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

##### 1 Answer
Mar 11, 2018

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{4 t - 1}{2 - 3 {e}^{t} - 3 t {e}^{t}}$

#### Explanation:

In a parametric equation $f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$, the derivative $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dy}}{\mathrm{dt}}}$

Here we have $f \left(t\right) = \left(2 t - 3 t {e}^{t} , 2 {t}^{2} - t\right)$

As $x \left(t\right) = 2 t - 3 t {e}^{t}$, $\frac{\mathrm{dx}}{\mathrm{dt}} = 2 - 3 {e}^{t} - 3 t {e}^{t}$

and as $y \left(t\right) = 2 {t}^{2} - t$, $\frac{\mathrm{dy}}{\mathrm{dt}} = 4 t - 1$

and hence $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{4 t - 1}{2 - 3 {e}^{t} - 3 t {e}^{t}}$