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

Jul 15, 2017

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

#### Explanation:

For a parametric function $f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$

derivative $\frac{\mathrm{dy}}{\mathrm{dx}}$ is defined as $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

Here $x \left(t\right) = \frac{t}{\sin} t$ hence $\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{\sin t - t \cos t}{\sin} ^ 2 t$

and as $y \left(t\right) = \cos \frac{t}{t} ^ 2$, $\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{- {t}^{2} \sin t - 2 t \cos t}{t} ^ 4$

Hence $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\frac{{t}^{2} \sin t + 2 t \cos t}{t} ^ 4}{\frac{\sin t - t \cos t}{\sin} ^ 2 t}$

= $\frac{{\sin}^{2} t \left({t}^{2} \sin t + 2 t \cos t\right)}{{t}^{4} \left(\sin t - t \cos t\right)}$

= $\frac{{\sin}^{2} t \left(t \sin t + 2 \cos t\right)}{{t}^{3} \left(\sin t - t \cos t\right)}$