# Differentiating Exponential Functions with Other Bases

## Key Questions

$f ' \left(x\right) = \ln 5 \cdot {5}^{x}$

#### Explanation:

Let $y = f \left(x\right) = {5}^{x}$

then $\ln y = x \ln 5$ and differentiating we get

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \ln 5$

or $\frac{\mathrm{dy}}{\mathrm{dx}} = \ln 5 \cdot y = \ln 5 \cdot {5}^{x}$

• The answer is $f ' \left(x\right) = - {3}^{- x} \ln 3$.

First, step is a change of base:

$f \left(x\right) = {3}^{- x}$
$= {e}^{\ln {3}^{- x}}$
$= {e}^{- x \ln 3}$

With the proper base $e$, we can just use the chain rule:

$f ' \left(x\right) = {e}^{- x \ln 3} \left(- \ln 3\right)$
$= {3}^{- x} \left(- \ln 3\right)$

rearrange and you will get the same answer as the first line.

The other option is to use the general exponential differentiation rule (if you can remember it):

$f \left(x\right) = {a}^{u}$
$f ' \left(x\right) = {a}^{u} \ln a \frac{\mathrm{du}}{\mathrm{dx}}$

• This is the exponential function of base $b$ (where $b > 0$ should be assumed). It can be thought of as ${b}^{x} = {e}^{x \ln \left(b\right)}$, so that, using the Chain Rule (See Chain Rule ) and the fact that $\left({e}^{x}\right) ' = {e}^{x}$ (see Exponentials with Base e ) yields $\left({b}^{x}\right) ' = {e}^{x \ln \left(b\right)} \setminus \times \ln \left(b\right) = {b}^{x} \setminus \times \ln \left(b\right)$
(see Exponential functions ).