# What is the derivative of 2^x?

Jul 28, 2015

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

#### Explanation:

In order to be able to calculate the derivative of ${2}^{x}$, you're going to need to use two things

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

• the chain rule

The idea here is that you can use the fact that you know what the derivative of ${e}^{x}$ is to try and determine what the derivative of another constant raised to the power of $x$, in this case equal to $2$, is.

To do that, you need to write $2$ as an exponential number that has the base equal to $e$.

Use the fact that

$\textcolor{b l u e}{{e}^{\ln \left(a\right)} = a}$

to write

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

This implies that ${2}^{x}$ will be equivalent to

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

Your derivative now looks like this

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

This is where the chain rule comes into play. You know that the derivative of a function $y = f \left(u\right)$ can be written as

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

In your case, $y = {e}^{x \cdot \ln 2}$, and $u = x \cdot \ln 2$, so that your derivative becomes

$\frac{d}{\mathrm{dx}} \left({e}^{u}\right) = {\underbrace{{e}^{u} / \left(\mathrm{du}\right)}}_{\textcolor{b l u e}{= {e}^{u}}} \cdot \frac{d}{\mathrm{dx}} \left(u\right)$

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

Now replace $u$ to calculate $\frac{d}{\mathrm{dx}} \left(u\right)$

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

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

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

Therefore,

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