Transcript Chapter 3 Section 2 - Columbus State University

DIFFERENTIATION RULES
3.2
The Product and
Quotient Rules
In this section, we will learn about:
 Formulas that enable us to differentiate new functions
 formed from old functions by multiplication or division.
THE PRODUCT RULE
 By analogy with the Sum and Difference Rules,
one might be tempted to guess—as Leibniz did
three centuries ago—that the derivative of a
product is the product of the derivatives.
 However, we can see that this guess is wrong
by looking at a particular example.
THE PRODUCT RULE
 Let f (x) = x and g(x) = x2.
 Then, the Power Rule gives f’(x) = 1 and g’(x) = 2x.
 However, (fg)(x) = x3.
 So, (fg)’(x) =3 x2.
 Thus, (fg)’ ≠ f’ g’.
THE PRODUCT RULE
 The correct formula was discovered by Leibniz
(soon after his false start) and is called the
Product Rule.
 Before stating the Product Rule, let us see how
we might discover it.
 We start by assuming that u = f (x) and v = g(x)
are both positive differentiable functions.
THE PRODUCT RULE
 Then, we can interpret the product uv as an area
of a rectangle.
THE PRODUCT RULE
 If x changes by an amount Δx, then the
corresponding changes in u and v are:
 Δu = f(x + Δx) - f(x)
 Δv = g(x + Δx) - g(x)
THE PRODUCT RULE
 The new value of the product, (u + Δu) (v + Δv),
can be interpreted as the area of the large
rectangle in the figure, provided that Δu and Δv
happen to be positive.
THE PRODUCT RULE
Equation 1
 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
THE PRODUCT RULE
 If we divide by ∆x, we get:
(uv )
v
u
v
u
v
 u
x
x
x
x
THE PRODUCT RULE
 If we let ∆x → 0, we get the derivative of uv:
d
(uv )
(uv )  lim
x 0
dx
x
u
v 
 v
 lim  u
v
 u

x 0
x
x 
 x
v
u
v 

 u lim
 v lim
 lim u  lim

x  0  x
x  0  x
x  0
x 0 x


dv
du
dv
u
v
 0.
dx
dx
dx


THE PRODUCT RULE
Equation 2
d
dv
du
(uv )  u
v
dx
dx
dx
 Notice that ∆u → 0 as ∆x → 0 since f is
differentiable and therefore continuous.
THE PRODUCT RULE
 Though we began by assuming, for the
geometric interpretation, that all quantities are
positive, we see Equation 1 is always true.
 The algebra is valid whether u, v, ∆u, and ∆v are
positive or negative.
 So, we have proved Equation 2, known as the
Product Rule, for all differentiable functions
u and v.
THE PRODUCT RULE
 If f and g are both differentiable, then:
d
d
d
 f ( x ) g ( x )  f ( x )  g ( x )   g ( x )  f ( x ) 
dx
dx
dx
 In words, the Product Rule says:
 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.
THE PRODUCT RULE
Example 1
a. If f(x) = xex, find f ’(x).
b. Find the nth derivative, f (n)(x)
THE PRODUCT RULE
Example 1 a
 By the Product Rule, we have:
d
x

f ( x )  ( xe )
dx
d x
x d
 x (e )  e
( x)
dx
dx
x
x
 xe  e  1
 ( x  1)e
x
THE PRODUCT RULE
Example 1 b
 Using the Product Rule again, we get:
d
x


f ( x )  ( x  1)e 
dx
d x
x d
 ( x  1) ( e )  e
( x  1)
dx
dx
x
x
 ( x  1)e  e  1
 ( x  2)e
x
THE PRODUCT RULE
Example 1 b
 Further applications of the Product Rule give:
f ( x )  ( x  3)e
f
(4)
x
( x )  ( x  4)e
x
THE PRODUCT RULE
Example 1 b
 In fact, each successive differentiation adds
another term ex.
 So:
f
(n)
( x )  ( x  n)e
x
THE PRODUCT RULE
Example 2
 Differentiate the function
f (t )  t ( a  bt )
THE PRODUCT RULE
Example 2—Solution 1
 Using the Product Rule, we have:
d
d
f (t )  t ( a  bt )  ( a  bt )
dt
dt
1 1 2
 t  b  ( a  bt )  2 t
( a  bt ) ( a  3bt )
b t 

