# How do you differentiate the following parametric equation:  x(t)=lnt-t, y(t)= tcos^2t ?

Apr 4, 2018

Find the derivative of the $x$ and $y$ equations.

$x ' \left(t\right) = \frac{1}{t} - 1$
$y ' \left(t\right) = 1 \left({\cos}^{2} t\right) - t \sin t \cos t = {\cos}^{2} t - t \sin t \cos t$

Therefore

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{\cos}^{2} t - 2 t \sin t \cos t}{\frac{1 - t}{t}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{t {\cos}^{2} t - 2 {t}^{2} \sin t \cos t}{1 - t}$

Hopefully this helps!