Answer
$$\frac{d}{dt} \left( \frac{xt-4}{t^2-x}\right)
=\frac{-xt^2+8t-x^2}{(t^2-x)^2}.$$
Work Step by Step
By the quotient rule, we have
$$\frac{d}{dt} \left( \frac{xt-4}{t^2-x}\right)=\frac{(t^2-x)(x )-(xt-4)(2t)}{(t^2-x)^2}\\
=\frac{-xt^2+8t-x^2}{(t^2-x)^2}.$$
