Differentiate the function $\log(x + \sqrt{x^2 + a})$ with respect to $x$.

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

The correct answer is Option (3) →$\frac{1}{\sqrt{x^2 + a}}$

Let $y = \log (x + \sqrt{x^2 + a})$

On differentiating w.r.t. $x$, we get

$\frac{dy}{dx} = \frac{d}{dx} \log (x + \sqrt{x^2 + a})$

$= \frac{1}{(x + \sqrt{x^2 + a})} \frac{d}{dx} [x + \sqrt{x^2 + a}] \quad \left[ ∵\frac{d}{dx} \log x = \frac{1}{x} \right]$

[applying function to function differentiation]

$= \frac{1}{(x + \sqrt{x^2 + a})} \left[ 1 + \frac{1}{2}(x^2 + a)^{-1/2} \cdot 2x \right]$

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

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