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

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

The function is a composite function with $\left(x + \sqrt{x^2+1}\right)$ raised to power $m$.

Using the chain rule where if $y = [f(x)]^m$, then $\frac{dy}{dx} = m \cdot [f(x)]^{m-1} \cdot f'(x)$

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Finding the derivative of the inside part:

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

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

$= 1 + \frac{d}{dx}(x^2+1)^{1/2}$

For the square root part:

$\frac{d}{dx}(x^2+1)^{1/2} = \frac{1}{2}(x^2+1)^{-1/2} \cdot 2x$

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

Therefore:

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

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

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Applying the chain rule:

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

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

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

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

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