Question

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

Answer 1

`f'(t) = (6t-1) / (1+lnt)`

Explanation:

Let `{ (x(t)=tlnt), (y(t)=3t^2-t) :}`

Then differentiating wrt `t` (using the product rule)

`{ (dx/dt=(t)(1/t)+(1)lnt,=1+lnt), (dy/dt,=6t-1) :}`

By the chain rule, we have
`dy/dx=dy/dt*dt/dx`, or `dy/dx=(dy/dt)/(dx/dt)`
`:. dy/dx = (6t-1) / (1+lnt)`