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

May 12, 2017

To differentiate $x \left(t\right)$ we will use the chain rule:

$x ' \left(t\right) = 1 - {e}^{{t}^{2} - t + 1} \left(\frac{d}{\mathrm{dt}} \left({t}^{2} - t + 1\right)\right)$

$\textcolor{w h i t e}{x ' \left(t\right)} = 1 - {e}^{{t}^{2} - t + 1} \left(2 t - 1\right)$

And to differentiate $y \left(t\right)$ we will need the product and chain rules:

$y ' \left(t\right) = \left(\frac{d}{\mathrm{dt}} t\right) {e}^{t - {t}^{2}} + t \left(\frac{d}{\mathrm{dt}} {e}^{t - {t}^{2}}\right)$

$\textcolor{w h i t e}{y ' \left(t\right)} = {e}^{t - {t}^{2}} + t {e}^{t - {t}^{2}} \left(\frac{d}{\mathrm{dt}} \left(t - {t}^{2}\right)\right)$

$\textcolor{w h i t e}{y ' \left(t\right)} = {e}^{t - {t}^{2}} + t {e}^{t - {t}^{2}} \left(1 - 2 t\right)$

$\textcolor{w h i t e}{y ' \left(t\right)} = {e}^{t - {t}^{2}} \left(1 + t \left(1 - 2 t\right)\right)$

$\textcolor{w h i t e}{y ' \left(t\right)} = {e}^{t - {t}^{2}} \left(1 + t - 2 {t}^{2}\right)$