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

Oct 21, 2016

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

#### Explanation:

Let's call $x \left(t\right) = {t}^{2} - \ln t$

then $x \left(' t\right) = \frac{\mathrm{dx}}{\mathrm{dt}} = 2 t - \frac{1}{t} = \frac{2 {t}^{2} - 1}{t}$

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

Then $y ' \left(t\right) = \frac{\mathrm{dy}}{\mathrm{dt}} = 2 t - 2 \cos t \cdot - \sin t = 2 t + \sin 2 t$

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

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