If $y=\begin{Bmatrix} log (x+\sqrt{x^2+1})\end{Bmatrix}^2$ then $(1+x^2)\frac{d^2y}{dx^2}+x\frac{dy}{dx}=$
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 2
$y'=\frac{2\log(x+\sqrt{x^2+1})}{x+\sqrt{x^2+1}}\left(1+\frac{x}{\sqrt{x^2+1}}\right)$
so $y'=\frac{2\log(x+\sqrt{x^2+1})}{\sqrt{x^2+1}}$
so $y'\sqrt{x^2+1}=2\log(x+\sqrt{x^2+1})$
differentiating
$y''(\sqrt{x^2+1})+\frac{xy'}{\sqrt{x^2+1}}=\frac{2}{x+\sqrt{x^2+1}}\left(1+\frac{x}{\sqrt{x^2+1}}\right)$
$⇒(1+x^2)\frac{d^2y}{dx^2}+x\frac{dy}{dx}=2$