Use the formula #f'(x) = lim_(h->0) (f(x + h) - f(x))/h#.
#f'(x) =lim_(h->0) (1/(x + h + 2) - 1/(x +2))/h#
#f'(x) = lim_(h-> 0) ((x + 2)/((x + 2)(x + h + 2)) - (x+ h + 2)/((x + 2)(x + h + 2)))/h#
#f'(x) = lim_(h->0) ((x + 2 - x - h - 2)/((x + 2)(x + h + 2)))/h#
#f'(x) = lim_(h->0) (-h)/((x+ 2)(x + h + 2)) xx 1/h#
#f'(x) = lim_(h->0) -1/((x + 2)(x + h + 2))#
We can now substitute directly.
#f'(x) = -1/((x+ 2)(x + 0 + 2))#
#f'(x) = -1/(x + 2)^2#
Verification with the quotient rule yields the same result:
Letting #f(x) = (g(x))/(h(x))# with #g(x) = 1# and #h(x) = x + 2#, #f'(x)# is given by #f'(x) = (g'(x) xx h(x) - g(x) xx h'(x))/(h(x))^2#.
Calculating, we have:
#f'(x) = (0(x + 2) - 1(1))/(x + 2)^2#
#f'(x) = -1/(x + 2)^2#, which is the answer we obtained previously.
Hopefully this helps!
