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

Apr 11, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 t - 1 + {e}^{t} \left(t + 1\right)}{3 {t}^{2} + {e}^{1 - t} \left(t - 1\right)}$

#### 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 $x \left(t\right) = {t}^{3} - t {e}^{1 - t}$ and $y \left(t\right) = {t}^{2} - t + t {e}^{t}$

therefore $\frac{\mathrm{dx}}{\mathrm{dt}} = 3 {t}^{2} - {e}^{1 - t} - t {e}^{1 - t} \cdot \left(- 1\right)$

= $3 {t}^{2} + {e}^{1 - t} \left(t - 1\right)$

and $\frac{\mathrm{dy}}{\mathrm{dt}} = 2 t - 1 + {e}^{t} + t {e}^{t} = 2 t - 1 + {e}^{t} \left(t + 1\right)$

Hence $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 t - 1 + {e}^{t} \left(t + 1\right)}{3 {t}^{2} + {e}^{1 - t} \left(t - 1\right)}$