Download 14 - Higher Order Derivartives

HIGHER ORDER DERIVATIVES
Consider the function 𝑓(𝑥) = 𝑥 3 + 𝑥 2 − 2𝑥 + 1. Differentiating this function gives the
first order derivative. Differentiating the first derivative gives the second derivative.
Differentiating the second derivative gives the third derivative, and so on. Refer to the table
below.
Given: 𝑓(𝑥) = 𝑥 3 + 𝑥 2 − 2𝑥 + 1
Prime Notation
1st Order Derivative
𝑓′(𝑥)
2nd Order Derivative
𝑓′′(𝑥)
3rd Order Derivative
𝑓′′′(𝑥)
4th Order Derivative
𝑓 (4) (𝑥)
5th Order Derivative
𝑓 (5) (𝑥)
.
.
.
.
.
.
.
.
𝑛𝑡ℎ 𝑜𝑟𝑑𝑒𝑟 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒
Ratio Notation
𝑑𝑓
𝑑
(𝑓(𝑥))
=
𝑑𝑥 𝑑𝑥
𝑑2𝑓
𝑑
(𝑓′(𝑥))
=
2
𝑑𝑥
𝑑𝑥
𝑑3𝑓
𝑑
(𝑓′′(𝑥))
=
3
𝑑𝑥
𝑑𝑥
𝑑4𝑓
𝑑𝑥 4
𝑑5𝑓
𝑑𝑥 5
.
.
.
.
𝑓 (𝑛) (𝑥)
𝑓 ′ (𝑥) = 3𝑥 2 + 2𝑥
−2
𝑓 ′′ (𝑥) = 6𝑥 + 2
𝑓 ′′′ (𝑥) = 6
𝑓 (4) (𝑥) = 0
𝑓 (5) (𝑥) = 0
.
.
.
.
𝑑𝑛 𝑓
𝑑𝑥 𝑛
𝑓 (𝑛) (𝑥) = 0
The 2nd, 3rd, 4th and so on up to 𝑛𝑡ℎ 𝑜𝑟𝑑𝑒𝑟 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 are known as Higher Order
Derivatives. Notice that in the given function 𝑓(𝑥), the 4th order, 𝑓 (4) (𝑥), up to the nth order
derivative, 𝑓 (𝑛) (𝑥), the derivative is zero.
QUICK MATH FACT:
If 𝒇(𝒙) is a polynomial of degree 𝒌, then, the 𝒏𝒕𝒉 𝒐𝒓𝒅𝒆𝒓 derivative is zero for 𝒏 ≥ 𝒌 + 𝟏. That is,
𝒇(𝒏) (𝒙) = 𝟎 𝒇𝒐𝒓 𝒏 ≥ 𝒌 + 𝟏
EXAMPLE 1: Find the second and third – order derivative of the function.
2
2
1.1. 𝑓(𝑥) = 3 𝑥 3 − 𝑥
SOLUTION:
APPLY POWER FORMULA
2
2
𝑓(𝑥) = 𝑥 3 −
3
𝑥
𝑓 ′(𝑥) = 2𝑥 2 +
2
𝑥2
𝑓 ′′(𝑥) = 4𝑥 −
𝑓 ′(𝑥) = 2𝑥 2 +
2
𝑥2
𝑓 ′′(𝑥) = 4𝑥 −
4
𝑥3
𝑓 ′′′(𝑥) = 4 +
4
𝑥3
12
𝑥4
1.2. 𝑦 = 𝑥 2 (𝑥 − 1)5
SOLUTION:
APPLY PRODUCT RULE
𝑦 ′ = 𝑥(𝑥 − 1)4 (7𝑥 − 2)
𝑦 = 𝑥 2 (𝑥 − 1)5
𝑦 ′ = (𝑥 − 1)4 (7𝑥 2 − 2𝑥)
𝑦 ′ = 𝑥 2 [5(𝑥 − 1)4 ] + (𝑥 − 1)5 (2𝑥)
