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

Jan 24, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{t} ^ 2 \left(\frac{t - 1}{t + 1}\right) , t \ne 0 , - 1$

#### Explanation:

$y \left(t\right) = {e}^{t} / t$

Diff.ing w.r.t. $t$, using the Quotient Rule, we have,

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

Similarly, we have, $\frac{\mathrm{dx}}{\mathrm{dt}} = t {e}^{t} + {e}^{t} = {e}^{t} \left(t + 1\right)$.

Finally, by the Rule of Parametric Diffn., we have,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}} = \frac{1}{t} ^ 2 \left(\frac{t - 1}{t + 1}\right) , t \ne 0 , - 1$