Section 2.3 The Product and Quotient rules and higher-order derivatives
Theorem 1: The Product Rule
The product of two differentiable functions is differentiable and
d
[ f ( x) g ( x)] = f ′( x) g ( x) + f ( x) g ′( x)
dx
Proof:
d
f ( x + ∆x) g ( x + ∆x) − f ( x) g ( x)
[ f ( x) g ( x)] = ∆lim
x
→
0
dx
∆x
f ( x + ∆x) g ( x + ∆x) − f ( x + ∆x) g ( x) + f ( x + ∆x) g ( x) − f ( x) g ( x)
= lim
∆x → 0
∆x
f ( x + ∆x) g ( x) − f ( x) g ( x)
f ( x + ∆x) g ( x + ∆x) − f ( x + ∆x) g ( x)
= lim
+ lim
∆x → 0
∆
x
→
0
∆x
∆x
f ( x + ∆x) − f ( x) 
g ( x + ∆x) − g ( x)

=  lim
 ⋅ g ( x) + lim f ( x + ∆x) ⋅ lim
∆x → 0
∆x → 0
∆x
∆x
 ∆x →0

= f ′( x) g ( x) + f ( x) g ′( x)
The product rule can easily be extended to
d
[ f ( x) g ( x)h( x)] = f ′( x) g ( x)h( x) + f ( x) g ′( x)h( x) + f ( x) g ( x)h′( x)
dx
Example 1: Find the derivative of each function.
a) f ( x) = ( x 3 + 2 x) sin x
f ( x) = ( x 3 + 2 x) sin x
(
)
d 3
d
x + 2 x ⋅ sin x + ( x 3 + 2 x) ⋅ (sin x )
dx
dx
2
3
= (3 x + 2) sin x + ( x + 2 x) cos x
f ′( x) =
b) g ( x) = x (cos x)
g ( x) = x (cos x)
g ′( x) =
=
1
2 x
(cos x) + x (− sin x)
cos x
− x (sin x)
2 x
cos x
x (sin x) ⋅ 2 x
−
2 x
1⋅ 2 x
cos x − 2 x sin x
=
2 x
=
Calculus I by Chinyoung Bergbauer at NHMCCD: 2.3 The product and quotient rules and higher order derivative
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Practice 1: Find the derivative of each function.
a) f ( x) = (4 x 2 + 3 x)(3 x + 5)
 2 
b) g ( x) = (3 sin x )

 x
Theorem 2: The Quotient Rule
The quotient f / g of two differentiable function f and g is differentiable at all values of x
d  f ( x)  f ′( x) g ( x) − f ( x) g ′( x)
for which g ( x) ≠ 0 and
=
dx  g ( x) 
[g ( x)]2
Proof:
f ( x + ∆x ) f ( x )
−


d f ( x)
g ( x + ∆x) g ( x)
= lim
dx  g ( x)  ∆x → 0
∆x
f ( x + ∆x ) g ( x ) − f ( x ) g ( x + ∆x )
= lim
∆x → 0
g ( x + ∆x ) g ( x ) ∆x
f ( x + ∆x ) g ( x ) − f ( x ) g ( x ) + f ( x ) g ( x ) − f ( x ) g ( x + ∆x )
= lim
∆x → 0
g ( x + ∆x ) g ( x ) ∆x
f ( x + ∆x ) g ( x ) − f ( x ) g ( x ) f ( x ) g ( x ) − f ( x ) g ( x + ∆x )
+
∆x
∆x
= lim
∆x → 0
g ( x + ∆x ) g ( x )
f ( x + ∆x ) g ( x ) − f ( x ) g ( x ) f ( x ) g ( x + ∆x ) − f ( x ) g ( x )
−
∆x
∆x
= lim
∆x → 0
g ( x + ∆x) g ( x)
f ( x + ∆x ) − f ( x )
g ( x + ∆x ) − g ( x )
g ( x) − f ( x)
∆x
∆x
= lim
∆x → 0
g ( x + ∆x) g ( x)
=
f ′( x) g ( x) − f ( x) g ′( x)
[g ( x)]2
d  1  − g ′( x)
=
dx  g ( x)  [g ( x)]2
Calculus I by Chinyoung Bergbauer at NHMCCD: 2.3 The product and quotient rules and higher order derivative
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Example 2: Find the derivative of each function.
5x − 2
a) f ( x) = 3
x + 2x
f ( x) =
f ′( x) =
=
=
5x − 2
x3 + 2 x
5( x 3 + 2 x) − (5 x − 2)(3 x 2 + 2)
(x
3
+ 2x
)
2
5 x 3 + 10 x − 15 x 3 + 6 x 2 − 10 x + 4
(x
3
+ 2x
)
2
− 10 x 3 + 6 x 2 + 4
(x
3
+ 2x
)
2
2 + 1/ x 2
f ( x) =
2x − 3
b)
2 + 1 / x 2 2 + x −2
=
2x − 3
2x − 3
−3
(−2 x )(2 x − 3) − (2 + 1 / x 2 )2
f ′( x) =
(2 x − 3) 2
f ( x) =
(−2 / x )(2 x − 3) − (2 + 1 / x )2
=
(2 x − 3) 2
3
=
−
=
=
2
−
2(2 x − 3) 
1 
−  2 + 2 2
3
x
x 

