How do you find the derivative of $f (x) = 2 x^{2} - x$ $f(x) = 2x^2-x$ using the limit definition?

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Answer 1 · 2016-07-12T20:08:45

$f'(x) = 4x-1$

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

$f'(x) =lim_(h->0) (2(x+h)^2 - x - h - 2x^2 + x)/h$

Expanding the bracket:

$f'(x) = lim_(h->0) (color(red)(2x^2)+4xh+2h^2color(blue)(-x)-hcolor(red)(-2x^2)+color(blue)(x))/h$

A few things cancel leaving:

$f'(x) = lim_(h->0)(4xh+2h^2-h)/h = lim_(h->0)(4x + 2h - 1) = 4x-1$

How do you find the derivative of f(x) = 2x^2-xf(x)=2x2−x f(x) = 2x^2-x using the limit definition?