Question

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

Answer 1

`(ds)/dt=-(t+cost-sint)/(t-1)^2`

Explanation:

`s=cost/(t-1)`

Use the quotient rule to differentiate `s`:

`d/dx((p(x))/(q(x)))=(q(x)p'(x)-p(x)q'(x))/([q(x)]^2)`

Let `p(x)=cost` and `q(x)=t-1`

Then `p'(x)=-sint` and `q'(x)=1` and `[q(x)]^2=(t-1)^2`

`(ds)/dt=(-(t-1)sint-cost)/(t-1)^2=-(tsint+cost-sint)/(t-1)^2`