Question

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

Answer 1

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}}#