nth Derivatives of standard functions and related problems

1. nth DERIVATIVES OF SOME
STANDARD RESULTS
Dr. Soya Mathew

2. I. Derivation for the nth derivative of 𝑦 = 𝑥𝑚
; where m is a positive integer
Proof:-
Let 𝑦 = 𝑥𝑚
𝑦1 = 𝑚 𝑥𝑚−1
.
𝑦2 = 𝑚(𝑚 − 1) 𝑥𝑚−2
.
𝑦3 = 𝑚(𝑚 − 1)(𝑚 − 2) 𝑥𝑚−3
.
……………………………………………………..
𝑦𝑛 = 𝑚 𝑚 − 1 𝑚 − 2 … (𝑚 − (𝑛 − 1)) 𝑥𝑚−𝑛
.

3. 𝑦𝑛 = 𝑚 𝑚 − 1 𝑚 − 2 … (𝑚 − 𝑛 + 1) 𝑥𝑚−𝑛.
⇒ 𝑦𝑛 =
𝑚 𝑚−1 𝑚−2 … 𝑚−𝑛+1 𝑚−𝑛 …2.1
𝑚−𝑛 𝑚−𝑛−1 …2.1
𝑥𝑚−𝑛.
⇒ 𝑦𝑛 =
𝑚!
𝑚−𝑛 !
𝑥𝑚−𝑛.
∴ 𝑫𝒏 𝒙𝒎 =
𝒎!
𝒎−𝒏 !
𝒙𝒎−𝒏.

4. Thus if 𝑦 = 𝑥𝑚, m is a positive integer and 𝑛 ≤ 𝑚, then
𝑦𝑛 =
𝑚!
𝑚−𝑛 !
𝑥𝑚−𝑛.
Note:
1. If 𝑦 = 𝑥𝑚, m is a positive integer and 𝑛 = 𝑚, then
𝑦𝑛 = 𝑛!
2. If 𝑦 = 𝑥𝑚
, m is a positive integer and 𝑛 > 𝑚, then
𝑦𝑛 = 0

5. II. Derivation for the nth derivative of 𝑦 = 𝑎𝑥 + 𝑏 𝑚
; where m is a positive
integer.
Proof:-
Let 𝑦 = 𝑎𝑥 + 𝑏 𝑚
𝑦1 = 𝑚 𝑎 𝑎𝑥 + 𝑏 𝑚−1.
𝑦2 = 𝑚(𝑚 − 1) 𝑎2 𝑎𝑥 + 𝑏 𝑚−2.
𝑦3 = 𝑚(𝑚 − 1)(𝑚 − 2) 𝑎3 𝑎𝑥 + 𝑏 𝑚−3.
………………………………………………………………………..
𝑦𝑛 = 𝑚 𝑚 − 1 𝑚 − 2 … (𝑚 − 𝑛 − 1 ) 𝑎𝑛 𝑎𝑥 + 𝑏 𝑚−𝑛.

6. ⇒ 𝑦𝑛 = 𝑚 𝑚 − 1 𝑚 − 2 … (𝑚 − 𝑛 + 1) 𝑎𝑛 𝑎𝑥 + 𝑏 𝑚−𝑛.
⇒ 𝑦𝑛 =
𝑚 𝑚−1 𝑚−2 … 𝑚−𝑛+1 𝑚−𝑛 …2.1
𝑚−𝑛 𝑚−𝑛−1 …2.1
𝑎𝑛 𝑎𝑥 + 𝑏 𝑚−𝑛.
⇒ 𝑦𝑛 =
𝑚!
𝑚−𝑛 !
𝑎𝑛 𝑎𝑥 + 𝑏 𝑚−𝑛.
∴ 𝑫𝒏(𝒂𝒙 + 𝒃)𝒎 =
𝒎!
𝒎−𝒏 !
𝒂𝒏 𝒂𝒙 + 𝒃 𝒎−𝒏.

