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

Nov 24, 2017

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

#### Explanation:

When a functiion is given in parametric form such as $f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$, its dervative is given by $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

Here we have $y \left(t\right) = {t}^{2} - t \sec t$ hence $\frac{\mathrm{dy}}{\mathrm{dt}} = 2 t - \sec t - t \sec t \tan t$

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

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

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