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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{1 + \setminus \sin 2 t}{\setminus \sin t + t \setminus \cos t}$

#### Explanation:

Given that

x(t)=t-\cos^2t\ \ &\ \ \ y(t)=t\sint

Now, differentiating w.r.t. $x$ as follows

$\frac{\mathrm{dy}}{\mathrm{dx}}$

$= \setminus \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

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

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

$= \setminus \frac{1 - 2 \setminus \cos t \left(- \setminus \sin t\right)}{t \setminus \cos t + \setminus \sin t}$

$= \setminus \frac{1 + \setminus \sin 2 t}{\setminus \sin t + t \setminus \cos t}$