How do you find the derivative of s=cost/(t-1)?

May 30, 2017

$\frac{\mathrm{ds}}{\mathrm{dt}} = - \frac{t + \cos t - \sin t}{t - 1} ^ 2$

Explanation:

$s = \cos \frac{t}{t - 1}$

Use the quotient rule to differentiate $s$:

$\frac{d}{\mathrm{dx}} \left(\frac{p \left(x\right)}{q \left(x\right)}\right) = \frac{q \left(x\right) p ' \left(x\right) - p \left(x\right) q ' \left(x\right)}{{\left[q \left(x\right)\right]}^{2}}$

Let $p \left(x\right) = \cos t$ and $q \left(x\right) = t - 1$

Then $p ' \left(x\right) = - \sin t$ and $q ' \left(x\right) = 1$ and ${\left[q \left(x\right)\right]}^{2} = {\left(t - 1\right)}^{2}$

$\frac{\mathrm{ds}}{\mathrm{dt}} = \frac{- \left(t - 1\right) \sin t - \cos t}{t - 1} ^ 2 = - \frac{t \sin t + \cos t - \sin t}{t - 1} ^ 2$