𝑦 ′′ = (𝑥 − 1)4 (14𝑥 − 2) + (7𝑥 2 − 2𝑥)(4)(𝑥 − 1)3
𝑦 ′ = 𝑥(𝑥 − 1)4 [5𝑥 + 2(𝑥 − 1)]
𝑦 ′′ = (𝑥 − 1)4 (2)(7𝑥 − 1) + (7𝑥 2 − 2𝑥)(4)(𝑥 − 1)3
𝑦 ′ = 𝑥(𝑥 − 1)4 [5𝑥 + 2𝑥 − 2]
𝑦 ′′ = 2(𝑥 − 1)3 [(𝑥 − 1)(7𝑥 − 1) + 2(7𝑥 2 − 2𝑥)]
𝑦 ′ = 𝑥(𝑥 − 1)4 (7𝑥 − 2)
𝑦 ′′ = 2(𝑥 − 1)3 [7𝑥 2 − 8𝑥 + 7 + 14𝑥 2 − 4𝑥]
𝑦 ′′ = 2(𝑥 − 1)3 (21𝑥 2 − 12𝑥 + 7)
𝑦 ′′ = 2(𝑥 − 1)3 (21𝑥 2 − 12𝑥 + 7)
𝑦 ′′′ = 2[(𝑥 − 1)3 (42𝑥 − 12) + (21𝑥 2 − 12𝑥 + 7)(3)(𝑥 − 1)2 ]
𝑦 ′′′ = 2[(𝑥 − 1)3 (6)(7𝑥 − 2) + (21𝑥 2 − 12𝑥 + 7)(3)(𝑥 − 1)2 ]
𝑦 ′′′ = 2(3)(𝑥 − 1)2 [(𝑥 − 1)(3)(7𝑥 − 2) + (21𝑥 2 − 12𝑥 + 7)]
𝑦 ′′′ = 6(𝑥 − 1)2 [(𝑥 − 1)(21𝑥 − 6) + 21𝑥 2 − 12𝑥 + 7]
𝑦 ′′′ = 6(𝑥 − 1)2 [21𝑥 2 − 27𝑥 + 6 + 21𝑥 2 − 12𝑥 + 7]
𝑦 ′′′ = 6(𝑥 − 1)2 (42𝑥 2 − 29𝑥 + 13)
1.3. 𝑦 =
𝑥
√𝑥−1
SOLUTION:
APPLY QUOTIENT RULE
𝑦=
𝑥
𝑦′ =
√𝑥 − 1
√𝑥 − 1(1) − 𝑥 (
𝑦′ =
1
)
2√𝑥 − 1
′′
𝑦 =
2
(√𝑥 − 1)
𝑥−2
2(𝑥 − 1)3/2
3
2(𝑥 − 1)3/2 (1) − (𝑥 − 2)(2) (2) (𝑥 − 1)1/2
3 2
[2(𝑥 − 1)2 ]
1
𝑥
[𝑥 − 1 − 2]
−1
𝑦 ′ = √𝑥
𝑥−1
𝑦 ′′ =
2(𝑥 − 1)3/2 − 3(𝑥 − 2)(𝑥 − 1)1/2
4(𝑥 − 1)3
𝑦 ′′ =
(𝑥 − 1)1/2 [2(𝑥 − 1) − 3(𝑥 − 2)]
4(𝑥 − 1)3
𝑥−2
2
𝑦 =
(𝑥 − 1)3/2
𝑦 ′′ =
[2𝑥 − 2 − 3𝑥 + 6]
4(𝑥 − 1)5/2
𝑥−2
𝑦 =
2(𝑥 − 1)3/2
𝑦 ′′ =
4−𝑥
4(𝑥 − 1)5/2
2𝑥 − 2 − 𝑥
2
𝑦 =
(𝑥 − 1)√𝑥 − 1
′
′
′
𝑦 ′′ =
𝑦 ′′′ =
4−𝑥
4(𝑥 − 1)5/2
5
5
4(𝑥 − 1)2 (−1) − (4 − 𝑥)(4) (2) (𝑥 − 1)3/2
5 2
[4(𝑥 − 1)2 ]
5
𝑦 ′′′
4(𝑥 − 1)2 (−1) − (4 − 𝑥)(10)(𝑥 − 1)3/2
=
16(𝑥 − 1)5
𝑦 ′′′ =
−2(𝑥 − 1)3/2 [2(𝑥 − 1) + 5(4 − 𝑥)]
16(𝑥 − 1)5
𝑦 ′′′ = −
(𝑥 − 1)3/2 [2𝑥 − 2 + 20 − 5𝑥]
8(𝑥 − 1)5
𝑦 ′′ = −
𝑦 ′′′ = −
[18 − 3𝑥]
8(𝑥 − 1)7/2
𝑦 ′′′ =
−3(𝑥 − 6)
8(𝑥 − 1)7/2
3(𝑥 − 6)
8(𝑥 − 1)7/2
1.4. 𝑓(𝑥) = sin2 (𝜋𝑥)
SOLUTION:
APPLY POWER RULE
𝑓(𝑥) = sin2 (𝜋𝑥)
𝑓 ′ (𝑥) = 𝜋 sin(2𝜋𝑥)
𝑓 ′ (𝑥) = 2 sin(𝜋𝑥) cos(𝜋𝑥) (𝜋)
𝑓 ′′ (𝑥) = 𝜋 cos(2𝜋𝑥) (2𝜋)
TRIGONOMETRIC IDENTITY:
