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How do you find f'(x) using the definition of a derivative for $f(x)= 2x^2-x$ ?

1 Answer

Sep 27, 2015

$4x-1$
See the explanation.

Explanation:

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

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

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

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