Question

A function $f\left(x\right)$ is defined for $x>0$ and satisfies $f\left(x^2\right)=x^3$ for all $x>0$. Then the value of $f^{\prime }\left(4\right)$ is

Answer 1

Note that it is not given that $f$ is a differentiable function We have
$f^{\prime }\left(4\right)=\underset{h\to 0}{lim}\frac{f(4+h)-f\left(4\right)}{h}$
$=\underset{h\to 0}{lim}\frac{f\left({\left(\sqrt{4+h}\right)}^2\right)-f\left(2^2\right)}{h}$
$=\underset{h\to 0}{lim}\frac{{(4+h)}^{3/2}-8}{h}=\underset{h\to 0}{lim}\frac{8[{(1+\frac{h}{4})}^{3/2}-1]}{h}$
$=\underset{h\to 0}{lim}\frac{8[1+\frac{3}{2}\frac{h}{4}+0\left(h^2\right)-1]}{h}=\underset{h\to 0}{lim}\frac{8[\frac{3}{8}h-0\left(h^2\right)]}{h}$
$=\underset{h\to 0}{lim}[3+0\left(h\right)]=3$