Question

If $y=x^{2x}$ then find $\frac{dy}{dx}$.

Answer 1

$y=x^{2x}$

$\log y=2x\log x$

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

$\Rightarrow \frac{dy}{dx}=2y[x\frac{d}{dx}\left(\log x\right)+\log x\frac{d}{dx}\left(x\right)]$

$=2y[x\times \frac{1}{x}+\log x]$

$=2y(1+\log x)$

$\therefore \frac{dy}{dx}=2x^{2x}(1+\log x)$ where $y=x^{2x}$