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

Dec 8, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{t + 1}$

#### Explanation:

To find the derivative of a parametric function, we use

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}} = \frac{y ' \left(t\right)}{x ' \left(t\right)}$

Applying that here, we use the product rule as well as that $\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$ to find

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

and

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

Thus

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