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

Oct 31, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {e}^{2 t} / {t}^{2}$

Explanation:

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

As $x \left(t\right) = t {e}^{- t}$, $\frac{\mathrm{dx}}{\mathrm{dt}} = {e}^{- t} + t \times \left(- {e}^{- t}\right) = {e}^{- t} \left(1 - t\right)$

and as $y = {e}^{t} / t$, $\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{t {e}^{t} - {e}^{t} \times 1}{t} ^ 2$

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

and $\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{t} / {t}^{2} \left(t - 1\right) \times \frac{1}{{e}^{- t} \left(1 - t\right)}$

= $- {e}^{2 t} / {t}^{2}$