How do I find #f'(x)# for #f(x)=3^-x# ?
1 Answer
Sep 7, 2014
The answer is
First, step is a change of base:
#f(x)=3^(-x)#
#=e^(ln 3^(-x))#
#=e^(-xln3)#
With the proper base
#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)#