Question

If $y=(x+\sqrt{1+x^2}{)}^n$ , 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. $2n^{2}y$

Answer 1

Answer: (A)

$y=\left(x+\sqrt{1+x^2}\right)$
$\therefore \frac{dy}{dx}$
$=n{\left(x+\sqrt{1+x^2}\right)}^{n-1}\left(1+\frac{1}{2\sqrt{1+x^2}}.2x\right)$
$=n{\left(x+\sqrt{1+x^2}\right)}^{n-1}\left(1+\frac{x}{\sqrt{1+x^2}}\right)$
$=\frac{n{\left(x+\sqrt{1+x^2}\right)}^{n-1}\left(x+\sqrt{1+x^2}\right)}{\sqrt{1+x^2}}=\frac{n{\left(x+\sqrt{1+x^2}\right)}^n}{\sqrt{1+x^2}}=\frac{ny}{\sqrt{1+x^2}}$
$\therefore \left(1+x^2\right){\left(\frac{dy}{dx}\right)}^2=n^2{y}^2$
$\Rightarrow {\left(1+x^2\right)}^2\frac{dy}{dx}.\frac{d^{2}y}{d{x}^2}+.2y{\left(\frac{dy}{dx}\right)}^2=n^2.2y\frac{dy}{dx}$
$\Rightarrow \left(1+x^2\right)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}=n^{2}y$