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How do I find #f'(x)# for #f(x)=4^sqrt(x)# ?

1 Answer

Aug 30, 2014

By Chain Rule, we can find
#f'(x)={(ln4)4^{sqrt{x}}}/{2sqrt{x}}#.

Remember:
#(b^x)'=(lnb)b^x#

By Chain Rule,
#f'(x)=(ln4)4^{sqrt{x}}cdot (sqrt{x})'=(ln4)4^{sqrt{x}}cdot{1}/{2sqrt{x}} ={(ln4)4^{sqrt{x}}}/{2sqrt{x}}#

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