The difference quotient is used to define the derivative as a limit.
Essentially, you find the slope of the secant line through a function at two points: #(x,f(x)), and (x+h, f(x+h))#. As #h->0#, we approach the slope of the line tangent to the function at only the point #(x, f(x))#
The formula for difference quotient is:
#lim_(h->0) = DQ = (f(x+h)-f(x))/(h)#
For this function, using #a# for our initial point...
#(1/sqrt(a+h+2)-1/sqrt(a+2))/h = DQ#
To give these factors in the numerator a common denominator, we will multiply the first factor, #1/sqrt(a+h+2)#, by #sqrt(a+2)/sqrt(a+2)#, and the second factor #1/sqrt(a+2)# by #sqrt(a+h+2)/sqrt(a+h+2)#
#DQ = ((1/sqrt(a+h+2))(sqrt(a+2)/sqrt(a+2)) - 1/sqrt(a+2) (sqrt(a+h+2)/sqrt(a+h+2)))/h #
Giving us...
# DQ= (sqrt(a+2)/(sqrt(a+h+2)sqrt(a+2))-sqrt(a+h+2)/(sqrt(a+h+2)sqrt(a+2)))/h#
We can now put this common denominator for these factors into the denominator proper:
#= (sqrt(a+2)-sqrt(a+h+2))/(h(sqrt(a+2))(sqrt(a+h+2)))#
Now to simplify the radicals, we multiply the numerator and denominator by the conjugate of the numerator. The conjugate of an expression #sqrt(m)-sqrt(n)# is simply #sqrt(m)+sqrt(n)#. Much like how the conjugate for #a-b# is #a+b#, multiplying the numerator by the conjugate will allow us to reach the equivalent of #a^2-b^2#, in our case #m-n#
Thus:
#(sqrt(a+2)+sqrt(a+h+2))/(sqrt(a+2)+sqrt(a+h+2)) * DQ#
#= ((a+2)-(a+h+2))/(h(sqrt(a+h+2))(sqrt(a+2))(sqrt(a+h+2) + sqrt(a+2)#
Simplifying the numerator...
#= -h/(h(sqrt(a+h+2))(sqrt(a+2))(sqrt(a+h+2) + sqrt(a+2)#
The h factors cancel...
#=-1/((sqrt(a+h+2))(sqrt(a+2))(sqrt(a+h+2) + sqrt(a+2)#
If we are doing this as #h->0..#
#lim_(x->0)DQ = -1/((sqrt(a+2))(sqrt(a+2))(2sqrt(a+2))) = -1/((a+2)2sqrt(a+2)) = -1/(2(a+2)^(3/2)#
Thus #f'(a) = -1/(2(a+2)^(3/2))#
