Question

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

Answer 1

`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)`