If a function f(x) is defined for all x > 0 and satisfies $f(x^2)=x^3$ for all x > 0, then f'(4) =

2 3 4 none of these

3

Using L'Hospital's rule $f'(4)=\lim\limits_{x \rightarrow 4} \frac{f(x)-f(4)}{x-4}=\lim\limits_{x \rightarrow 4} \frac{x^{3 / 2}-4^{3 / 2}}{x-4}=\frac{3}{2}(4)^{3 / 2-1}=3$