# How do you differentiate the following parametric equation:  x(t)=t-e^(t^2-t+1)/t, y(t)= t^2-e^(t-t^2)?

Nov 24, 2017

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

#### Explanation:

When a functiion is given in parametric form such as $f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$, its dervative is given by $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

Here we have $y \left(t\right) = {t}^{2} - {e}^{t - {t}^{2}}$ hence $\frac{\mathrm{dy}}{\mathrm{dt}} = 2 t - {e}^{t - {t}^{2}} \left(1 - 2 t\right)$

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

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

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

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