# How do you find the derivative of f(t)=-2t^2+3t-6?

Jan 17, 2017

$f \left(x\right) = - 2 {t}^{2} + 3 t - 6 \implies f ' \left(t\right) = - 4 t + 3$

#### Explanation:

$f \left(x\right) = - 2 {t}^{2} + 3 t - 6$ is a polynomial

So, we must use the fact that the derivative of sums equals the sum of derivatives.

$f \left(x\right) = {\sum}_{k = 0}^{n} {a}_{k} {x}^{k} \implies f ' \left(x\right) = \left({\sum}_{k = 0}^{n} {a}_{k} {x}^{k}\right) ' = {\sum}_{k = 0}^{n} \left({a}_{k} {x}^{k}\right) '$

in this case

$f \left(x\right) = - 2 {t}^{2} + 3 t - 6 \implies f ' \left(x\right) = \left(- 2 {t}^{2}\right) ' + \left(3 t\right) ' - \left(6\right) '$

Since 6 is a constant its derivative is zero. This is because the derivative of any constant is zero.

So,

$f ' \left(x\right) = \left(- 2 {t}^{2}\right) ' + \left(3 t\right) '$

Next we can use the Power rule $\left({x}^{n}\right) ' = n {x}^{n - 1}$
and the fact that a derivative times a constant equals constant times derivative $\left(c f \left(x\right)\right) ' = c f ' \left(x\right)$

In this case $n = 2$ and $c = - 2$

So

$\left(- 2 {t}^{2}\right) ' = - 2 \left({t}^{2}\right) ' = - 2 \left(2\right) \left({t}^{2 - 1}\right) = - 4 {t}^{1} = - 4 t$

and

$\left(3 t\right) ' = 3 \left({t}^{1}\right) ' = 3 \left({t}^{1 - 1}\right) = 3 {t}^{0} = 3 \left(1\right) = 3$

Then we put it together and get

$f ' \left(x\right) = - 4 t + 3$