The derivative of $x^{2x}$ w.r.t. $x$ is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $2x^{2x}(1 + \log x)$ ##
Let $y = x^{2x}$
$\log y = 2x \log x$
$\frac{d}{dx} \log y = \frac{d}{dx} (2x \log x)$
$\frac{1}{y} \frac{dy}{dx} = 2 \left[ x \cdot \frac{d}{dx} \log x + \log x \cdot \frac{d}{dx} x \right]$
$\frac{dy}{dx} = 2y \left[ x \cdot \frac{1}{x} + \log x \right]$
$\frac{dy}{dx}= 2x^{2x} [1 + \log x]$