7. Thus if 𝑦 = 𝑎𝑥 + 𝑏 𝑚, m is a positive integer and 𝑛 ≤ 𝑚, then
𝑦𝑛 =
𝑚!
𝑚−𝑛 !
𝑎𝑛 𝑎𝑥 + 𝑏 𝑚−𝑛.
Note:
1. If 𝑦 = 𝑎𝑥 + 𝑏 𝑚, m is a positive integer and 𝑛 = 𝑚, then
𝑦𝑛 = 𝑛! 𝑎𝑛
2. If 𝑦 = 𝑎𝑥 + 𝑏 𝑚
, m is a positive integer and 𝑛 > 𝑚, then
𝑦𝑛 = 0

8. III. Derivation for the nth derivative of 𝑦 =
1
𝑎𝑥+𝑏
.
Proof:-
Let 𝑦 =
1
𝑎𝑥+𝑏
⇒ 𝑦 = 𝑎𝑥 + 𝑏 −1
𝑦1 = (−1) . 𝑎 𝑎𝑥 + 𝑏 −2.
𝑦2 = −1 −2 . 𝑎2 𝑎𝑥 + 𝑏 −3 = (−1)2 2! 𝑎2 𝑎𝑥 + 𝑏 −3.
𝑦3 = (−1)(−2)(−3) . 𝑎3 𝑎𝑥 + 𝑏 −4 = (−1)3 3! 𝑎3 𝑎𝑥 + 𝑏 −4
………………………………………………………………………………………………………

9. ⇒ 𝑦𝑛 = (−1)𝑛
𝑛! 𝑎𝑛
𝑎𝑥 + 𝑏 −(𝑛+1)
.
∴ 𝑫𝒏 𝟏
𝒂𝒙+𝒃
=
(−𝟏)𝒏 𝒏! 𝒂𝒏
𝒂𝒙+𝒃 𝒏+𝟏 .
Note:
𝐷𝑛 1
𝑎𝑥+𝑏 𝑚 = −1 𝑛 (𝑚+𝑛−1)!
𝑚−1 !
1
𝑎𝑥+𝑏 𝑚+𝑛 𝑎𝑛

10. IV. Derivation for the nth derivative of 𝑦 = 𝑙𝑜𝑔(𝑎𝑥 + 𝑏).
Proof:-
Let 𝑦 = 𝑙𝑜𝑔(𝑎𝑥 + 𝑏)
𝑦1 = 𝑎.
1
𝑎𝑥+𝑏
= 𝑎. 𝑎𝑥 + 𝑏 −1
𝑦2 = (−1) . 𝑎2 𝑎𝑥 + 𝑏 −2.
𝑦3 = −1 −2 . 𝑎3 𝑎𝑥 + 𝑏 −3 = (−1)2 2! 𝑎3 𝑎𝑥 + 𝑏 −3.
𝑦4 = (−1)(−2)(−3) . 𝑎4 𝑎𝑥 + 𝑏 −4 = (−1)3 3! 𝑎4 𝑎𝑥 + 𝑏 −4.
………………………………………………………………………………………………………

11. ⇒ 𝑦𝑛 = (−1)𝑛−1 (𝑛 − 1)! 𝑎𝑛 𝑎𝑥 + 𝑏 −𝑛.
∴ 𝑫𝒏 𝒍𝒐𝒈(𝒂𝒙 + 𝒃) =
(−𝟏)𝒏−𝟏 (𝒏−𝟏)! 𝒂𝒏
𝒂𝒙+𝒃 𝒏

12. Recall
1. 𝐷𝑛
𝑥𝑚
=
𝑚!
𝑚−𝑛 !
𝑥𝑚−𝑛
2. 𝐷𝑛
(𝑎𝑥 + 𝑏)𝑚
=
𝑚!
𝑚−𝑛 !
𝑎𝑛
𝑎𝑥 + 𝑏 𝑚−𝑛
.
3. 𝐷𝑛 1
𝑎𝑥+𝑏
=
(−1)𝑛 𝑛! 𝑎𝑛
𝑎𝑥+𝑏 𝑛+1
4. 𝐷𝑛 1
𝑎𝑥+𝑏 𝑚 =
−1 𝑛 𝑚+𝑛−1 ! 𝑎𝑛
𝑚−1 ! 𝑎𝑥+𝑏 𝑚+𝑛
5. 𝐷𝑛
𝑙𝑜𝑔(𝑎𝑥 + 𝑏) =
(−1)𝑛−1 (𝑛−1)! 𝑎𝑛
𝑎𝑥+𝑏 𝑛

