How do you find f'(x) using the definition of a derivative for $f (x) = \sqrt{2 x}$ $f(x)=sqrt(2x)$ ?

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Answer 1 · 2015-10-06T16:48:42

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

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

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

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

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

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

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

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

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

$f'(x)=2/(2sqrt(2x))$

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

How do you find f'(x) using the definition of a derivative for f(x)=sqrt(2x)f(x)=√2xf(x)=sqrt(2x)?