dx}$
$\Rightarrow \left(1+x^2\right)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}$
$=n^{2}y$