How do you differentiate #y=ax^2+bx+c#?

1 Answer
Jim G.
Oct 17, 2016

#dy/dx=2ax+b#

Explanation:

differentiate using the #color(blue)"power rule"#

#color(red)(bar(ul(|color(white)(a/a)color(black)(d/dx(ax^n)=nax^(n-1))color(white)(a/a)|)))#

and #color(red)(bar(ul(|color(white)(a/a)color(black)(d/dx("constant")=0)color(white)(a/a)|)))#

#y=ax^2+bx+clarr" c is a constant"#

#rArrdy/dx=2ax^(2-1)+bx^(1-1)+0#

#=2ax^1+bx^0+0=2ax+b#

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