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How do you find f'(x) using the definition of a derivative $f (x) = x^{2} - 1$ $f(x) =x^2 - 1$ ?

1 Answer

Apr 24, 2018

$f'(x)=2x$

Explanation:

The definition of a derivative is:
$f'(x)=lim_(hto0)(f(x+h)-f(x))/h$

Here, $f(x)=x^2-1$

$f'(x)=lim_(hto0)((x+h)^2-1-x^2+1)/h$

$f'(x)=lim_(hto0)(x^2+2xh+h^2-x^2)/h$

$f'(x)=lim_(hto0)(2xh+h^2)/h$

$f'(x)=lim_(hto0)2x+h=2x+0=2x$

$f'(x)=2x$ for $f(x)=x^2-1$

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