# What is the derivative of ln e^(2x)?

Oct 3, 2016

$y = \ln \left(u\right)$

$\therefore \frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{u} = \frac{1}{{e}^{2 x}}$

$u = {e}^{2 x}$, $\therefore \frac{\mathrm{du}}{\mathrm{dx}} = 2 {e}^{2 x}$

Now, using the chain rule...

$\frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{{e}^{2 x}} \cdot 2 {e}^{2 x} = 2$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = 2$

This all worked because:

When $y = {e}^{f \left(x\right)}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right) {e}^{f \left(x\right)}$