We are going to use the formula (u*v)'=u'v+uv'
u=2^x
u'=2^x ln2 (See below)
v=ln2
v'=0 (Note that v is a constant.)
Thus f'(x)=(2^x ln2)(ln2)+(2^x ln2)(0)
f'(x)=2^x (ln2)^2
Note !:
If you have not memorized d/dx(a^x) = a^x lna, then you'll need to use:
2^x = e^(xln2), whose derivative can be found by the chain rule:
d/dx(2^x) = d/dx(e^(xln2)) = e^(xln2) * d/dx(xln2)
= e^(xln2) ln2 = 2^x ln2
Note2:
Because ln2 is a constant, we don't really need the product rule.
d/dx(2^x ln2) =( ln2) d/dx(2^x) = (ln2) (2^x ln2) = 2^x (ln2)^2