The derivative of $x^{2x}$ w.r.t. $x$ is:

$2x^{2x}(1 + \log x)$

The correct answer is Option (3) → $2x^{2x}(1 + \log x)$ ##

Let $y = x^{2x}$

$\log y = 2x \log x$

$\frac{d}{dx} \log y = \frac{d}{dx} (2x \log x)$

$\frac{1}{y} \frac{dy}{dx} = 2 \left[ x \cdot \frac{d}{dx} \log x + \log x \cdot \frac{d}{dx} x \right]$

$\frac{dy}{dx} = 2y \left[ x \cdot \frac{1}{x} + \log x \right]$

$\frac{dy}{dx}= 2x^{2x} [1 + \log x]$