# How do you differentiate the following parametric equation:  x(t)=t-lnt, y(t)= cost-sint ?

Feb 7, 2016

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

#### Explanation:

Differentiate each function individually to obtain:

$\frac{\mathrm{dx}}{\mathrm{dt}}$ and $\frac{\mathrm{dy}}{\mathrm{dt}}$

Then we can say:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dt}} \cdot {\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{-} 1$

$\frac{\mathrm{dx}}{\mathrm{dt}} = 1 - \frac{1}{t} = \frac{t - 1}{t} \to {\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{-} 1 = \frac{t}{t - 1}$
$\frac{\mathrm{dy}}{\mathrm{dt}} = - \sin t - \cos t$

Thus$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dt}} \cdot {\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{-} 1 = \left(- \sin t - \cos t\right) \cdot \frac{t}{t - 1}$

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