How do you find f'(2) using the limit definition given $f (x) = \sqrt{x} + 2$ $f(x) = sqrt x + 2$ ?

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Answer 1 · 2017-01-25T12:00:52

$f'(2) = 1/(2sqrt2)$

By the definition of the derivative we have:

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

For $f(x) = sqrt(x) +2$ , this is:

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

Rationalize the numerator multiplying by $(sqrt(2+h) + sqrt(2) )$ above and below the line:

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

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

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

$f'(2) = lim_(h->0) 1/((sqrt(2+h) + sqrt(2) )) = 1/(2sqrt(2))$

How do you find f'(2) using the limit definition given f(x) = sqrt x + 2f(x)=√x+2f(x) = sqrt x + 2?