Answer
$y'= x^{2x+1}(2ln(x) + \frac{2x+1}{x})$
Work Step by Step
$y=x^{2x+1}$
Taking log to the base $e$ of both sides gives,
$ln(y)=(2x+1)ln(x)$
Differentiating both sides gives,
$\frac{y'}{y}=(2x+1)'ln(x) + (ln(x))'(2x+1)$
Or, $\frac{y'}{y}=2ln(x) + \frac{2x+1}{x}$
Therefore, $y'=y(2ln(x) + \frac{2x+1}{x}) = x^{2x+1}(2ln(x) + \frac{2x+1}{x})$
