How do you find the derivative of f(x)=ax^2+bx+cf(x)=ax2+bx+c?

1 Answer
Jan 21, 2017

f(x)=ax^2+bx+c => f'(x)=2ax+b

Explanation:

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

(y(x)+g(x)+z(x))'=y'(x)+g'(x)+z'(x)

In this case

f(x)=y(x)+g(x)+z(x)

where

y(x)=ax^2

g(x)=bx

and

z(x)=c

First remember the derivative of a constant is zero

Therefore z(x)=c => z'(x)=0

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

and by the power rule (x^n)'=nx^(n-1)

g(x)=bx=bx^1 => g'(x)=(bx^1)'=b(x^1)'=b(1x^(1-1))=bx^0=b(1)=b

and

y(x)=ax^2 => y'(x)=(ax^2)'=a(x^2)'=2ax^(2-1)=2ax^1=2ax

Then we plug in

f'(x)=y'(x)+g'(x)+z'(x)=2ax+b+0=ul(2ax+b)