Comment on Activity 2

We differentiate both sides of `u^(-1)=t^s` with respect to `t`. On the left side we use the result of Power Rule for the `-1` power and the Chain Rule:

`d/(dt)u^(-1)=-u^(-2)(du)/(dt).`

On the right side we use the Power Rule for positive rational powers:

`d/(dt)t^s=st^(s-1).`

Equating the two sides, we find `-u^(-2)(du)/(dt)=st^(s-1)`or

`(du)/(dt)=-su^2t^(s-1).`

When we substitute `t^(-s)` for `u`, we get

`(du)/(dt)=-s(t^(-s))^2t^(s-1)=-st^(-2s+s-1)=-st^(-s-1).`

Now, since `r=-s`,

`(du)/(dt)=rt^(r-1).`