How do you differentiate #f(x)=2^x#?

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Answer 1 · 2016-08-08T11:32:07

#f'(x) =2^xln(2)#

#f(x) = y = 2^x#

Take natural logs of both sides:

#ln(y) = ln(2^x) = xln(2)#

Implicitly differentiate both sides:

#1/y * (dy)/(dx) = ln(2)#

#(dy)/(dx) = yln(2)#

#y = 2^x implies (dy)/(dx) = 2^xln(2)#