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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 t - {e}^{t - 1}}{{e}^{t} + t {e}^{t}}$
Derivative of parametric equation $\left(x \left(t\right) , y \left(t\right)\right)$ is given by $\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$
Here as $y = {t}^{2} - {e}^{t - 1}$, $\frac{\mathrm{dy}}{\mathrm{dt}} = 2 t - {e}^{t - 1}$ and as $x = t {e}^{t}$, $\frac{\mathrm{dx}}{\mathrm{dt}} = {e}^{t} + t {e}^{t}$.
Hence, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 t - {e}^{t - 1}}{{e}^{t} + t {e}^{t}}$