# How do you differentiate y=2^xe^x using the product rule?

Dec 24, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x} \left({2}^{x - 1} + {2}^{x}\right)$

#### Explanation:

If $y = u v$

Then $\frac{\mathrm{dy}}{\mathrm{dx}} = u v ' + v u '$

Another way to write $\frac{\mathrm{dy}}{\mathrm{dx}}$ is using the prime symbol - an apostrophe - so that will be used instead. It looks like this:

$\frac{\mathrm{dy}}{\mathrm{dx}} = y '$

$y = {2}^{x} {e}^{x}$
$u = {2}^{x} , v = {e}^{x}$

$u ' = x {2}^{x - 1}$
$v ' = {e}^{x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = v u ' + u v ' = {e}^{x} {2}^{x - 1} + {2}^{x} {e}^{x} = {e}^{x} \left({2}^{x - 1} + {2}^{x}\right)$

Dec 26, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(1 + \ln 2\right) {2}^{x} {e}^{x}$

#### Explanation:

As $y = {2}^{x} {e}^{x}$

$\ln y = x \ln 2 + x$

Hence $\frac{1}{y} \times \frac{\mathrm{dy}}{\mathrm{dx}} = \ln 2 + 1$

and $\frac{\mathrm{dy}}{\mathrm{dx}} = \left(1 + \ln 2\right) y = \left(1 + \ln 2\right) {2}^{x} {e}^{x}$

Using Product Rule - Using the product rule let us assume $u = {2}^{x}$ and $v = {e}^{x}$

Hence, while $\frac{\mathrm{dv}}{\mathrm{dx}} = {e}^{x}$, we have $\ln u = x \ln 2$ and hence

$\frac{1}{u} \frac{\mathrm{du}}{\mathrm{dx}} = \ln 2$ i.e. $\frac{\mathrm{du}}{\mathrm{dx}} = {2}^{x} \ln 2$

As $y = u v$, we have $\frac{\mathrm{dy}}{\mathrm{dx}} = u \frac{\mathrm{dv}}{\mathrm{dx}} + v \frac{\mathrm{du}}{\mathrm{dx}}$

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

= $\left(1 + \ln 2\right) {2}^{x} {e}^{x}$