Question

What is the derivative of `f(t) = (1/(3t+2) , t/(t^2-1) )`?

Answer 1

`f'(t)=(-3/(3t+2)^2, -(t^2+1)/(t^2-1)^2)`

Explanation:

Take the derivative component-by-component:

`f'(t)=(d/dt(1/(3t+2)),d/dt(t/(t^2-1)))`

`d/dt(1/(3t+2))=-3/(3t+2)^2` by the chain rule
`d/dt(t/(t^2-1))=((t^2-1)(1)-t(2t))/(t^2-1)^2`
`=(-t^2-1)/(t^2-1)^2=-(t^2+1)/(t^2-1)^2`

So,

`f'(t)=(-3/(3t+2)^2, -(t^2+1)/(t^2-1)^2)`