# How do you find the derivative of 2^xln2?

Jun 12, 2015

We are going to use the formula $\left(u \cdot v\right) ' = u ' v + u v '$

$u = {2}^{x}$
$u ' = {2}^{x} \ln 2$ (See below)
$v = \ln 2$
$v ' = 0$ (Note that $v$ is a constant.)

Thus $f ' \left(x\right) = \left({2}^{x} \ln 2\right) \left(\ln 2\right) + \left({2}^{x} \ln 2\right) \left(0\right)$

$f ' \left(x\right) = {2}^{x} {\left(\ln 2\right)}^{2}$

Note !:

If you have not memorized $\frac{d}{\mathrm{dx}} \left({a}^{x}\right) = {a}^{x} \ln a$, then you'll need to use:

${2}^{x} = {e}^{x \ln 2}$, whose derivative can be found by the chain rule:

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

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

Note2:

Because $\ln 2$ is a constant, we don't really need the product rule.

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