# How do you differentiate the following parametric equation:  x(t)=cos^2t, y(t)=sin2t ?

Nov 21, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 \cot 2 t$

#### Explanation:

$x \left(t\right) = {\cos}^{2} t$
differentiate wrt to t

$\frac{\mathrm{dx}}{\mathrm{dt}} = 2 \left(\cos t\right) \frac{d}{\mathrm{dt}} \left(\cos t\right)$
$\frac{\mathrm{dx}}{\mathrm{dt}} = 2 \left(\cos t\right) \left(- \sin t\right)$
$\frac{\mathrm{dx}}{\mathrm{dt}} = - \sin 2 t$

$y \left(t\right) = \sin 2 t$
Differentiate wrt to t
$\frac{\mathrm{dy}}{\mathrm{dt}} = 2 \cos 2 t$

Using chain rule $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dt}}$ x $\frac{\mathrm{dt}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dt}} = 2 \cos t 2 t$ x $\frac{1}{- \sin 2 t}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2}{\tan 2 t}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 \cot 2 t$