=
2
(2 x − 3)
2(2 x − 3) 2(2 x 2 + 1)
2(2 x − 3) 3 2(2 x 2 + 1) 3
−
−
⋅x −
⋅x
3
2
x3
x2
x
x
=
(2 x − 3) 2
x 3 (2 x − 3) 2
− 2(2 x − 3) − 2 x(2 x 2 + 1) − 4 x + 6 − 4 x 3 − 2 x − 4 x 3 − 6 x + 6
=
= 3
x 3 (2 x − 3) 2
x 3 (2 x − 3) 2
x (2 x − 3) 2


2x2 +1
 Alternate method : f ( x) = 2
: Use the quotient rule 
x (2 x − 3)


Practice 2: Find the derivative of each function.
Calculus I by Chinyoung Bergbauer at NHMCCD: 2.3 The product and quotient rules and higher order derivative
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a)
b)
c)
4x − 3
5x 2 + 4 x
3x − 2 x 2
f ( x) =
x3
9
f ( x) = 3
7x
f ( x) =
Modified Quotient Rule: When the numerator is just 1, then
d  1  − g ′( x)
=
dx  g ( x)  [g ( x)]2
Example 3: Use the quotient rule to get the derivative of f ( x) = x −5 .
1
f ( x ) = x −5 = 5
x
4
− 5x
5
f ′( x) = 10 = − 6
x
x
Practice 3: Use the quotient rule to get the derivative of f ( x) = x −9 .
Example 4: Use the quotient rule to get the derivative of f ( x) = csc x .
1
f ( x) = csc x =
sin x
cos x
cos x 1
f ′( x) = − 2 = −
⋅
= − cot x csc x
sin x
sin x sin x
Practice 4: Use the quotient rule to get the derivative of f ( x) = cot x and f ( x) = sec x
Theorem 3: Derivative of Trigonometric Functions
d
d
[tan x] = sec2 x;
[cot x] = − csc2 x
dx
dx
d
[sec x] = sec x tan x;
dx
d
[csc x] = − csc x cot x
dx
Example 5: Find the derivative of each function.
a) f ( x) = x tan x
f ( x) = x tan x
f ′( x) = tan x + x(sec 2 x)
b) g ( x) = (sec x ) x
Calculus I by Chinyoung Bergbauer at NHMCCD: 2.3 The product and quotient rules and higher order derivative
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g ( x) = (sec x ) x
g ′( x) = (sec x tan x) x + (sec x)
1
2 x
=
2 x(sec x tan x) + sec x
2 x
=
sec x(2 x tan x + 1)
2 x
cos x
1 + sin x
cos x
h( x) =
1 + sin x
− sin x(1 + sin x) − cos x(cos x) − sin x − sin 2 x − cos 2 x − sin x − (sin 2 x + cos 2 x)
h′( x) =
=
=
(1 + sin x) 2
(1 + sin x) 2
(1 + sin x) 2
− sin x − 1 − (1 + sin x)
1
=
=
=−
2
2
(1 + sin x)
(1 + sin x)
1 + sin x
c) h( x) =
Practice 5: Find the derivative of each function.
a) f ( x) = 3 x 2 cot x
b) g ( x) = 2 x + csc x
1 − sin x
c) h( x) =
1 + sin x
Higher-Order Derivatives
dy
d
,
[ f ( x)], Dx [ y ]
dx dx
d2y
d2
2
Second derivative: y′′, f ′′( x),
,
[ f ( x)], Dx [ y ]
dx 2
dx 2
d3y
d3
3
Third derivative: y′′′, f ′′′( x),
,
[ f ( x)], Dx [ y ]
3
3
dx
dx
4
d y
d4
4
( 4)
( 4)
Fourth derivative: y , f ( x),
,
[ f ( x)], Dx [ y ]
4
4
dx
dx
.
.
.
dny
dn
n
nth derivative: y ( n ) , f ( n ) ( x),
,
[ f ( x)], Dx [ y ]
dx n
dx n
First derivative: y′,
f ′( x),
Calculus I by Chinyoung Bergbauer at NHMCCD: 2.3 The product and quotient rules and higher order derivative
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Example 6: Find the second derivative of the function: f ( x) =
x +1
x−2
The second derivative of the position function is the acceleration function a (t ) = s′′(t )
Example 7: Finding the acceleration due to gravity
Because the moon has no atmosphere, a falling object on the moon encounters no air
resistance. In 1971, astronaut David Scott demonstrated that a feather and a hammer fall
at the same rate on the moon.
The position function of a falling object on the moon is s (t ) = −0.81t 2 + 2 where the
height in meters is s (t ) and t is the time in seconds. What is the acceleration due to
gravity on the moon? Note that the acceleration due to gravity on earth is – 9.8 m / sec 2 .
s (t ) = −0.81t 2 + 2
v(t ) = s ′(t ) = −1.62t
a (t ) = s′′(t ) = −1.62m / sec 2
Calculus I by Chinyoung Bergbauer at NHMCCD: 2.3 The product and quotient rules and higher order derivative
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