How do you use the definition of a derivative to find the derivative of $f (x) = \sqrt{x} + 2$ $f(x) = sqrtx + 2$ to calculate f'(2)?

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Answer 1 · 2017-02-05T15:57:11

${[\frac{d}{d x} (\sqrt{x} + 2)]}_{x = 2} = \frac{1}{2 \sqrt{2}}$ $[d/dx (sqrt(x) + 2)]_(x=2) = 1/(2sqrt2)$

By definition of derivative:

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

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

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

simplify:

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

rationalize the numerator using the identity $(a+b)(a-b) = (a^2-b^2)$ :

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

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

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

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

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

How do you use the definition of a derivative to find the derivative of f(x) = sqrtx + 2f(x)=√x+2f(x) = sqrtx + 2 to calculate f'(2)?