Question

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

Answer 1

`f'(t) = (2t^2cost - t^3sint) / (2lnt - t )`

Explanation:

Define `x(t)` and `y(t)` such that

`{(x(t) = ln^2t-t), (y(t) = t^2cost) :}`

Using the chain rule and product rule we can differentiate `x` and `y` wrt `t` as follows:

`dx/dt = 2(lnt)1/t - 1`

`:. dx/dt = 2/tlnt - 1`

And,

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

`:. dy/dt = 2tcost - t^2sint`

And Also, By the chain rule:

`dy/dx = dy/dt * dt/dx => (dy/dt) / (dx/dt)`

`:. dy/dx = (2tcost - t^2sint) / (2/tlnt - 1 )`

`:. dy/dx = (2tcost - t^2sint) / (1/t(2lnt - t ))`

`:. dy/dx = (2t^2cost - t^3sint) / (2lnt - t )`