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

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Answer 1 · 2014-08-30T14:33:48

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

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

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

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