# How do I find f'(x) for f(x)=3^-x ?

##### 1 Answer
Sep 7, 2014

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):

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