# How do you find the derivative of ln((x^2)(e^x))?

$\ln \left({x}^{2} \cdot {e}^{x}\right) = \ln \left({x}^{2}\right) + \ln \left({e}^{x}\right) = 2 \cdot \ln x + x$
$d \frac{2 \cdot \ln x + x}{\mathrm{dx}} = \frac{2}{x} + 1$
for values of $x > 0$