How do you differentiate y=1/(t-1)^2?

Mar 5, 2017

$\frac{\mathrm{dy}}{\mathrm{dt}} = - \frac{2}{t - 1} ^ 3$

Explanation:

We have:

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

So then by the chain rule, we have:

$\frac{\mathrm{dy}}{\mathrm{dt}} = \left(- 2\right) {\left(t - 1\right)}^{- 3} \cdot \left(1\right)$
$\text{ } = - 2 {\left(t - 1\right)}^{- 3}$
$\text{ } = - \frac{2}{t - 1} ^ 3$