# What is the derivative of f(t) = (e^(t^2-1)-t, 2t^3-4t ) ?

Apr 8, 2017

($2 t {e}^{{t}^{2} - 1}$, $6 {t}^{2} - 4$)

#### Explanation:

To find the derivative of a parametric you just find the derivative of each component separately.

Generally for something in the form ${e}^{u}$ the derivative is $u ' {e}^{u}$
${e}^{{t}^{2} - 1}$ derivative is $2 t {e}^{{t}^{2} - 1}$

For $2 {t}^{3} - 4 t$ the derivative is in a log way
$3 \cdot 2 {t}^{3 - 1} - 4 {t}^{1 - 1}$
$6 {t}^{2} - 4$

All together it becomes:
($x ' , y '$)
($2 t {e}^{{t}^{2} - 1}$, $6 {t}^{2} - 4$)

Apr 9, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 {t}^{2} - 4}{2 t {e}^{{t}^{2} - 1} - 1}$

#### Explanation:

This is an equation given in parametric form, where

$x = {e}^{{t}^{2} - 1} - t$ and $y = 2 {t}^{3} - 4 t$

In such cases $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

Here $\frac{\mathrm{dx}}{\mathrm{dt}} = {e}^{{t}^{2} - 1} \times 2 t - 1 = 2 t {e}^{{t}^{2} - 1} - 1$

and $\frac{\mathrm{dy}}{\mathrm{dt}} = 6 {t}^{2} - 4$

Hence $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 {t}^{2} - 4}{2 t {e}^{{t}^{2} - 1} - 1}$