Question

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

Answer 1

`dy/dt = -2/(t-1)^3`

Explanation:

We have:

`y = 1/(t-1)^2 = (t-1)^(-2)`

So then by the chain rule, we have:

`dy/dt = (-2)(t-1)^(-3)*(1)`
`" " = -2(t-1)^(-3)`
`" " = -2/(t-1)^3`