3 differentiation rules. 3.2 the product and quotient rules differentiation rules in this section,...
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3DIFFERENTIATION RULESDIFFERENTIATION RULES
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3.2The Product and
Quotient Rules
DIFFERENTIATION RULES
In this section, we will learn about:
Formulas that enable us to differentiate new functions
formed from old functions by multiplication or division.
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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
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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
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The correct formula was discovered by
Leibniz (soon after his false start) and is
called the Product Rule.
THE PRODUCT RULE
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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.
THE PRODUCT RULE
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Then, we can interpret the product uv
as an area of a rectangle.
THE PRODUCT RULE
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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
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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
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The change in the area of the rectangle is:
( ) ( )( )
the sum of the three shaded areas
uv u u v v uv
u v v u u v
Δ = + Δ + Δ −= Δ + Δ + Δ Δ=
THE PRODUCT RULE Equation 1
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If we divide by ∆x, we get:
( )uv v u vu v u
x x x x
Δ Δ Δ Δ= + +Δ
Δ Δ Δ Δ
THE PRODUCT RULE
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If we let ∆x → 0, we get the derivative of uv:
( )
0
0
0 0 0 0
( )( ) lim
lim
lim lim lim lim
0.
x
x
x x x x
d uvuv
dx xv u v
u v ux x x
v u vu v u
x x x
dv du dvu vdx dx dx
Δ →
Δ →
Δ → Δ → Δ → Δ →
Δ=
ΔΔ Δ Δ⎛ ⎞= + +Δ⎜ ⎟Δ Δ Δ⎝ ⎠Δ Δ Δ⎛ ⎞= + + Δ ⎜ ⎟Δ Δ Δ⎝ ⎠
= + +
THE PRODUCT RULE
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Notice that ∆u → 0 as ∆x → 0 since f is differentiable and therefore continuous.
( )d dv duuv u v
dx dx dx= +
THE PRODUCT RULE Equation 2
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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
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If f and g are both differentiable, then:
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.
[ ] [ ] [ ]( ) ( ) ( ) ( ) ( ) ( )d d df x g x f x g x g x f x
dx dx dx= +
THE PRODUCT RULE
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a. If f(x) = xex, find f ’(x).
b. Find the nth derivative, f (n)(x)
THE PRODUCT RULE Example 1
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By the Product Rule, we have:
THE PRODUCT RULE Example 1 a
'( ) ( )
( ) ( )
1
( 1)
x
x x
x x
x
df x xe
dxd d
x e e xdx dx
xe e
x e
=
= +
= + ⋅
= +
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Using the Product Rule again, we get:
''( ) ( 1)
( 1) ( ) ( 1)
( 1) 1
( 2)
x
x x
x x
x
df x x e
dxd d
x e e xdx dx
x e e
x e
⎡ ⎤= +⎣ ⎦
= + + +
= + + ⋅
= +
THE PRODUCT RULE Example 1 b
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Further applications of the Product Rule give:
4
'''( ) ( 3)
( ) ( 4)
x
x
f x x e
f x x e
= +
= +
THE PRODUCT RULE Example 1 b
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In fact, each successive differentiation adds another term ex.
So:( ) ( )n xf x x n e= +
THE PRODUCT RULE Example 1 b
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Differentiate the functionTHE PRODUCT RULE Example 2
( ) ( )f t t a bt= +
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Using the Product Rule, we have:
THE PRODUCT RULE E. g. 2—Solution 1
( )1 21
2
'( ) ( ) ( )
( )
( ) ( 3 )
2 2
d df t t a bt a bt t
dt dt
t b a bt t
a bt a btb t
t t
−
= + + +
= ⋅ + + ⋅
+ += + =
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If we first use the laws of exponents to
rewrite f(t), then we can proceed directly
without using the Product Rule.
This is equivalent to the answer in Solution 1.
THE PRODUCT RULE E. g. 2—Solution 2
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Example 2 shows that it is sometimes easierto simplify a product of functions than to usethe Product Rule.
In Example 1, however, the Product Rule isthe only possible method.
THE PRODUCT RULE
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If , where
g(4) = 2 and g’(4) = 3, find f’(4).