𝑓 ′′ (𝑥) = 2𝜋 2 cos(2𝜋𝑥)
sin(2𝑥) = 2 sin(𝑥) cos(𝑥)
𝑓 ′′ (𝑥) = 2𝜋 2 cos(2𝜋𝑥)
𝑓 ′ (𝑥) = sin(2𝜋𝑥) (𝜋)
𝑓 ′′′ (𝑥) = −2𝜋 2 sin(2𝜋𝑥) (2𝜋)
𝑓 ′ (𝑥) = 𝜋 sin(2𝜋𝑥)
𝑓 ′′′ (𝑥) = −4𝜋 3 sin(2𝜋𝑥)
1.5. 𝑎2 𝑥 2 + 𝑦 2 𝑏 2 = 𝑎2 𝑏 2 , find (𝑎)𝑦 ′ (𝑏)𝑦 ′′ (𝑐)𝑦′′′
SOLUTION:
APPLY IMPLICIT DIFFERENTIATION
𝑎2 𝑥 2 + 𝑦 2 𝑏 2 = 𝑎2 𝑏 2
2𝑎2 𝑥 + 2𝑏 2 𝑦𝑦 ′ = 0
2𝑏 2 𝑦𝑦 ′ = −2𝑎2 𝑥
−2𝑎2 𝑥
𝑦 =
2𝑏 2 𝑦
′
𝑦′ =
−𝑎2 𝑥
𝑏2𝑦
From 𝑎2 𝑥 2 + 𝑦 2 𝑏2 = 𝑎2 𝑏 2
𝑎2 𝑏 2 − 𝑎2 𝑥 2
𝑦=√
𝑏2
𝑦=
𝑎√𝑏 2 − 𝑥 2
𝑏
So,
𝑦′ =
−𝑎2 𝑥
𝑏2𝑦
−𝑎2 𝑥
𝑦′ =
𝑏2 (
𝑦′ =
𝑎√𝑏 2 − 𝑥 2
)
𝑏
−𝑎𝑥
𝑏√𝑏 2 − 𝑥 2
𝑦′ =
−𝑎𝑥
𝑦 ′′ = −
𝑏√𝑏 2 − 𝑥 2
𝑏√𝑏 2 − 𝑥 2 (−𝑎) − (−𝑎𝑥)(𝑏) (
𝑦 ′′ =
[𝑏√𝑏 2 − 𝑥 2 ]
−𝑎𝑏√𝑏 2 − 𝑥 2 + 𝑎𝑏𝑥 (
𝑦 ′′ =
−2𝑥
)
2√𝑏 2 − 𝑥 2
2
−𝑥
)
− 𝑥2
(𝑏 2
𝑎𝑏
− 𝑥 2 )3/2
Since, 𝑎 𝑎𝑛𝑑 𝑏 are constants, rewrite 𝑦′′ so
that we can use POWER FORMULA instead
of QUOTIENT RULE in finding 𝑦 ′′′ .
√𝑏 2
𝑏 2 (𝑏2 − 𝑥 2 )
−𝑎𝑏
[𝑏 2 − 𝑥 2 + 𝑥 2 ]
2 − 𝑥2
√𝑏
𝑦 ′′ =
𝑏 2 (𝑏 2 − 𝑥 2 )
𝑦 ′′ = −
𝑦 ′′ = −
𝑎𝑏(𝑏 2 )
5
3
𝑦 ′′′ = −𝑎𝑏 (− ) (𝑏 2 − 𝑥 2 )−2 (−2𝑥)
2
𝑦 ′′′ =
𝑏 2 (𝑏2 − 𝑥 2 )√𝑏 2 − 𝑥 2
(𝑏 2
𝑦 ′′ = −𝑎𝑏(𝑏 2 − 𝑥 2 )−3/2
−3𝑎𝑏𝑥
(𝑏 2 − 𝑥 2 )5/2
𝑎𝑏
− 𝑥 2 )3/2
PRACTICE PROBLEMS TO WORKOUT!
MASTERY TEST. Find the second and third – order derivative of the function.
1
1. 𝑦 = 3 𝑥 3 − 2𝑥 2 + 𝑥 − 10
2. 𝑦 = 5𝑒 𝑥
3. 𝑦 = √4 − 𝑥 2
𝑥2
4. 𝑦 = 𝑥+1
5. 𝑥 2 + 𝑦 2 = 1
ANSWER TO MASTERY TEST:
ANSWER TO MASTERY TEST:
Second – Order Derivative.
Third – Order Derivative.
1. 𝑦 ′′ = 2(𝑥 − 2)
2. 𝑦 ′′ = 5𝑒 𝑥
4
3. 𝑦 ′′ = − (4−𝑥2 )3/2
2
1. 𝑦 ′′′ = 2
2. 𝑦 ′′′ = 5𝑒 𝑥
12𝑥
3. 𝑦 ′′′ = − (4−𝑥2 )5/2
−6
4. 𝑦 ′′ = (𝑥+1)3
4. 𝑦 ′′′ = (𝑥+1)4
1
5. 𝑦 ′′ = − (1−𝑥2 )3/2
3𝑥
5. 𝑦 ′′′ = − (1−𝑥2 )5/2
*****END OF DISCUSSION*****
ANSWER TO MASTERY TEST
1. 0 2. 0 3. 0 4. −∞ 5. −∞ 6. +∞ 7. 0 8. −
2
9. −∞ 10. −1 11. −∞ 12. −1
5