In the last lecture we developed a set of simple rules that help us to more quickly and easily compute the derivative
without having to do a limit calculation. Here are those rules.
The next two rules that we are going to develop have to do with derivatives of products and quotients. The bad news here is that these rules do not work in an obvious way.
Before we get started on the derivative rules for products and quotients, we are going to need this useful theorem:
Theorem (Differentiable Functions are Continuous) If f(x) is differentiable at some point x = a then f(x) is also continuous at x = a.
Proof As usual, we check the three conditions for continuity.
1. f(a) is defined at a. Because the function is differentiable at a, we know that the limit
must exist. If f(a) were not defined, there would be no chance of the limit existing. Since the limit exists, f(a) must be defined.
2. The limit as x approaches a of f(x) must exist. An essential pre-condition for the limit
to exist is that the numerator in the fraction must go to 0 as h goes to 0. Thus
Substituting x for a + h turns that limit into
or
3. The limit as x approaches a of f(x) must equal f(a). We just showed this in step 2.
Theorem (Product Rule) If f(x) and g(x) are differentiable functions, then their product f(x)g(x) is differentiable and
Proof Once again, we introduce an auxiliary function
u(x) = f(x) g(x)
and evaluate the limit
First we substitute in for u(x)
At this point we have to apply a clever trick to be able to proceed. We add and subtract a factor of f(x) g(x+h) in the numerator
Rearranging terms and splitting into two limits gives
The first limit on the left exists and equals g(x), because if a function is differentiable it is automatically continuous.
The next result we prove is the quotient rule, which shows how to compute the derivative of the quotient of two functions.
Theorem If f(x) and g(x) are differentiable functions, then f(x)/g(x) is differentiable where 
and
Proof Once again, we introduce an auxiliary function
u(x) = f(x) / g(x)
and evaluate the limit
Substituting for u(x) gives
The obvious next step is to combine the two fractions in the numerator into one.
Next, we introduce two new terms into the sum. The reason for this is not at all obvious at first, but as we will see this is the right thing to do.
Note that the first two terms in the numerator have a common factor, and the last two terms have a different common factor. Pulling these out gives
The thing to do now is to pull the limit of the fraction apart into two limits and then regroup further.
Because f(x) and g(x) are both differentiable, and hence continuous, we can do each of the limits above. Note that two of these limits look just like the definitions of the derivatives of f(x) and -g(x).
This is the result, although a little bit of additional algebraic manipulation will clean it up.
A special case of the quotient rule is the derivative of one over a function.
This result makes it possible to differentiate xn when n is a negative integer.
Section 3.2: 6, 7, 9, 10, 40, 41, 43