Differentiating Exponential Functions with Other Bases
Key Questions

Answer:
#f'(x)=ln5*5^x# Explanation:
Let
#y=f(x)=5^x# then
#lny=xln5# and differentiating we get#1/y(dy)/(dx)=ln5# or
#(dy)/(dx)=ln5*y=ln5*5^x# 
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)# 
This is the exponential function of base
#b# (where#b>0# should be assumed). It can be thought of as#b^x=e^(xln(b))# , so that, using the Chain Rule (See Chain Rule ) and the fact that#(e^x)'=e^x# (see Exponentials with Base e ) yields#(b^x)'=e^(xln(b))\times ln(b)=b^x\times ln(b)#
(see Exponential functions ).