Question

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

Answer 1

The answer is `f'(x)=-3^(-x)ln3`.

First, step is a change of base:

`f(x)=3^(-x)`
`=e^(ln 3^(-x))`
`=e^(-xln3)`

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

`f'(x)=e^(-xln3)(-ln3)`
`=3^(-x)(-ln3)`

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(x)=a^u`
`f'(x)=a^u ln a (du)/(dx)`