# How do you differentiate the following parametric equation:  x(t)=tlnt, y(t)= t^3cost-tsin^2t ?

Feb 28, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- {t}^{3} \sin t + 3 {t}^{2} \cos t - t \sin 2 t - {\sin}^{2} t}{1 + \ln t}$

#### Explanation:

We have two parametric equations:

$x = t \ln t$
$y = {t}^{3} \cos t - t {\sin}^{2} t$

We differentiate wrt $t$ (and apply the product rule):

$\frac{\mathrm{dx}}{\mathrm{dt}} = \left(t\right) \left(\frac{1}{t}\right) + \left(1\right) \left(\ln t\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus = 1 + \ln t$

And:

$\frac{\mathrm{dy}}{\mathrm{dt}} = {t}^{3} \cos t - t {\sin}^{2} t$
$\setminus \setminus \setminus \setminus \setminus \setminus = \left({t}^{3}\right) \left(- \sin t\right) + \left(3 {t}^{2}\right) \left(\cos t\right) - \left\{\left(t\right) \left(2 \sin t \cos t\right) + \left(1\right) \left({\sin}^{2} t\right)\right\}$
$\setminus \setminus \setminus \setminus \setminus \setminus = - {t}^{3} \sin t + 3 {t}^{2} \cos t - t \sin 2 t - {\sin}^{2} t$

And from the chain rule we have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dy}}{\mathrm{dx}}}$
$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{- {t}^{3} \sin t + 3 {t}^{2} \cos t - t \sin 2 t - {\sin}^{2} t}{1 + \ln t}$