If \(y=\left(x+\sqrt{x^2+1}\right)^n\), then the value of \((x^2+1)\displaystyle\frac{d^2y}{dx^2}+x\frac{dy}{dx}\) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(n^2y\)
By the chain rule:
\( \displaystyle \frac{dy}{dx} = n\left(x+\sqrt{x^2+1}\right)^{n-1}\left(1 + \frac{x}{\sqrt{x^2+1}}\right) \)
\( \displaystyle = n\left(x+\sqrt{x^2+1}\right)^{n-1} \cdot \frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}} \)
\( \displaystyle = \frac{ny}{\sqrt{x^2+1}} \)
So
\( \displaystyle \sqrt{x^2+1}\,\frac{dy}{dx} = ny \)
Squaring removes the root:
\( \displaystyle (x^2+1)\left(\frac{dy}{dx}\right)^2 = n^2y^2 \)
Differentiating again with respect to x:
\( \displaystyle (x^2+1)\cdot 2\frac{dy}{dx}\cdot\frac{d^2y}{dx^2} + 2x\left(\frac{dy}{dx}\right)^2 = 2n^2y\frac{dy}{dx} \)
Dividing by \(2\frac{dy}{dx}\):
\( \displaystyle (x^2+1)\frac{d^2y}{dx^2} + x\frac{dy}{dx} = n^2y \)