Differentiation rules for learning Calculus

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3 Differentiation Rules

Copyright © Cengage Learning. All rights reserved.
3.2 The Product and Quotient Rules

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The Product Rule
By analogy with the Sum and Difference Rules, one might
be tempted to guess, that the derivative of a product is the
product of the derivatives.
We can see, however, that this guess is wrong by looking
at a particular example.
Let f(x) = x and g(x) = x2
. Then the Power Rule gives
f(x) = 1 and g(x) = 2x.
But (fg)(x) = x3
, so (fg)(x) = 3x2
. Thus (fg)  fg.

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The Product Rule
The correct formula was discovered by Leibniz and is
called the Product Rule.
Before stating the Product Rule, let’s see how we might
discover it.
We start by assuming that u = f(x) and v = g(x) are both
positive differentiable functions. Then we can interpret the
product uv as an area of a rectangle (see Figure 1).
Figure 1
The geometry of the Product
Rule

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The Product Rule
If x changes by an amount x, then the corresponding
changes in u and v are
u = f(x + x) – f(x) v = g(x + x) – g(x)
and the new value of the product, (u + u)(v + v), can be
interpreted as the area of the large rectangle in Figure 1
(provided that u and v happen to be positive).
The change in the area of the rectangle is
(uv) = (u + u)(v + v) – uv = u v + v u + u v
= the sum of the three shaded areas

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The Product Rule
If we divide by x, we get
If we now let x  0, we get the derivative of uv:

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The Product Rule
(Notice that u  0 as x  0 since f is differentiable and
therefore continuous.)
Although we started by assuming (for the geometric
interpretation) that all the quantities are positive, we notice
that Equation 1 is always true. (The algebra is valid
whether u, v, u, v and are positive or negative.)

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The Product Rule
So we have proved Equation 2, known as the Product Rule,
for all differentiable functions u and v.
In words, the Product Rule says that the derivative of a
product of two functions is the first function times the
derivative of the second function plus the second function
times the derivative of the first function.

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Example 1
(a) If f(x) = xex
, find f(x).
(b) Find the nth derivative, f(n)
(x).
Solution:
(a) By the Product Rule, we have

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Example 1 – Solution
(b) Using the Product Rule a second time, we get
cont’d

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Example 1 – Solution
Further applications of the Product Rule give
f(x) = (x + 3)ex
f(4)
(x) = (x + 4)ex
In fact, each successive differentiation adds another term
ex
, so
f(n)
(x) = (x + n)ex
cont’d

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The Quotient Rule
We find a rule for differentiating the quotient of two
differentiable functions u = f(x) and v = g(x) in much the
same way that we found the Product Rule.
If x, u, and v change by amounts x, u, and v, then the
corresponding change in the quotient uv is

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The Quotient Rule
so
As x  0, v  0 also, because v = g(x) is differentiable
and therefore continuous.
Thus, using the Limit Laws, we get

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The Quotient Rule
In words, the Quotient Rule says that the derivative of a
quotient is the denominator times the derivative of the
numerator minus the numerator times the derivative of the
denominator, all divided by the square of the denominator.

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The Quotient Rule
Table of Differentiation Formulas

Differentiation rules for learning Calculus

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