Answer
$f'(x)=\frac{1}{8}t^2$
Work Step by Step
Given:
$f(2x)=x^2$
Taking derivative on the L.H.S.
$\frac{d}{dx}[f(x)]=x^2$
$f'(2x).2=x^2$
Let suppose, $t=2x$, then, we get
$f'(t)=\frac{1}{2}(\frac{1}{2}t)^2$
$=\frac{1}{2}(\frac{1}{4}t^2)$
$f'(t)=\frac{1}{8}t^2$
so,
$f'(x)=\frac{1}{8}t^2$
