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

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Answer 1 · 2014-09-07T07:50:54

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)#