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

Jul 8, 2016

Knowing the chain rule for derivatives, you can see that

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

and

$\frac{\mathrm{dy}}{\mathrm{dt}} = 4 t - 4$

Thus, $f ' \left(t\right) = \left(2 t {e}^{{t}^{2} - 1} - {e}^{t} , 4 t - 4\right)$

#### Explanation:

In this case, the "trick" is to apply the chain rule to the the first term of the expression

${e}^{{t}^{2} - 1} - {e}^{t}$

to obtain the derivative.

The chain rule formula used in this problem comes from

$\frac{d}{\mathrm{dx}} \left[{e}^{z}\right] = {e}^{z} \frac{\mathrm{dz}}{\mathrm{dx}}$

where

$z = {t}^{2} - 1$