Higher Derivatives
Higher order derivatives – the derivatives other than the first derivative.
For the function 𝑦 = 𝑓(𝑥)
𝑑𝑦
𝑑𝑥
= 𝑦 ′ = 𝑓′(𝑥) – first derivative
𝑑2 𝑦
𝑑𝑥 2
= 𝑦 ′′ = 𝑓′′(𝑥) – second derivative
𝑑3 𝑦
= 𝑦 ′′′ = 𝑓′′′(𝑥) – third derivative
𝑑𝑥 3
𝑑4 𝑦
𝑑𝑥 4
= 𝑦 (4) = 𝑓 (4) (𝑥) – fourth derivative
1. 𝑦 = 𝑥 5 − 𝑥 −3 − 5𝑥 + 7 , find 𝑦′′′ .
𝑦 ′ = 5𝑥 4 + 3𝑥 −4 − 5
𝑦 ′′ = 5(4)𝑥 3 + 3(−4)𝑥 −5 − 0
𝑦 ′′ = 20𝑥 3 − 12𝑥 −5
𝑦 ′′′ = 20(3)𝑥 2 − 12(−5)𝑥 −6
𝑦 ′′′ = 60𝑥 2 + 60𝑥 −6
2. 𝑦 = (𝑥 2 − 2)3 , find 𝑦′′ .
𝑛=3
𝑢 = 𝑥2 − 2
𝑑𝑢
= 2𝑥
𝑑𝑥
𝑑 𝑑𝑢
(𝑢𝑛 ) = 𝑛𝑢𝑛−1
𝑑𝑥 𝑑𝑥
′ 2 3−1
𝑦 = 3(𝑥 − 2) (2𝑥)
𝑦 ′ = 6𝑥(𝑥 2 − 2)2
𝑑𝑢
𝑢 = 6𝑥, 𝑑𝑥 = 6
𝑑𝑣
𝑣 = (𝑥 2 − 2)2 , 𝑑𝑥
= 2(𝑥 2 − 2)2−1 (2𝑥)
𝑑𝑣
𝑑𝑥
= 4𝑥(𝑥 2 − 2)
𝑑 𝑑𝑣 𝑑𝑢
(𝑢𝑣) = 𝑢 𝑑𝑥 + 𝑣 𝑑𝑥
𝑑𝑥
′′ 2 (𝑥 2 2
𝑦 = 6𝑥[4𝑥(𝑥 − 2)] + − 2) (6)
𝑦 ′′ 2 2
= 6(𝑥 − 2)[𝑥(4𝑥) + (𝑥 − 2)]
𝑦 ′′ = 6(𝑥 2 − 2)(4𝑥 2 + 𝑥 2 − 2)
𝑦 ′′ = 6(𝑥 2 − 2)(5𝑥 2 − 2)
4𝑥+5
3. 𝑦 = 3−2𝑥 find 𝑦 ′′
𝑑𝑢
𝑢 = 4𝑥 + 5, 𝑑𝑥
=4
𝑑𝑣
𝑣 = 3 − 2𝑥, = −2
𝑑𝑥
𝑢 𝑑𝑢 𝑑𝑣
𝑑( ) 𝑣 −𝑢
𝑣 𝑑𝑥 𝑑𝑥
𝑑𝑥
= 𝑣2
′ (3−2𝑥)(4)−(4𝑥+5)(−2)
𝑦 = (3−2𝑥)2
Prepared by:
Engr. Rosalinda B. Mole𝐧̃o
Engr. Ma. Yvonne C. Bangero
Engr. Cristina A. Vale
1
12−8𝑥+8𝑥+10
𝑦′ = (3−2𝑥)2
22
𝑦′ = (3−2𝑥)2
𝑐 = 22
𝑑𝑣
𝑣 = (3 − 2𝑥)2 , 𝑑𝑥 = 2(3 − 2𝑥)(−2)
𝑑𝑣
𝑑𝑥
= −4(3 − 2𝑥)
𝑐 𝑑𝑣
𝑑( ) −𝑐
𝑣 𝑑𝑥
𝑑𝑥
= 𝑣2
−(22)(−4)(3−2𝑥)
𝑦 ′′ =
[(3−2𝑥)2 ]2
88(3−2𝑥)
𝑦 ′′ = (3−2𝑥)4
88
𝑦 ′′ = (3−2𝑥)3
4. 𝑦 2 = 𝑥 2 + 2𝑥, find 𝑦 ′′ .
