Given: #f(x) = y = sqrt(x−3)#
Then:
#f(x+h) = sqrt(x+h−3)#
Using the limit definition:
#f'(x) = lim_(h to 0) (f(x+h)-f(x))/h#
Substitute in the functions:
#f'(x) = lim_(h to 0) (sqrt(x+h−3)-sqrt(x−3))/h#
We know that, if we multiply the numerator by #sqrt(x+h−3)+sqrt(x−3)#, we will eliminate the radicals but we must, also, multiply the denominator by the same thing:
#f'(x) = lim_(h to 0) (sqrt(x+h−3)+sqrt(x−3))/(sqrt(x+h−3)+sqrt(x−3))(sqrt(x+h−3)-sqrt(x−3))/h#
Please observe that we have not changed the value of anything, because we have multiplied by 1 in a very special form.
Now, the radicals in the numerator disappear, because of the property #(a+b)(a-b) = a^2-b^2#, and the denominators just multiply:
#f'(x) = lim_(h to 0) ((x+h−3)-(x−3))/(h(sqrt(x+h−3)+sqrt(x−3)))#
Combine like terms in the numerator:
#f'(x) = lim_(h to 0) h/(h(sqrt(x+h−3)+sqrt(x−3)))#
The h in the numerator cancels with the h in the denominator:
#f'(x) = lim_(h to 0) 1/((sqrt(x+h−3)+sqrt(x−3)))#
Now we can allow h to become 0:
#f'(x) = 1/((sqrt(x−3)+sqrt(x−3)))#
Combine like terms:
#f'(x) = 1/(2sqrt(x−3))#
