How do you find f'(x) using the definition of a derivative $f(x) =1/(x-3)$ ?

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Answer 1 · 2018-03-20T03:23:15

$f'(x) = -(1)/(x-3)^2$

We are given:

$f(x) = 1/(x-3)$

The limit definition of the derivative is:

$f^(')(x) = lim_(\Deltax->0) (f(x+\Deltax)-f(x))/(\Deltax)$

$=lim_(\Deltax->0)(1/(x+\Deltax-3)-1/(x-3))/(\Deltax)$

$=lim_(\Deltax->0)((x-3)/((x+\Deltax-3)(x-3))-(x+\Deltax-3)/((x+\Deltax-3)(x-3)))/(\Deltax)$

$=lim_(\Deltax->0)((x-3-(x+\Deltax-3))/((x+\Deltax-3)(x-3)))/(\Deltax)$

$=lim_(\Deltax->0)(cancelxcancel(-3)cancel(-x)-\Deltaxcancel(+3))/(\Deltax(x+\Deltax-3)(x-3))$

$=lim_(\Deltax->0)-(cancel(\Deltax))/(cancel(\Deltax)(x+\Deltax-3)(x-3))$

$=lim_(\Deltax->0)-(1)/((x+\Deltax-3)(x-3))$

$=-(1)/((x+(0)-3)(x-3))$

$=-(1)/((x-3)(x-3))$

$=-(1)/(x-3)^2$

Hence:

$=> color(green)(f'(x) = -(1)/(x-3)^2)$

How do you find f'(x) using the definition of a derivative f(x) =1/(x-3)f(x) =1/(x-3)?