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

Mar 1, 2015

The answer is: $y ' = 2 {e}^{2 x \ln x} \left(\ln x + 1\right)$.

Instead of remembering a complicate formula, we use these logarithmic properties:

$a = {e}^{\ln} a$ or, better: ${a}^{b} = {e}^{\ln} \left({a}^{b}\right) = {e}^{b \ln a}$.

So our function becomes:

$y = {e}^{2 x \ln x}$

and

$y ' = {e}^{2 x \ln x} \left(2 \cdot 1 \cdot \ln x + 2 x \cdot \frac{1}{x}\right) = 2 {e}^{2 x \ln x} \left(\ln x + 1\right)$.