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)#