Calculus Of One Real Variable – By Pheng Kim Ving Chapter 3: Rules Of Differentiation – Section 3.2: Differentiation Of Products And Quotients 3.2 Differentiation Of Products And Quotients

 1. Differentiation Of Products Of Functions

Recall from algebra that the product  of functions f and g is the function denoted by fg and defined by ( fg)(x) =
f(x)g(x). The value of the product fg at x is the product of the values of f and g at x. We want to find the derivative of
fg. Let's consider an example. Let f(x) = 2x and g(x) = x3. So ( fg)(x) = (2x)(x3) = 2x4. We have f '(x) = 2, g'(x) =
3x2, and ( fg)'(x) = 8x3. Now, f '(x)g'(x) = (2)(3x2) = 6x2. Thus, ( fg)'(x) is not  equal to f '(x)g'(x).

We'll see in the theorem below that the derivative ( fg)' of fg is f 'g + fg', not  f 'g'! The derivative of the product is not
the product of the derivatives!

Theorem 1.1 – The Product Rule

 If f and g are differentiable, then fg is also differentiable, and:   ( fg)' = f 'g + fg'.

Proof
Let x be an arbitrary point where both f and g are differentiable. Then:  EOP

Recall from Section 2.4 Theorem 1.1 that if a function is differentiable at a point, then it's continuous there. Note also that The General Product Rule

The product rule can be extended to more than two factors. If f, g, and h are differentiable, then fgh is also differentiable,
and:

( fgh)' = f '( gh) + f( gh)' = f 'gh + f( g'h + gh' ) = f 'gh + fg'h + fgh'.

In general, if f1, f2, ..., fn are differentiable, then f1 f2 ...fn is also differentiable, and:

( f1 f2...fn)' = f1'f2...fn + f1 f2'...fn + ... + f1 f2...fn'.

Example 1.1

Find the derivative of y = (x2 + 1)(x3 – 2).

Solution
y' = 2x(x3 – 2) + (x2 + 1)(3x2) = 2x4 – 4x + 3x4 + 3x2 = 5x4 + 3x2 – 4x.
EOS

 2. Differentiation Of The Square Root Function

Corollary 2.1

 We have: for all x > 0.

Proof
Using the product rule we get: EOP

Example 2.1 Solution EOS

 3. Differentiation Of Reciprocals Of Functions Theorem 3.1 – The Reciprocal Rule Proof   EOP

At any point x where f(x) = 0, f is differentiable, but 1/f isn't, because 1/f isn't defined there.

## A Special Case

A special case is the derivative of 1/x. Since (d/dx) x = 1 we have: Example 3.1

Find: Solution EOS

 4. The Power Rule – The Derivative Of xn

In Section 3.1 Theorem 4.1 we see that (d/dx) xn = nxn–1 for all positive integer n. We now extend this formula to all
integers.

Corollary 4.1 – The Power Rule

 We have: for all integer n.

Proof EOP

We'll see in Section 6.3 Eq. [4.1] that (d/dx) xa = axa–1 for any real  number a. That's called the general power rule.

Example 4.1

Find the derivative of f(t) = 2t3 + 4t–3.

Solution 1
f '(t) = 6t2 – 12t–4.

EOS

In this solution we use the power rule on both terms 2t3 and 4t–3 of f(t). The reciprocal rule can also be utilized on 4t–3,
as done in Solution 2 below.

Solution 2 EOS

We see that the power rule is much simpler than the reciprocal rule.

 5. Differentiation Of Quotients Of Functions ## Theorem 5.1 – The Quotient Rule Proof EOP

Example 5.1

Evaluate: Solution EOS

Remark that we substitute the value x = 2 without first simplifying. It's simpler to do numerical calculations than to handle
algebraic symbols like the letter x.

# Problems & Solutions

1.  Differentiate the following functions. Solution  2.  Evaluate: Solution Note

We evaluate the derivative immediately after it's calculated, before any simplification takes place. That's because it's
easier to simplify an expression with numbers than with algebraic symbols. 3.  Find the tangent and normal lines to the curve y = (x + 1)/(x – 1) at x = 2.

Solution  4.  Find all points on the curve y = x + (1/x) where the tangent line is horizontal.

Solution   5.  Let b be a non-zero constant. Find the line that passes thru the point (0, b) and is tangent to the curve y = 1/x.

Solution

Suppose the line is tangent to the curve at x = a. Then the slope of the line is: 