How do you use the limit definition to find the derivative of $y = - 3 x^{2} + x + 4$ $y=-3x^2+x+4$ ?

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Answer 1 · 2017-01-05T18:11:47

-6x+1

Using limit definition derivative of y, that is f(x) would be

$f'(x)= Lim_(h->0) (f(x+h) -f(x))/h$

= $Lim_(h->0) (-3(x+h)^2 +x+h +4 - (-3x^2 +x+4))/h$

= $Lim_(h->0) (-3x^2 -6xh -3h^2 +x+h+4 +3x^2 -x-4)/h$

= $Lim_(h->0) (-6xh-3h^2 +h)/h$

= $Lim_(h->0) -6x-3h+1$

=-6x+1

How do you use the limit definition to find the derivative of y=-3x^2+x+4y=−3x2+x+4y=-3x^2+x+4?