Question

What is the derivative of `f(t) = (tcos^2t , t^2-cost )`?

Answer 1

`f'(t) = (2t + sint)/ (cos^2t - tsin2t)`

Explanation:

`f(t) = ( tcos^2t , t^2 - cost)`
Let `x(t) = tcos^2t` and `y(t) = t^2 - cost`

The derivative of x(t) is
`x'(t) = cos^2t + t (2 cost)(- sint)`

`x'(t) = cos^2t - t sin2t`

The derivative of y(t) is
`y'(t) = 2t -(-sint)`
`y'(t) = 2t +sint`

Using chain rule `f'(t) = y'(t)` x `1/(x'(t) )`

`f'(t) =( 2t + sin t) / (cos^2t - t sin2t)`