Question

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

• A. $n^2y$
• B. $ - n^2y$
• 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}{dx^{2}} + . 2y \left(\frac{dy}{dx}\right)^{2} = n^{2} . 2y \frac{dy}{dx} $
$\Rightarrow \left(1+x^{2}\right) \frac{d^{2}y}{dx^{2}} + x \frac{dy}{dx} = n^{2}y$