How do you use the definition of a derivative to find the derivative of #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))#