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

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

#### Explanation:

Start from the given $x \left(t\right) = \ln t - 2 t$ and $y \left(t\right) = {\cos}^{2} t$

to find $\frac{\mathrm{dy}}{\mathrm{dx}}$ the equivalent of $\frac{\mathrm{dy}}{\mathrm{dt}}$ and $\frac{\mathrm{dx}}{\mathrm{dt}}$ must be obtained then proceed to determine $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}$

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

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

Next, find $\frac{\mathrm{dx}}{\mathrm{dt}}$

$\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left(\ln t - 2 t\right) = \frac{1}{t} - 2 = \frac{1 - 2 t}{t}$

$\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{1 - 2 t}{t}$

Finally, divide $\frac{\mathrm{dy}}{\mathrm{dt}}$ by $\frac{\mathrm{dx}}{\mathrm{dt}}$

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

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

from Plane Trigonometry, $2 \cdot \sin t \cdot \cos t = \sin 2 t$

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

placing the negative sign (-) at the bottom so that the final answer becomes:

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

I hope the explanation helps ...