13. Problems:
1. Find the nth derivative of
2𝑥−1
(𝑥 − 2)(𝑥 + 1)
Solution:
Let 𝑦 =
2𝑥−1
(𝑥 − 2)(𝑥 + 1)
Resolving into partial fraction, we get
𝑦 =
2𝑥−1
(𝑥 − 2)(𝑥 + 1)
=
𝐴
(𝑥 − 2)
+
𝐵
(𝑥 + 1)
… … … … … … (1)
⇒ 2𝑥 − 1 = 𝐴 𝑥 + 1 + 𝐵 𝑥 − 2
When 𝑥 = −1, 𝐵 = 1
When 𝑥 = 2, 𝐴 = 1

14. ∴ (1) ⇒ 𝑦 =
1
𝑥 − 2
+
1
𝑥 + 1
∴ 𝑦𝑛 = 𝐷𝑛 1
𝑥 −2
+ 𝐷𝑛 1
𝑥 + 1
⇒ 𝑦𝑛 =
−1 𝑛 𝑛!
𝑥 −2 𝑛+1 +
−1 𝑛 𝑛!
𝑥 + 1 𝑛+1
⇒ 𝑦𝑛 = −1 𝑛
𝑛!
1
𝑥 −2 𝑛+1 +
1
𝑥 + 1 𝑛+1

15. 2. Find the nth derivative of
1
(𝑥 + 2)(𝑥 − 1)
Solution:
Let 𝑦 =
1
(𝑥 + 2)(𝑥 − 1)
Resolving into partial fraction, we get
𝑦 =
1
(𝑥 + 2)(𝑥 − 1)
=
1
(𝑥 + 2)(−2 − 1)
+
1
(1 + 2)(𝑥 − 1)
⇒ 𝑦 = −
1
3
1
(𝑥 + 2)
+
1
3
1
(𝑥 − 1)

16. ⇒ 𝑦 =
1
3
1
(𝑥 − 1)
−
1
(𝑥 + 2)
∴ 𝑦𝑛 =
1
3
𝐷𝑛 1
𝑥 − 1
− 𝐷𝑛 1
𝑥 + 2
⇒ 𝑦𝑛 =
1
3
[
−1 𝑛 𝑛!
𝑥 − 1 𝑛+1 −
−1 𝑛 𝑛!
𝑥 + 2 𝑛+1]
⇒ 𝑦𝑛 =
−1 𝑛 𝑛!
3
[
1
𝑥 − 1 𝑛+1 −
1
𝑥 + 2 𝑛+1]

17. 3. Find the nth derivative of
𝑥2
𝑥 − 1 2(𝑥 − 2)
Solution:
Let 𝑦 =
𝑥2
𝑥 − 1 2(𝑥 − 2)
⇒ y =
𝑥2
𝑥 − 1 2(𝑥 − 2)
=
𝐴
𝑥 − 1
+
𝐵
𝑥 − 1 2 +
𝐶
𝑥 − 2
… … … … … (1)
⇒ 𝑥2
= 𝐴 𝑥 − 1 𝑥 − 2 + 𝐵 𝑥 − 2 + 𝐶 𝑥 − 1 2
When 𝑥 = 1, 𝐵 = −1
When 𝑥 = 2, 𝐶 = 4
When 𝑥 = 0, 2𝐴 − 2𝐵 + 𝐶 = 0 ⇒ 𝐴 = −3