𝑑 𝑑 𝑑
(𝑦 2 ) = (𝑥 2 ) + (2𝑥)
𝑑𝑥 𝑑𝑥 𝑑𝑥
2𝑦𝑦 ′ = 2𝑥 + 2
2𝑥+2
𝑦 ′ = 2𝑦
2(𝑥+1)
𝑦′ =
2𝑦
𝑥+1
𝑦′ =
𝑦
𝑑𝑢
𝑢 = 𝑥 + 1, 𝑑𝑥
=1
𝑑𝑣
𝑣 = 𝑦, = 𝑦′
𝑑𝑥
𝑢 𝑑𝑢 𝑑𝑣
𝑑( ) 𝑣 −𝑢
𝑣 𝑑𝑥 𝑑𝑥
𝑑𝑥
= 𝑣2
′′ (𝑦)(1)−(𝑥+1)(𝑦 ′) 𝑥+1
𝑦 = but 𝑦 ′ =
(𝑦)2 𝑦
𝑥+1
𝑦−(𝑥+1)( )
′′ 𝑦
𝑦 = 𝑦2
𝑦2 −(𝑥+1)2
𝑦
𝑦 ′′ = 𝑦2
𝑦 2 −(𝑥+1)2 1
𝑦 ′′ = 𝑦
∙ 𝑦2
𝑦 2 −(𝑥+1)2
𝑦 ′′ = 𝑦3
5. 𝑥 = 𝑦 5 (3𝑦 2 − 7), find 𝑥′′
𝑥 = 3𝑦 7 − 7𝑦 5
𝑥 ′ = 3(7)(𝑦 7−1 ) − 7(5)(𝑦 5−1 )
𝑥 ′ = 21𝑦 6 − 35𝑦 4
𝑥 ′′ = 21(6)𝑦 6−1 − 35(4)𝑦 4−1
𝑥 ′′ = 126𝑦 5 − 140𝑦 3
Prepared by:
Engr. Rosalinda B. Mole𝐧̃o
Engr. Ma. Yvonne C. Bangero
Engr. Cristina A. Vale
2
1
6. 𝑥 = (2𝑦 − 1)2 , find 𝑥′′′.
1 𝑑𝑢
𝑛 = , 𝑢 = 2𝑦 − 1, =2
2 𝑑𝑦
𝑑 𝑑𝑢
(𝑢𝑛 ) = 𝑛𝑢𝑛−1
𝑑𝑦 𝑑𝑦
1
′ 1 −1
𝑥 = 2 (2𝑦 − 1)2 (2)
1
′ −
𝑥 = (2𝑦 − 1) 2
1
1
𝑥 ′′ = − 2 (2𝑦 − 1)−2−1 (2)
3
𝑥 ′′ = −(2𝑦 − 1)−2
3
3
𝑥 ′′′ = −(− 2)(2𝑦 − 1)−2−1 (2)
5
𝑥 ′′′ = 3(2𝑦 − 1)−2
7. 𝑥 2 − 𝑦 2 = 6, find 𝑥′′.
𝑑 𝑑 𝑑
(𝑥 2 ) − (𝑦 2 ) = (6)
𝑑𝑦 𝑑𝑦 𝑑𝑦
2𝑥𝑥 ′ − 2𝑦 = 0
2𝑥𝑥 ′ = 2𝑦
2𝑦
𝑥 ′ = 2𝑥
𝑦
𝑥′ = 𝑥
𝑑𝑢
𝑢=𝑦 =1
𝑑𝑦
𝑑𝑣
𝑣=𝑥 𝑑𝑦
= 𝑥′
𝑢 𝑑𝑢 𝑑𝑣
𝑑( ) 𝑣 −𝑢
𝑣 𝑑𝑦 𝑑𝑦
𝑑𝑦
= 𝑣2
𝑥(1)−𝑦(𝑥 ′ ) 𝑦
𝑥 ′′ = but 𝑥 ′ =
𝑥2 𝑥
𝑦
𝑥−𝑦( )
𝑥 ′′ = 𝑥
𝑥2
𝑥2 −𝑦2
𝑥 ′′ = 𝑥
𝑥2
𝑥 2 −𝑦 2 1
𝑥 ′′ = 𝑥
∙ 𝑥2
𝑥 2 −𝑦 2
𝑥 ′′ = 𝑥3
; 𝑥 − 𝑦2 = 6
2
6
𝑥 ′′ = 𝑥3
𝑑2 𝑦
8. 𝑦 = (𝑥 3 + 4𝑥 − 5)2 find 𝑑𝑥 2
𝑑𝑦
= 2(𝑥 3 + 4𝑥 − 5)(3𝑥 2 + 4 − 0)
𝑑𝑥
