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

$\frac{my}{\sqrt{x^2+1}}$

The correct answer is Option (4) → $\frac{my}{\sqrt{x^2+1}}$

Given:

$y = \left(x + \sqrt{x^2 + 1} \right)^m$

Step 1: Let

$u = x + \sqrt{x^2 + 1}$

Then, $y = u^m$

Step 2: Differentiate using the chain rule:

$\frac{dy}{dx} = m u^{m-1} \cdot \frac{du}{dx}$

Step 3: Compute $\frac{du}{dx}$

$\frac{du}{dx} = \frac{d}{dx}\left(x + \sqrt{x^2 + 1}\right) = 1 + \frac{1}{2\sqrt{x^2 + 1}} \cdot 2x = 1 + \frac{x}{\sqrt{x^2 + 1}}$

Step 4: Final Answer

$\frac{dy}{dx} = m \left(x + \sqrt{x^2 + 1} \right)^{m-1} \left(1 + \frac{x}{\sqrt{x^2 + 1}} \right)$