# What is the derivative of f(t) = (te^t , e^tcost ) ?

Sep 25, 2017

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

#### Explanation:

Derivative of a parametric equation of type $f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$

is given by $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

Here $\frac{\mathrm{dy}}{\mathrm{dt}} = {e}^{t} \cos t - {e}^{t} \sin t = {e}^{t} \left(\cos t - \sin t\right)$

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

Hence $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{t} \left(\cos t - \sin t\right)}{{e}^{t} \left(t + 1\right)} = \frac{\cos t - \sin t}{t + 1}$