Question

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

Answer 1

`dy/dx=(t(2t+sint))/(2t^2-1)`

Explanation:

Let's call `x(t)=t^2-lnt`

then `x('t)=dx/dt=2t-1/t=(2t^2-1)/t`

and `y(t)=t^2-cos^2t`

Then `y'(t)=dy/dt=2t-2cost*-sint=2t+sin2t`

And finally
`dy/dx=(dy/dt)/(dx/dt)=(2t +sint)/((2t^2-1)/t)`

`dy/dx=(t(2t+sint))/(2t^2-1)`