How do you use the definition of a derivative to find the derivative of $g (x) = \sqrt{9 - x}$ $g(x) = sqrt(9 − x)$ ?

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Answer 1 · 2016-07-12T03:08:31

$g'(x) = 1/(2sqrt(9-x))$

From first principles, $g'(x) = Lim_"h->0" (g(x+h) - g(x))/h$

In this example $g(x) = sqrt(9-x)$

Therefore $g'(x) = Lim_"h->0" (sqrt(9-(x+h)) -sqrt(9-x))/h$

Multiply top and bottom by $(sqrt(9-(x+h)) +sqrt(9-x)) ->$

$g'(x) = Lim_"h->0" (9-(x+h) -(9-x))/ (h (sqrt(9-(x+h)) +sqrt(9-x))$

$=Lim_"h->0" h/ (h (sqrt(9-(x+h)) +sqrt(9-x))$

$=Lim_"h->0" cancel h/ (cancel h (sqrt(9-(x+h)) +sqrt(9-x))$

$= 1/ ( (sqrt(9-(x+0)) +sqrt(9-x))$

$=1/(2sqrt(9-x))$

How do you use the definition of a derivative to find the derivative of g(x) = sqrt(9 − x)g(x)=√9−xg(x) = sqrt(9 − x)?