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

Nov 13, 2016

$f ' \left(t\right) = \frac{2 {t}^{2} \cos t - {t}^{3} \sin t}{2 \ln t - t}$

#### Explanation:

Define $x \left(t\right)$ and $y \left(t\right)$ such that

$\left\{\begin{matrix}x \left(t\right) = {\ln}^{2} t - t \\ y \left(t\right) = {t}^{2} \cos t\end{matrix}\right.$

Using the chain rule and product rule we can differentiate $x$ and $y$ wrt $t$ as follows:

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

$\therefore \frac{\mathrm{dx}}{\mathrm{dt}} = \frac{2}{t} \ln t - 1$

And,

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

$\therefore \frac{\mathrm{dy}}{\mathrm{dt}} = 2 t \cos t - {t}^{2} \sin t$

And Also, By the chain rule:

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

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

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 t \cos t - {t}^{2} \sin t}{\frac{1}{t} \left(2 \ln t - t\right)}$

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