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

Mar 29, 2018

Derivative is $- \frac{2 t + {e}^{t}}{3 {t}^{2} - 3 {e}^{3 t - 1}}$

#### Explanation:

The derivative of a function $f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$ is given by $\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

Here $x \left(t\right) = {t}^{3} - {e}^{3 t - 1}$ and hence

$\frac{\mathrm{dx}}{\mathrm{dt}} = 3 {t}^{2} - {e}^{3 t - 1} \cdot 3 = 3 {t}^{2} - 3 {e}^{3 t - 1}$

and as $y \left(t\right) = - {t}^{2} - {e}^{t}$, $\frac{\mathrm{dy}}{\mathrm{dt}} = - 2 t - {e}^{t}$

Hence derivative is $\frac{- 2 t - {e}^{t}}{3 {t}^{2} - 3 {e}^{3 t - 1}} = - \frac{2 t + {e}^{t}}{3 {t}^{2} - 3 {e}^{3 t - 1}}$