How do you find the derivative of # 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#