#### Answer

The answers:
$y=f(x)=x^2-2x$.
$\frac{\text{dy}}{\text{dx}}=2 x-2$.

#### Work Step by Step

First way:
Using Eq. (8) we have
$\frac{\text{dy}}{\text{dx}}=\frac{y'(t)}{x'(t)}=\frac{\frac{t}{2}-1}{\frac{1}{2}}=t-2$.
Second way:
Since $x=(1/2) t$, so $t=2x$. Substituting $t$ into $y=(1/4) t^2-t$ gives
$y=(1/4) (2x)^2-2x=x^2-2x$,
$y=f(x)=x^2-2x$.
Differentiating $f(x)$ we get
$\frac{\text{dy}}{\text{dx}}=2 x-2$.
Since $t=2x$, this is the same as $t-2$.