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

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Answer 1 · 2017-01-21T18:55:27

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

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

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

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