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

Mar 29, 2015

Anna A. shows the "easy way" to answer this question, but it is often helpful to see other ways of answering a question.

What if you "messed up" and didn't simplify first?

No problem, you'll still get the correct answer.

We'll use $\frac{d}{\mathrm{dx}} \left(\ln \left(g \left(x\right)\right)\right) = \frac{1}{g} \left(x\right) g ' \left(x\right)$

And, because, in this problem, $g \left(x\right) = {e}^{2 x}$, we'll also use:
$\frac{d}{\mathrm{dx}} \left({e}^{2 x}\right) = 2 {e}^{2 x}$

So, here we go:

$\frac{d}{\mathrm{dx}} \left(\ln \left({e}^{2 x}\right)\right) = \frac{1}{e} ^ \left(2 x\right) \left[2 {e}^{2 x}\right] = \frac{2 {e}^{2 x}}{e} ^ \left(2 x\right)$,

$\frac{d}{\mathrm{dx}} \left(\ln \left({e}^{2 x}\right)\right) = \frac{2 {e}^{2 x}}{e} ^ \left(2 x\right) = 2$,