2 t
2 t
 t
THE PRODUCT RULE
Example 2—Solution 2
 If we first use the laws of exponents to rewrite
f(t), then we can proceed directly without using
the Product Rule.
f (t )  a t  bt t  at
f '(t )  at
1
2
1 2
 bt
3
2
12
 bt
32
12
 This is equivalent to the answer in Solution 1.
THE PRODUCT RULE
 Example 2 shows that it is sometimes easier to
simplify a product of functions than to use the
Product Rule.
 In Example 1, however, the Product Rule is the
only possible method.
THE PRODUCT RULE
 If
Example 3
f ( x )  xg ( x ) , where g(4) = 2 and g’(4) = 3,
find f’(4).
THE PRODUCT RULE
Example 3
 Applying the Product Rule, we get:
d
d
d


 x
f ( x ) 
xg
(
x
)

x
g
(
x
)

g
(
x
)



dx 
dx
dx  
g( x)
1 2
1
 xg ( x )  g ( x )  2 x  xg ( x ) 
2 x
 So,
g (4)
2
f (4)  4g (4) 
 23
 6.5
2 2
2 4
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.
THE QUOTIENT RULE
 If x, u, and v change by amounts Δx, Δu, and Δv,
then the corresponding change in the quotient u
/ v is:
 u  u  u u
  

 v  v  v v
u  u  v  u  v  v  vu  uv



v  v  v 
v  v  v 
THE QUOTIENT RULE
 So,
 u v 
d u
   lim
x 0
dx  v 
x
u
v
v
u

x

x
 lim
x 0 v  v  v 
THE QUOTIENT RULE
 As ∆x → 0, ∆v → 0 also—because v = g(x) is
differentiable and therefore continuous.
 Thus, using the Limit Laws, we get:
u
v
du
dv
v lim
 u lim
v
u
d u
x  0  x
x  0 x
dx
dx


 
dx  v 
v lim  v  v 
v2
x  0
THE QUOTIENT RULE
 If f and g are differentiable, then:
d
d
f
(
x
)
g
(
x
)

f
(
x
)
g
(
x
)




d  f ( x )  dx
dx

dx  g ( x ) 
 g ( x )2 
 In words, the Quotient Rule says:
 The derivative of a quotient is the numerator times the
derivative of the denominator minus the numerator times
the derivative of the denominator, all divided by the
square of the denominator.
THE QUOTIENT RULE
 The Quotient Rule and the other differentiation
formulas enable us to compute the derivative of
any rational function, as the next example
illustrates.
THE QUOTIENT RULE
Example 4
 Let
x  x2
y
3
x 6
2
THE QUOTIENT RULE
 Then,
Example 4
d 2
d 3
2
 x  6  dx  x  x  2    x  x  2  dx  x  6 
y' 
2
3
 x  6



3
3
2
2
x

6
2
x

1

x

x

2
3
x
   
 
x
3
 6
2
4
3
4
3
2
2
x

x

12
x

6

3
x

3
x

6
x

 

x
3
 6
2
 x 4  2 x 3  6 x 2  12 x  6
x
3
 6
2
THE QUOTIENT RULE
Example 5
 Find an equation of the tangent line to the curve
y = ex / (1 + x2) at the point (1, e/2).
THE QUOTIENT RULE
Example 5
 According to the Quotient Rule, we have:
d
x
x d
2
1 x 
e e
1 x 



dy
dx
dx

2
2
dx
1  x 
2
1  x  e  e  2 x  e 1  x 



1  x 
1  x 
2
x
x
2
2
x
2
2
2
THE QUOTIENT RULE
Example 5
 So, the slope of the tangent line at (1, e/2) is:
dy
dx
0
x 1
 This means that the tangent line at (1, e/2) is
horizontal and its equation is y = e/2.
THE QUOTIENT RULE
Example 5
 In the figure, notice that the function is
increasing and crosses its tangent line at (1, e/2).
NOTE
 Do not use the Quotient Rule every time you see
a quotient.
 Sometimes, it is easier to rewrite a quotient first
to put it in a form that is simpler for the purpose
of differentiation.
NOTE
 For instance, though it is possible to differentiate
the function
3x  2 x
F ( x) 
x
2
using the Quotient Rule, it is much easier to
perform the division first and write the function
as F ( x )  3x  2 x 1 2 before differentiating.
DIFFERENTIATION FORMULAS
 Here is a summary of the differentiation formulas
we have learned so far.
d
c   0
dx
d n
n 1
x   nx

dx
d x
x
e e

dx
 cf  '  cf '
 f  g  '  f ' g '
 f  g  '  f ' g '
 fg  '  fg ' gf '
'
 f  gf ' fg '
 g   g2
 