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How do you differentiate #y= 3^(2x+7)#?

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Mar 30, 2015

Use #d/(dx)(a^x)=a^x ln a# together with the chain rule:

#d/(dx)(f(u))=f'(u) (du)/(dx)#

For #y=3^(2x+7)# we get:

#(dy)/(dx) = 3^(2x+7) ln3 (2) =2(3^(2x+7) ) ln3#

Or use #2ln3=ln9# to write:

#(dy)/(dx) = 3^(2x+7) ln9 #

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