CHAPTER 3. RULES FOR DERIVATIVES
3.2
69
Product and Quotient Rules
Rule (Product Rule:).
�
�
�
�
d
d
d
[f (x)g(x)] =
f (x) g(x) + f (x)
g(x)
dx
dx
dx
�
�
�
(f · g) = f · g + f · g
(product of two functions)� = (deriv. of first)(second) + (first)(deriv. of second)
Comments. We have stated the product rule three times. The first time has too
much notation: too many x’s, too many parentheses, etc. This version will only be
useful for the proof of the product rule we give presently. The second version is far
better for memorizing and using. The third version is how people actually do the
product rule once they’ve gotten pretty good at it: they just keep track of which
function they’re looking at and write down the derivative of each part as they go.
You should only try to do it this way once you can write down a derivative of a
simple function without stopping and using formulas,
Proof.
d
f (x + h)g(x + h) − f (x)g(x)
[f (x)g(x)] = lim
h→0
dx
h
f (x + h)g(x + h) − f (x)g(x + h) + f (x)g(x + h) − f (x)g(x)
= lim
h→0
h
f (x + h)g(x + h) − f (x)g(x + h) f (x)g(x + h) − f (x)g(x)
= lim
+
h→0
h
h
f (x + h) − f (x) g(x + h) f (x) g(x + h) − g(x)
= lim
+
h→0
h
1
1
h
Note that when you take the limit the first fraction becomes f � (x), the second
fraction becomes g(x), the third becomes f (x) and the last becomes g � (x), so we
get
f � (x) · g(x) + f (x) · g � (x).
Example 1. Find
d
(3x + ex )(x2 − 4ex ).
dx
Solution. To use the formula (f g)� = f � g + f g � we start by labelling the parts of
the function we were given:
(3x + ex ) (x2 − 4ex ) .
� �� � � �� �
f
Now we calculate f � g + f g �
g
d
(3x + ex )(x2 − 4ex ) = (3x + ex )� · (x2 − 4ex ) + (3x + ex ) · (x2 − 4ex )�
dx
= (3 + ex )(x2 − 4ex ) + (3x + ex )(2x − 4ex ).
CHAPTER 3. RULES FOR DERIVATIVES
70
Figure 3.1: Intuitive, graphical, understanding of product rule
f
g
g
�
g
g
f
f
f�
Picture the product f g as the area of a In a similar way, we picture f � and g � as small
rectangle with sides f and g. Interpret changes in f and g. Then f � · g and f · g �
(f g)� as a small change in the area, pic- are the areas of two skinny rectangles (again
tured here in red.
pictured in red)
Comparing the two previous pictures (and ignoring the very small white square
in the corner that we’ve left out) we see that the red area in the first picture is
the same as the red area in the second picture. In other words,
(f g)� = f � · g + f · g �
d √
Example 2. Find
(5 x + 2ex )
dx
�
�
10 x5
+
.
x5 10
Solution. To use the formula (f g)� = f � g + f g � we start by labelling the parts of
the function we were given:
�
�
√
10 x5
x
(5 x + 2e )
+
�
��
� x5 10
�
��
�
f
g
Now we calculate f � g + f g �
�
�
�
�
�
��
√
√
d √
10 x5
x5
10 x5
x
x � 10
x
(5 x + 2e )
+
= (5 x + 2e )
+
+ (5 x + 2e )
+
dx
x5 10
x5 10
x5 10
�
��
�
�
�
√
1
10 x5
5 4
x
x
−6
= 5 √ + 2e
+
+ (5 x + 2e ) 10(−5)x + x
x5 10
10
2 x
�
��
�
�
�
5
√
5
10 x
−50 1 4
√ + 2ex
=
+
+ (5 x + 2ex )
+ x
5
x
10
x6
2
2 x
This is where we ended on Wednesday, October 9
Rule (Quotient Rule:).
� ��
f
f � · g − f · g�
=
g
g2
(deriv. of top)(bottom) − (top)(deriv. of bottom)
(fraction)� =
(bottom)2
CHAPTER 3. RULES FOR DERIVATIVES
Example 3. Find the derivative of
Solution. Let f (t) = 2t +
71
2t + 1t
√ .
2+ t
√
1
and g(t) = 2 + t. Then the derivative is
t
d 2t + 1t
f �g − f g�
√ =
dt 2 + t
g2
√
√
�
��
�
�
2t + 1t · (2 + t) − 2t + 1t · (2 + t)�
√
=
(2 + t)2
√
(2 − t−2 )(2 + t) − (2t + 1t )( 12 t−1/2 )
√
=
(2 + t)2
ex
.
x2
(b) Find the equation of the tangent line of f (x) at x = 1, and graph this with
the original function.
(c) Graph the original function and it’s derivative on the same axes. Compare
the graphs.
Example 4.
Solution.
(a) Find the derivative of f (x) =
(a)
�
ex
x2
��
(ex )� x2 − ex (x2 )�
(x2 )2
ex x2 − ex 2x
=
x4
x
xe (x − 2)
=
x4
x
e (x − 2)
=
x3
=
(b)
y = m(x − a) + y0
a=1
e1
=e
12
e1 (1 − 2)
m = f � (1) =
= −e
13
y = −e(x − 1) + e
y0 = f (1) =
The graph is shown below
CHAPTER 3. RULES FOR DERIVATIVES
72
(c) The two graphs are shown below, with f (x) in black and f � (x) in red.
Note that we can verify, a little bit, that the derivative is correct. For instance,
at x = 2 the slope of f (x) is 0, the slope is negative just to the left and the
slope is positive just to the right. This translates into: at x = 2 the y-value
of f � (x) is 0, the y-values are negative just to the left, and the y-values are
positive just to the right.