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

Jul 12, 2016

$\frac{\mathrm{df}}{\mathrm{dt}} = \left(2 t {e}^{{t}^{2} - 1} - 1 , - 3 {t}^{2} - 4\right)$

#### 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:
$\frac{d}{\mathrm{dt}} \left({e}^{{t}^{2} - 1} - t\right) = 2 t {e}^{{t}^{2} - 1} - 1$
(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:
$\frac{\mathrm{df}}{\mathrm{dt}} = \left(2 t {e}^{{t}^{2} - 1} - 1 , - 3 {t}^{2} - 4\right)$