# How do you differentiate the following parametric equation:  x(t)=sqrt(t^2+tcos2t), y(t)=t^2sint ?

Jun 5, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(4 t \sin t + 2 {t}^{2} \cos t\right) \sqrt{{t}^{2} + t \cos 2 t}}{2 t - 2 t \sin 2 t + \cos 2 t}$

#### Explanation:

The derivative of a parametric equation is given by the following formula:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dt}} \div i \mathrm{de} \frac{\mathrm{dx}}{\mathrm{dt}}$

$y \left(t\right) = {t}^{2} \sin t$

$\frac{\mathrm{dy}}{\mathrm{dt}} = 2 t \sin t + {t}^{2} \cos t$

$x \left(t\right) = \sqrt{{t}^{2} + t \cos 2 t}$

$\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{2 t - 2 t \sin 2 t + \cos 2 t}{2 \sqrt{{t}^{2} + t \cos 2 t}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 t \sin t + {t}^{2} \cos t \div i \mathrm{de} \frac{2 t - 2 t \sin 2 t + \cos 2 t}{2 \sqrt{{t}^{2} + t \cos 2 t}}$

$= \frac{\left(4 t \sin t + 2 {t}^{2} \cos t\right) \sqrt{{t}^{2} + t \cos 2 t}}{2 t - 2 t \sin 2 t + \cos 2 t}$