18. (1) ⇒ y =
−3
𝑥 − 1
+
−1
𝑥 − 1 2 +
4
𝑥 − 2
∴ 𝑦𝑛 = −3 𝐷𝑛 1
𝑥 − 1
− 𝐷𝑛 1
𝑥 − 1 2 + 4 𝐷𝑛 1
𝑥 − 2
⇒ 𝑦𝑛 = −3
−1 𝑛 𝑛!
𝑥 − 1 𝑛+1 −
−1 𝑛 2 + 𝑛 − 1 !
𝑥 − 1 𝑛+2 + 4
−1 𝑛 𝑛!
𝑥 − 2 𝑛+1
⇒ 𝑦𝑛 = −1 𝑛𝑛!
−3
𝑥 − 1 𝑛+1 −
𝑛+1
𝑥 − 1 𝑛+2 +
4
𝑥 − 2 𝑛+1
Note: 𝐷𝑛 1
𝑎𝑥+𝑏
=
(−1)𝑛 𝑛! 𝑎𝑛
𝑎𝑥+𝑏 𝑛+1 , 𝐷𝑛 1
𝑎𝑥+𝑏 2 =
−1 𝑛 𝑛+1 ! 𝑎𝑛
1! 𝑎𝑥+𝑏 𝑛+2 ,
𝐷𝑛 1
𝑎𝑥+𝑏 3 =
−1 𝑛 𝑛+2 ! 𝑎𝑛
2! 𝑎𝑥+𝑏 𝑛+3 …

19. 4. Find the nth derivative of log
2𝑥 + 1
𝑥 − 2
Solution:
Let 𝑦 = log
2𝑥 + 1
𝑥 − 2
⇒ 𝑦 =
1
2
log
2𝑥 + 1
𝑥 − 2
(since, log 𝑎 = log(𝑎)
1
2 =
1
2
log 𝑎)
⇒ 𝑦 =
1
2
log 2𝑥 + 1 − log(𝑥 − 2)
∴ 𝑦𝑛 =
1
2
𝐷𝑛 log 2𝑥 + 1 − 𝐷𝑛 log(𝑥 − 2)

20. ⇒ 𝑦𝑛 =
1
2
[
−1 𝑛−1(𝑛−1)! 2𝑛
2𝑥 + 1 𝑛 −
−1 𝑛−1(𝑛−1)!
𝑥 − 2 𝑛 ]
⇒ 𝑦𝑛 =
−1 𝑛−1(𝑛−1)!
2
[
2𝑛
2𝑥 + 1 𝑛 −
1
𝑥 − 2 𝑛]

21. 5. Find the nth derivative of 𝑦 = log (𝑎𝑥 + 𝑥2)
Solution:
Let 𝑦 = log [𝑥 𝑎 + 𝑥 ]
⇒ 𝑦 = log 𝑥 + log(𝑎 + 𝑥) (since, log 𝑎𝑏 = log 𝑎 + log 𝑏)
∴ 𝑦𝑛 = 𝐷𝑛
log 𝑥 + 𝐷𝑛
log 𝑎 + 𝑥
⇒ 𝑦𝑛 =
−1 𝑛−1(𝑛−1)!
𝑥 𝑛 +
−1 𝑛−1(𝑛−1)!
𝑎 + 𝑥 𝑛
⇒ 𝑦𝑛 = −1 𝑛−1(𝑛 − 1)!
1
𝑥𝑛 +
1
𝑎 + 𝑥 𝑛

22. Questions for practice
Find the nth derivatives of
1. 𝑦 =
1
6𝑥2 −5𝑥+1
2. 𝑦 =
4𝑥
(𝑥 −1)2 (𝑥+1)
3. 𝑦 = log(𝑥2
− 𝑎2
)
4. 𝑦 = log(𝑎𝑥 − 𝑥2)

23. 1. Find the nth derivative of
1
6𝑥2 −5𝑥+1
Solution:
Let 𝑦 =
1
6𝑥2 −5𝑥+1
⇒ 𝑦 =
1
2𝑥−1 (3𝑥−1)
Resolving into partial fraction, we get
𝑦 =
1
2𝑥−1 (3𝑥−1)
=
𝐴
(2𝑥 −1)
+
𝐵
(3𝑥 − 1)
… … … … … … (1)
⇒ 1 = 𝐴 3𝑥 − 1 + 𝐵 2𝑥 − 1
When 𝑥 =
1
3
, 𝐵
2
3
− 1 = 1 ⇒
−1
3
𝐵 = 1 ⇒ 𝐵 = −3
When 𝑥 =
1
, 𝐴
3
− 1 = 1 ⇒
1
𝐴 = 1 ⇒ 𝐴 = 2

