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

##### 1 Answer
Feb 18, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 {t}^{2}}{\ln} t \left(4 + 2 {t}^{2} {e}^{t} \left(t + 3\right)\right)$

#### Explanation:

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

$x = {\left(\ln t\right)}^{2}$

$\frac{\mathrm{dx}}{\mathrm{dt}} = 2 \left(\ln t\right) \left(\frac{1}{t}\right) = \frac{2}{t} \ln t$

$y = - 2 {t}^{2} - 2 {t}^{3} {e}^{t}$

d/dt(e^tf(t)=e^t(f(t)+f'(t))

$f \left(t\right) = {t}^{3} , f ' \left(t\right) = 3 {t}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dt}} = - 2 \times 2 t - 2 \times \left({e}^{t} \left({t}^{3} + 3 {t}^{2}\right)\right) = - 4 t - 2 {e}^{t} \left({t}^{3} + 3 {t}^{2}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 4 t - 2 {e}^{t} \left({t}^{3} + 3 {t}^{2}\right)}{\frac{2}{t} \ln t}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \left(t \frac{4 + 2 {t}^{2} {e}^{t} \left(t + 3\right)}{\frac{2}{t} \ln t}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 {t}^{2}}{\ln} t \left(4 + 2 {t}^{2} {e}^{t} \left(t + 3\right)\right)$