What is the derivative of f(t) = (t^2-sint , 1/(t-1) ) ?

Jan 14, 2016

Integrate each part seperately, since they are in a different axis each.

$f ' \left(t\right) = \left(2 t - \cos t , - \frac{1}{t - 1} ^ 2\right)$

Explanation:

1st part

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

2nd part

$\left(\frac{1}{t - 1}\right) ' = \left({\left(t - 1\right)}^{-} 1\right) ' = - 1 \cdot {\left(t - 1\right)}^{- 1 - 1} \cdot \left(t - 1\right) ' =$

$= - {\left(t - 1\right)}^{- 2} \cdot 1 = - \frac{1}{t - 1} ^ 2$

Result

$f ' \left(t\right) = \left(2 t - \cos t , - \frac{1}{t - 1} ^ 2\right)$

Jan 14, 2016

$- \frac{1}{\left(2 t - \cos t\right) {\left(t - 1\right)}^{2}}$

Explanation:

$x \left(t\right) = {t}^{2} - \sin t$
$y \left(t\right) = \frac{1}{t - 1}$

$x ' \left(t\right) = 2 t - \cos t$
$y ' \left(t\right) = - \frac{1}{t - 1} ^ 2$

To find the derivative of a parametric function, find

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