How do you find the derivative of y=sqrt(x−3) using the limit definition?

1 Answer
Sep 18, 2017

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))