Question #99bbe

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Answer 1 · 2017-01-26T02:17:19

$d/(dx) sqrt(x+4) = 1/( 2sqrt(x+4))$

Using the definition of derivative:

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

we have:

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

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

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

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

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

$d/(dx) sqrt(x+4) = lim_(h->0) 1/( (sqrt(x+4+h)+sqrt(x+4))) = 1/( 2sqrt(x+4))$