= 2(𝑥 3 + 4𝑥 − 5)(3𝑥 2 + 4)
𝑑𝑢
𝑢 = (𝑥 3 + 4𝑥 − 5) 𝑑𝑥
= 3𝑥 2 + 4
Prepared by:
Engr. Rosalinda B. Mole𝐧̃o
Engr. Ma. Yvonne C. Bangero
Engr. Cristina A. Vale
3
𝑑𝑣
𝑣 = (3𝑥 2 + 4) 𝑑𝑥
= 6𝑥
𝑑(𝑢𝑣) 𝑑𝑣 𝑑𝑢
𝑑𝑥
= 𝑢 𝑑𝑥 + 𝑣 𝑑𝑥
𝑑2 𝑦
𝑑𝑥 2
= 2[(𝑥 3 + 4𝑥 − 5)(6𝑥) + (3𝑥 2 + 4)(3𝑥 2 + 4)]
= 2(6𝑥 + 24𝑥 − 30𝑥 + 9𝑥 4 + 24𝑥 2 + 16)
4 2
𝑑2 𝑦
𝑑𝑥 2
= 2(15𝑥 4 + 48𝑥 2 − 30𝑥 + 16)
9. 𝑥 2 = 4𝑎𝑦 find 𝑦′′
2𝑥 = 4𝑎𝑦′
2𝑥
𝑦′ = 4𝑎
𝑥 Numerical coefficient
= 2𝑎
1
𝑦 ′ = 2𝑎 𝑥
1
𝑦′′ = (1)
2𝑎
1
= 2𝑎
10. 𝑥 2 = 4𝑎𝑦 find 𝑥′′
𝑑 𝑑
(𝑥 2 ) = (4𝑎𝑦)
𝑑𝑦 𝑑𝑦
2𝑥𝑥′ = 4𝑎
4𝑎
𝑥′ = 2𝑥
2𝑎
𝑥′ = 𝑥
𝑐 = 2𝑎 𝑣=𝑥
𝑑𝑣
𝑑𝑦
= 𝑥′
𝑐 𝑑𝑣
𝑑( ) −𝑐
𝑣 𝑑𝑦
=
𝑑𝑦 𝑣2
2𝑎(𝑥′)
𝑥′′ = −
𝑥2
2𝑎
but 𝑥′ = 𝑥
2𝑎
2𝑎( )
𝑥
𝑥′′ = − 𝑥2
′′ 4𝑎 2 1
𝑥 =− 𝑥 ∙ 𝑥2
4𝑎 2
=− 3
𝑥
Prepared by:
Engr. Rosalinda B. Mole𝐧̃o
Engr. Ma. Yvonne C. Bangero
Engr. Cristina A. Vale
4
Exercises:
Find the indicated derivative.
𝑑4 𝑦
1. 𝑦 = 2𝑦 −1 + 3𝑦 4 , 𝑑𝑥 4
2. 𝑦 = (5𝑥 3 − 4)2 , 𝑦′′′
3. 𝑦 = (4𝑥 + 1)(𝑥 − 2)2 , 𝑦′′
𝑑2 𝑦
4. 𝑥 2 + 2𝑦 2 − 4 = 0, 𝑑𝑥 2
5. 𝑦 = (3𝑥 + 2)(2𝑥 − 3)−1 , 𝑦′′
7 5 3
6. 𝑥 = 4𝑦 −4 + 3𝑦 −3 − 4𝑦 −2 + 10, 𝑥′′
𝑑2 𝑥
7. 𝑥 = 4 4√𝑦 − 2𝑦√𝑦 , 𝑑𝑦 2
3𝑦−4
8. 𝑥 = 1+2𝑦 , 𝑥′′
9. 𝑥 2 = 𝑦 2 − 4𝑦, 𝑥′′
𝑑2𝑠
10. 𝑠 = 𝑡 3 (𝑡 − 1)2 ,
𝑑𝑡 2
Prepared by:
Engr. Rosalinda B. Mole𝐧̃o
Engr. Ma. Yvonne C. Bangero
Engr. Cristina A. Vale
5
