What is the derivative of e^xln2?

Mar 10, 2016

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

Explanation:

First, note that

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

Next, recall that $\ln 2$ is a constant, just like $2$ or $- 5$. This means it gets "carried along" with the differentiation and won't be modified. Thus,

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

Mar 10, 2016

If there is a typo in the question and the intended question was for ${e}^{x \ln 2}$

Explanation:

In this case, we are differentiating a function of the form ${e}^{u}$.

We'll use $\frac{d}{\mathrm{dx}} \left({e}^{u}\right) = {e}^{u} \frac{\mathrm{du}}{\mathrm{dx}}$.

Because $u = x \ln 2$ we need the same observation mason m made in the answer to the question as typed, that
$\ln 2$ is simply some constant, so $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(x \ln 2\right) = \ln 2$

We get

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