THE PRODUCT RULE Example 3
( ) ( )f x xg x=
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Applying the Product Rule, we get:
So,
THE PRODUCT RULE Example 3
[ ]
1 212
'( ) ( ) ( ) ( )
( )'( ) ( ) '( )
2
d d df x xg x x g x g x x
dx dx dxg x
xg x g x x xg xx
−
⎡ ⎤ ⎡ ⎤= = +⎣ ⎦ ⎣ ⎦
= + ⋅ = +
(4) 2'(4) 4 '(4) 2 3 6.5
2 22 4
gf g= + = ⋅ + =
⋅
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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
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If x, u, and v change by amounts Δx, Δu,
and Δv, then the corresponding change
in the quotient u / v is:
THE QUOTIENT RULE
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So,
( )
( )
0
0
lim
lim
x
x
u vd u
dx v x
u vv ux x
v v v
Δ →
Δ →
Δ⎛ ⎞=⎜ ⎟ Δ⎝ ⎠Δ Δ−
Δ Δ=+ Δ
THE QUOTIENT RULE
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As ∆x → 0, ∆v → 0 also—because v = g(x)
is differentiable and therefore continuous.
Thus, using the Limit Laws, we get:
( )0 0
2
0
lim lim
lim
x x
x
u v du dvv u v u
d u x x dx dxdx v v v v v
Δ → Δ →
Δ →
Δ Δ− −
⎛ ⎞ Δ Δ= =⎜ ⎟ + Δ⎝ ⎠
THE QUOTIENT RULE
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If f and g are differentiable, then:
In words, the Quotient Rule says: 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.
[ ] [ ]2
( ) ( ) ( ) ( )( )
( ) ( )
d dg x f x f x g xd f x dx dx
dx g x g x
−⎡ ⎤=⎢ ⎥ ⎡ ⎤⎣ ⎦ ⎣ ⎦
THE QUOTIENT RULE
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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
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Let
2
3
2
6
x xy
x
+ −=
+
THE QUOTIENT RULE Example 4
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Then,
THE QUOTIENT RULE Example 4
( ) ( ) ( ) ( )( )
( )( ) ( )( )( )
( ) ( )( )
( )
3 2 2 3
23
3 2 2
23
4 3 4 3 2
23
4 3 2
23
6 2 2 6'
6
6 2 1 2 3
6
2 12 6 3 3 6
6
2 6 12 6
6
d dx x x x x x
dx dxyx
x x x x x
x
x x x x x x
x
x x x x
x
+ + − − + − +=
+
+ + − + −=
+
+ + + − + −=
+
− − + + +=
+
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Find an equation of the tangent
line to the curve y = ex / (1 + x2)
at the point (1, ½e).
THE QUOTIENT RULE Example 5
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According to the Quotient Rule,
we have:
THE QUOTIENT RULE Example 5
( ) ( ) ( )( )
( ) ( )( )
( )( )
2 2
22
22
2 22 2
1 1
1
1 2 1
1 1
x x
x x x
d dx e e x
dy dx dxdx x
x e e x e x
x x
+ − +=
+
+ − −= =
+ +
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So, the slope of the tangent line at
(1, ½e) is:
This means that the tangent line at (1, ½e) is horizontal and its equation is y = ½e.
1
0
x
dy
dx =
=
THE QUOTIENT RULE Example 5
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In the figure, notice that the function
is increasing and crosses its tangent line
at (1, ½e).
THE QUOTIENT RULE Example 5
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Don’t use the Quotient Rule every time you see a quotient.
Sometimes, it’s easier to rewrite a quotient first to put it in a form that is simpler for the purpose of differentiation.
NOTE
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For instance, though it is possible to
differentiate the function
using the Quotient Rule, it is much easier
to perform the division first and write
the function as
before differentiating.
23 2( )
x xF x
x
+=
1 2( ) 3 2F x x x−= +
NOTE
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Here’s a summary of the differentiation
formulas we have learned so far.
( ) ( ) ( )
( ) ( ) ( )
( )
1
'
2
0
' ' ' ' ' ' ' '
' '' ' '
n n x xd d dc x nx e e
dx dx dx
cf cf f g f g f g f g
f gf fgfg fg gf
g g
−= = =
= + = + − = −
⎛ ⎞ −= + =⎜ ⎟
⎝ ⎠
DIFFERENTIATION FORMULAS