Question

What is the derivative of `f(t) = (e^(t^2-1)-t, -t^3-4t )`?

Answer 1

`(df)/dt = (2t e^(t^2 - 1) - 1, -3t^2 - 4)`

Explanation:

Since we are taking the derivative of a vector-valued function parameterized over a single variable, `t`, we need only to take the individual derivative of each component:
`(d/dt f(t) = d/dt(e^(t^2-1)-t, -t^3-4t) = (d/dt(e^(t^2-1)-t), d/dt(-t^3-4t))`

So ultimately this is simply two very simple derivatives:
`d/dt(e^(t^2-1)-t) = 2t e^(t^2 - 1) - 1`
(by applying the chain rule and standard polynomial differentiation)
`d/dt(-t^3-4t)) = -3t^2 - 4`
(by standard polynomial differentiation)

These are combined into the final answer:
`(df)/dt = (2t e^(t^2 - 1) - 1, -3t^2 - 4)`