# How do you find the derivative of f(x)=ax^2+bx+c?

Jan 21, 2017

$f \left(x\right) = a {x}^{2} + b x + c \implies f ' \left(x\right) = 2 a x + b$

#### Explanation:

Remember that the derivative of a sum is the sum of the derivatives.

$\left(y \left(x\right) + g \left(x\right) + z \left(x\right)\right) ' = y ' \left(x\right) + g ' \left(x\right) + z ' \left(x\right)$

In this case

$f \left(x\right) = y \left(x\right) + g \left(x\right) + z \left(x\right)$

where

$y \left(x\right) = a {x}^{2}$

$g \left(x\right) = b x$

and

$z \left(x\right) = c$

First remember the derivative of a constant is zero

Therefore $z \left(x\right) = c \implies z ' \left(x\right) = 0$

By the fact that (cf(x))'=cf'(x))

and by the power rule $\left({x}^{n}\right) ' = n {x}^{n - 1}$

$g \left(x\right) = b x = b {x}^{1} \implies g ' \left(x\right) = \left(b {x}^{1}\right) ' = b \left({x}^{1}\right) ' = b \left(1 {x}^{1 - 1}\right) = b {x}^{0} = b \left(1\right) = b$

and

$y \left(x\right) = a {x}^{2} \implies y ' \left(x\right) = \left(a {x}^{2}\right) ' = a \left({x}^{2}\right) ' = 2 a {x}^{2 - 1} = 2 a {x}^{1} = 2 a x$

Then we plug in

$f ' \left(x\right) = y ' \left(x\right) + g ' \left(x\right) + z ' \left(x\right) = 2 a x + b + 0 = \underline{2 a x + b}$