24. ∴ (1) ⇒ 𝑦 ==
2
(2𝑥 −1)
+
−3
(3𝑥 − 1)
∴ 𝑦𝑛 = 2𝐷𝑛 1
2𝑥 −1
− 3𝐷𝑛 1
3𝑥− 1
⇒ 𝑦𝑛 = 2
−1 𝑛 𝑛! 2𝑛
2𝑥 −1 𝑛+1 − 3
−1 𝑛 𝑛! 3𝑛
3𝑥 − 1 𝑛+1
⇒ 𝑦𝑛 = −1 𝑛 𝑛!
2𝑛+1
2𝑥 −1 𝑛+1 −
3𝑛+1
3𝑥 − 1 𝑛+1

25. 2. Find the nth derivative of
4𝑥
(𝑥 −1)2 (𝑥+1)
Solution:
Let 𝑦 =
4𝑥
(𝑥 −1)2 (𝑥+1)
⇒ y =
4𝑥
(𝑥 −1)2 (𝑥+1)
=
𝐴
𝑥 − 1
+
𝐵
𝑥 − 1 2 +
𝐶
𝑥 + 1
… … … … … (1)
⇒ 4𝑥 = 𝐴 𝑥 − 1 𝑥 + 1 + 𝐵 𝑥 + 1 + 𝐶 𝑥 − 1 2
When 𝑥 = 1, 𝐵 = 2
When 𝑥 = −1, 𝐶 = −1
When 𝑥 = 0, −𝐴 + 𝐵 + 𝐶 = 0 ⇒ 𝐴 = 1

26. (1) ⇒ y =
1
𝑥 − 1
+
2
𝑥 − 1 2 −
1
𝑥 + 1
∴ 𝑦𝑛 = 𝐷𝑛 1
𝑥 − 1
+ 2𝐷𝑛 1
𝑥 − 1 2 − 𝐷𝑛 1
𝑥 + 1
⇒ 𝑦𝑛 =
−1 𝑛 𝑛!
𝑥 − 1 𝑛+1 + 2
−1 𝑛 2 + 𝑛 − 1 !
𝑥 − 1 𝑛+2 −
−1 𝑛 𝑛!
𝑥+1 𝑛+1
⇒ 𝑦𝑛 = −1 𝑛
𝑛!
1
𝑥 − 1 𝑛+1 + 2
𝑛+1
𝑥 − 1 𝑛+2 −
1
𝑥 + 1 𝑛+1

27. 3. Find the nth derivative of 𝑦 = log(𝑥2
− 𝑎2
)
Solution:
Let 𝑦 = log(𝑥2 − 𝑎2)
⇒ 𝑦 = log [(𝑥 + 𝑎) (𝑥 − 𝑎)]
⇒ 𝑦 = log (𝑥 + 𝑎) + log (𝑥 − 𝑎)
∴ 𝑦𝑛 = 𝐷𝑛 log(𝑥 + 𝑎) + 𝐷𝑛 log 𝑥 − 𝑎
⇒ 𝑦𝑛 =
−1 𝑛−1(𝑛−1)!
𝑥 + 𝑎 𝑛 +
−1 𝑛−1(𝑛−1)!
𝑥 − 𝑎 𝑛
⇒ 𝑦𝑛 = −1 𝑛−1
(𝑛 − 1)!
1
(𝑥+𝑎)𝑛 +
1
𝑥 − 𝑎 𝑛

28. 4. Find the nth derivative of 𝑦 = log(𝑎𝑥 − 𝑥2
)
Solution:
Let y = log(𝑎𝑥 − 𝑥2)
⇒ 𝑦 = log [𝑥 𝑎 − 𝑥 ]
⇒ 𝑦 = log 𝑥 + log(𝑎 − 𝑥)
∴ 𝑦𝑛 = 𝐷𝑛 log 𝑥 + 𝐷𝑛 log 𝑎 − 𝑥
⇒ 𝑦𝑛 =
−1 𝑛−1(𝑛−1)!
𝑥 𝑛 +
−1 𝑛−1 𝑛−1 ! −1 𝑛
𝑎 − 𝑥 𝑛
⇒ 𝑦𝑛 = −1 𝑛−1(𝑛 − 1)!
1
𝑥𝑛 +
−1 𝑛
𝑎 − 𝑥 𝑛