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

Jun 3, 2016

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

#### Explanation:

This is parametric form of equation. In parametric form of equation,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dt}} / \frac{\mathrm{dx}}{\mathrm{dt}}$

Now as $y = {t}^{2} + 1 - {e}^{2 {t}^{2} - 2}$

$\frac{\mathrm{dy}}{\mathrm{dt}} = 2 t - \left({e}^{2 {t}^{2} - 2} \times 4 t\right)$

= $2 t \left(1 - 2 {e}^{2 {t}^{2} - 2}\right)$

As $x = {t}^{2} - 1$

$\frac{\mathrm{dx}}{\mathrm{dt}} = 2 t$

Hence, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 t \left(1 - 2 {e}^{2 {t}^{2} - 2}\right)}{2 t}$

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