# What is the derivative of f(t) = (t-lnt^2, t^2sint ) ?

Feb 22, 2018

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

#### Explanation:

Derivative $\frac{\mathrm{dy}}{\mathrm{dx}}$ of $f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$ is given by $\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

As $y \left(t\right) = {t}^{2} \sin t$, $\frac{\mathrm{dy}}{\mathrm{dt}} = 2 t \sin t + {t}^{2} \cos t$

and as $x \left(t\right) = t - \ln {t}^{2}$, $\frac{\mathrm{dx}}{\mathrm{dt}} = 1 - \frac{2 t}{t} ^ 2 = 1 - \frac{2}{t}$

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

= $\frac{2 {t}^{2} \sin t + {t}^{3} \cos t}{t - 2}$