How do you use the definition of a derivative to find f' given $f (x) = \sqrt{4 x + 3}$ $f(x)=sqrt(4x+3)$ at x>-3/4?

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Answer 1 · 2017-06-24T13:42:42

$f'=2/sqrt(4x+3)$

See below for derivation

The first principle defines the derivative of a function $f$ as such:
$f'=lim_(h->0)(f(x+h)-f(x))/(h)$

We let $f=sqrt(4x+3)$
Then
$f'=lim_(h->0)(sqrt(4(x+h)+3)-sqrt(4x+3))/(h)$

$=lim_(h->0)1/h*(sqrt(4x+4h+3)-sqrt(4x+3))$

$=lim_(h->0)1/h*(sqrt(4x+4h+3)-sqrt(4x+3))*(sqrt(4x+4h+3)+sqrt(4x+3))/(sqrt(4x+4h+3)+sqrt(4x+3))$

$=lim_(h->0)1/h*((sqrt(4x+4h+3))^2-(sqrt(4x+3))^2)/(sqrt(4x+4h+3)+sqrt(4x+3))$

$=lim_(h->0)1/h*(4x+4h+3-4x-3)/(sqrt(4x+4h+3)+sqrt(4x+3))$

$=lim_(h->0)1/h*(4h)/(sqrt(4x+4h+3)+sqrt(4x+3))$

$=lim_(h->0)4/(sqrt(4x+4h+3)+sqrt(4x+3))$

$=4/(sqrt(4x+0+3)+sqrt(4x+3))$

$=4/(2sqrt(4x+3))$

$=2/sqrt(4x+3)$

How do you use the definition of a derivative to find f' given f(x)=sqrt(4x+3)f(x)=√4x+3f(x)=sqrt(4x+3) at x>-3/4?