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

Mar 24, 2018

$\frac{\mathrm{dx}}{\mathrm{dt}} = t {e}^{t} \cos t + t {e}^{t} \sin t + {e}^{t} \sin t$

$\frac{\mathrm{dy}}{\mathrm{dt}} = - t \sin t + \cos t - 2 \sin t \cos t$

#### Explanation:

We have parametric equations:

$x \left(t\right) = t {e}^{t} \sin t$
$y \left(t\right) = t \cos t - {\sin}^{2} t$

So, using the product rule and differentiating wrt $t$ we have:

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

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(t {e}^{t}\right) \left(\cos t\right) + \left(t\right) \left({e}^{t}\right) \left(\sin t\right) + \left(1\right) \left({e}^{t} \sin t\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = t {e}^{t} \cos t + t {e}^{t} \sin t + {e}^{t} \sin t$

And:

$y ' \left(t\right) = \left(t\right) \left(\frac{d}{\mathrm{dt}} \cos t\right) + \left(\frac{d}{\mathrm{dt}} t\right) \left(\cos t\right) - \frac{d}{\mathrm{dt}} {\sin}^{2} t$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(t\right) \left(- \sin t\right) + \left(1\right) \left(\cos t\right) - 2 \sin t \left(\cos t\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = - t \sin t + \cos t - 2 \sin t \cos t$

This technically is the solution, but more likely we seek the full derivative:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy} / \mathrm{dt}}{\mathrm{dx} / \mathrm{dt}}$
$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{\cos t - t \sin t - 2 \sin t \cos t}{t {e}^{t} \cos t + t {e}^{t} \sin t + {e}^{t} \sin t}$