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

Aug 6, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\cos t - t \sin t - 2 t \sin t \cos t - {\sin}^{2} t}{{e}^{t} \left(\sin t + \cos t\right)}$

#### Explanation:

We have:

$x = {e}^{t} \sin t$
$y = t \cos t - t {\sin}^{2} t$

Differentiating wrt $t$ we get:

$\frac{\mathrm{dx}}{\mathrm{dt}} = \left({e}^{t}\right) \left(\frac{d}{t} \sin t\right) + \left(\frac{d}{\mathrm{dt}} {e}^{t}\right) \left(\sin t\right)$
$\text{ } = \left({e}^{t}\right) \left(\cos t\right) + \left({e}^{t}\right) \left(\sin t\right)$
$\text{ } = {e}^{t} \left(\sin t + \cos t\right)$

 dy/dt = (t)(d/dtcost) + (d/dtt)(cost) - {(t)(d/dtsin^2t)+(d/dtt)(sin^2t)
 " " = (t)(-sint) + (1)(cost) - {(t)(2sintcost)+(1)(sin^2t)
$\text{ } = - t \sin t + \cos t - 2 t \sin t \cos t - {\sin}^{2} t$

By the chain rule we have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy} / \mathrm{dt}}{\mathrm{dx} / \mathrm{dt}}$
$\text{ } = \frac{\cos t - t \sin t - 2 t \sin t \cos t - {\sin}^{2} t}{{e}^{t} \left(\sin t + \cos t\right)}$