# How do you differentiate the following parametric equation:  x(t)=e^t-t, y(t)=tant-sect ?

Jun 7, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{t} - 1}{{\sec}^{2} t - \sec t \tan t}$

#### Explanation:

When we have parametric equations, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

As $y = {e}^{t} - t$, $\frac{\mathrm{dy}}{\mathrm{dt}} = {e}^{t} - 1$

and as $x = \tan t - \sec t$, $\frac{\mathrm{dx}}{\mathrm{dt}} = {\sec}^{2} t - \sec t \tan t$

and $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{t} - 1}{{\sec}^{2} t - \sec t \tan t}$