Question

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

Answer 1

Integrate each part seperately, since they are in a different axis each.

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

Explanation:

1st part

`(t^2-sint)'=2t-cost`

2nd part

`(1/(t-1))'=((t-1)^-1)'=-1*(t-1)^(-1-1)*(t-1)'=`

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

Result

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

Answer 2

`-1/((2t-cost)(t-1)^2)`

Explanation:

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

`x'(t)=2t-cost`
`y'(t)=-1/(t-1)^2`

To find the derivative of a parametric function, find

`dy/dx=(dy/dt)/(dx/dt)=(y'(t))/(x'(t))=(-1/(t-1)^2)/(2t-cost)=-1/((2t-cost)(t-1)^2)`