if f in sqrt(f(x)) is differentiable, would it equal to
f'(x)
------
2sqrt[f(x)] ?
Not quite right. You should have:
$\displaystyle
f(x) = \sqrt {x^3}
$
$\displaystyle
f'(x) = \frac {3x^2} {2 \sqrt{x^3}} = \frac 3 2 \sqrt x
$
An easier way is to realize that
$\displaystyle \sqrt {x^3} = x^{3/2}
$
So you get right to:
$\displaystyle
f'(x) = \frac 3 2 x^{1/2} = \frac 3 2 \sqrt x
$