How do you find f'(x) using the definition of a derivative $f(x) =sqrt(x−3)$ ?

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Answer 1 · 2016-03-28T06:25:19

Just take advantage of the $a^2-b^2=(a-b)(a+b)$

Answer is:

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

$f(x)=sqrt(x-3)$

$f'(x)=lim_(h->0)(sqrt(x+h-3)-sqrt(x-3))/h=$

$=lim_(h->0)((sqrt(x+h-3)-sqrt(x-3))*(sqrt(x+h-3)+sqrt(x-3)))/(h(sqrt(x+h-3)+sqrt(x-3)))=$

$=lim_(h->0)(sqrt(x+h-3)^2-sqrt(x-3)^2)/(h(sqrt(x+h-3)+sqrt(x-3)))=$

$=lim_(h->0)(x+h-3-x-3)/(h(sqrt(x+h-3)+sqrt(x-3)))=$

$=lim_(h->0)h/(h(sqrt(x+h-3)+sqrt(x-3)))=$

$=lim_(h->0)cancel(h)/(cancel(h)(sqrt(x+h-3)+sqrt(x-3)))=$

$=lim_(h->0)1/((sqrt(x+h-3)+sqrt(x-3)))=$

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

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

How do you find f'(x) using the definition of a derivative f(x) =sqrt(x−3)f(x) =sqrt(x−3)?