# What is the derivative of f(t) = (tcos^2t , t^2-t /sect ) ?

Oct 17, 2017

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

#### Explanation:

To find derivative of a parmetric equation such as $f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$, we use formula $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

As $y \left(t\right) = t {\cos}^{2} t$, we have $\frac{\mathrm{dy}}{\mathrm{dt}} = {\cos}^{2} t + t \times 2 \cos t \times \left(- \sin t\right)$

= ${\cos}^{2} t - t \times 2 \sin t \cos t = {\cos}^{2} t - t \sin 2 t$

and as $x \left(t\right) = {t}^{2} - \frac{t}{\sec} t = {t}^{2} - t \cos t$

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

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