Question

What is the derivative of `f(t) = (e^(t^2-1)+3t, -t^3+t )`?

Answer 1

`dy/dx=(-3t^2+1)/(2te^(t^2-1)+3)`

Explanation:

`x(t)=e^(t^2-1)+3t`
`y(t)=-t^3+t`

`x'(t)=2te^(t^2-1)+3`
`y'(t)=-3t^2+1`

The derivative of a parametric function can be found through

`dy/dx=(dy/dt)/(dx/dt)=(y'(t))/(x'(t))=(-3t^2+1)/(2te^(t^2-1)+3)`