What is the derivative of #ln e^(2x)#?

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Answer 1 · 2016-10-03T00:33:40

#y=ln(u)#

#:. dy/(du)=1/u=1/(e^(2x))#

#u=e^(2x)#, #:. (du)/(dx)=2e^(2x)#

Now, using the chain rule...

#(dy)/(du)*(du)/(dx)=1/(e^(2x))*2e^(2x)=2#

#:. (dy)/(dx)=2#

This all worked because:

When #y=e^(f(x))#

#(dy)/(dx)=f'(x)e^(f(x))#