How do I use the limit definition of derivative to find $f'(x)$ for $f(x)=sqrt(2+6x)$ ?

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Answer 1 · 2014-10-03T04:01:33

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

$f(x)=sqrt(2+6x)$

$f(x+h)=sqrt(2+6(x+h))=sqrt(2+6x+6h$

Make the substitutions for $f(x)$ and $f(x+h)$

$f'(x)=lim_(h->0) (sqrt(2+6x+6h)-sqrt(2+6x))/h$

Rationalize the numerator

$=lim_(h->0) (sqrt(2+6x+6h)-sqrt(2+6x))/h*(sqrt(2+6x+6h)+sqrt(2+6x))/(sqrt(2+6x+6h)+sqrt(2+6x))$

Remember the difference of perfect squares for the numerator

$=lim_(h->0) ((2+6x+6h)-(2+6x))/(h*sqrt(2+6x+6h)+sqrt(2+6x))$

Distribute the negative

$=lim_(h->0) (2+6x+6h-2-6x)/(h*sqrt(2+6x+6h)+sqrt(2+6x))$

Simplify numerator

$=lim_(h->0) (6h)/(h*sqrt(2+6x+6h)+sqrt(2+6x))$

Cancel the factors of $h$

$=lim_(h->0) (6)/(sqrt(2+6x+6h)+sqrt(2+6x))$

Substitute in the value of 0 for $h$ and then simplify

$=(6)/(sqrt(2+6x+6(0))+sqrt(2+6x))$

$=(6)/(sqrt(2+6x)+sqrt(2+6x))$

$=(6)/(2sqrt(2+6x))$

$=3/sqrt(2+6x)$

How do I use the limit definition of derivative to find f'(x)f'(x) for f(x)=sqrt(2+6x)f(x)=sqrt(2+6x) ?