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

Nov 21, 2016

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

#### Explanation:

$f \left(t\right) = \left(t {\cos}^{2} t , {t}^{2} - \cos t\right)$
Let $x \left(t\right) = t {\cos}^{2} t$ and $y \left(t\right) = {t}^{2} - \cos t$

The derivative of x(t) is
$x ' \left(t\right) = {\cos}^{2} t + t \left(2 \cos t\right) \left(- \sin t\right)$

$x ' \left(t\right) = {\cos}^{2} t - t \sin 2 t$

The derivative of y(t) is
$y ' \left(t\right) = 2 t - \left(- \sin t\right)$
$y ' \left(t\right) = 2 t + \sin t$

Using chain rule $f ' \left(t\right) = y ' \left(t\right)$ x $\frac{1}{x ' \left(t\right)}$

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