# What is the derivative of f(t) = (e^(t^2-1)-t, 2t^2-4t ) ?

Jul 31, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 t - 4}{2 t {e}^{{t}^{2} - 1} - 1}$

#### Explanation:

Derivative of a parametric function $f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$ is given by $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$
As $y \left(t\right) = 2 {t}^{2} - 4 t$, $\frac{\mathrm{dy}}{\mathrm{dt}} = 4 t - 4$

and as $x \left(t\right) = {e}^{{t}^{2} - 1} - t$,, $\frac{\mathrm{dx}}{\mathrm{dt}} = {e}^{{t}^{2} - 1} \times 2 t - 1 = 2 t {e}^{{t}^{2} - 1} - 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 t - 4}{2 t {e}^{{t}^{2} - 1} - 1}$