cbse ncert solutions for class 11 mathematics chapter 13 · 2019. 11. 29. ·...
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Class–XI–CBSE-Mathematics Limits and Derivatives
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CBSE NCERT Solutions for Class 11 Mathematics Chapter 13
Back of Chapter Questions
Exercise 13.1
1. Evaluate the Given limit:lim𝑥→3
𝑥 + 3
Hint: Put 𝑥 = 3
Solution:
Solution step 1: lim𝑥→3
𝑥 + 3 = 3 + 3 = 6
2. Evaluate the Given limit: lim𝑥→𝜋
(𝑥 −22
7)
Hint: Put 𝑥 = 𝜋
Solution:
Solution step 1: lim𝑥→𝜋
(𝑥 −22
7) = (𝜋 −
22
7)
3. Evaluate the Given limit: lim𝑟→1
𝜋𝑟2
Hint: Put 𝑟 = 1
Solution:
Solution step 1: lim𝑟→1
𝜋𝑟2 = 𝜋(1)2 = 𝜋
4. Evaluate the Given limit: lim𝑥→4
4𝑥+3
𝑥−2
Hint: Put 𝑥 = 4
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Solution:
Solution step 1: lim𝑥→4
4𝑥+3
𝑥−2=
4(4)+3
4−2=
16+3
2=
19
2
5. Evaluate the Given limit: lim𝑥→−1
𝑥10+𝑥5+1
𝑥−1
Hint: Put 𝑥 = −1
Solution:
Solution step 1: lim𝑥→−1
𝑥10+𝑥5+1
𝑥−1=
(−1)10+(−1)5+1
−1−1=
1−1+1
−2= −
1
2
6. Evaluate the Given limit: lim𝑥→0
(𝑥+1)5−1
𝑥
Hint: At the time when we put variable value make sure function won’t take undefined form.
Solution:
Solution step 1: lim𝑥→0
(𝑥+1)5−1
𝑥
Put 𝑥 + 1 = 𝑦so that 𝑦 → 1 as 𝑥 → 0.
Accordingly, lim𝑥→0
(𝑥+1)5−1
𝑥= lim
𝑦→1
𝑦5−1
𝑦−1
= lim𝑦→1
𝑦5 − 15
𝑦 − 1
= 5.15−1 [lim𝑥→𝑎
𝑥𝑛−𝑎𝑛
𝑥−𝑎= 𝑛𝑎𝑛−1]
= 5
∴ lim𝑥→0
(𝑥 + 1)5 − 1
𝑥= 5
7. Evaluate the Given limit: lim𝑥→2
3𝑥2−𝑥−10
𝑥2−4
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Hint: At the time when we put variable value make sure function won’t take undefined form.
Solution:
Solution step 1: At 𝑥 = 2, the value of the given rational function takes the form0
0
∴ lim𝑥→2
3𝑥2 − 𝑥 − 10
𝑥2 − 4= lim
𝑥→2
(𝑥 − 2)(2𝑥 + 5)
(𝑥 − 2)(𝑥 + 2)
lim𝑥→2
3𝑥 + 5
𝑥 + 2
=3(2) + 5
2 + 2
=11
4
8. Evaluate the Given limit: lim𝑥→3
𝑥4−81
2𝑥2−5𝑥−3
Hint: At the time when we put variable value make sure function won’t take undefined form.
Solution:
Solution step 1: At 𝑥 = 2, the value of the given rational function takes the form 0
0
∴ lim𝑥→3
𝑥4 − 81
2𝑥2 − 5𝑥 − 3= lim
𝑥→3
(𝑥 − 3)(𝑥 + 3)(𝑥2 + 9)
(𝑥 − 3)(2𝑥 + 1)
= lim𝑥→3
(𝑥 + 3)(𝑥2 + 9)
2𝑥 + 1
=(3 + 3)(32 + 9)
2(3) + 1
=6 × 18
7
=108
7
9. Evaluate the Given limit: lim𝑥→0
𝑎𝑥+𝑏
𝑐𝑥+1
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Hint: At the time when we put variable value make sure function won’t take undefined form.
Solution:
Solution step 1: lim𝑥→0
𝑎𝑥+𝑏
𝑐𝑥+1=
𝑎(0)+𝑏
𝑐(0)+1= 𝑏
10. Evaluate the Given limit: lim𝑧→1
𝑧13−1
𝑧16−1
Hint: At the time when we put variable value make sure function won’t take undefined form.
Solution:
Solution step 1: lim𝑧→1
𝑧13−1
𝑧16−1
At 𝑧 = 1, the value of the given function takes the form 0
0
Put 𝑧1
6 = 𝑥so that 𝑧 → 1 as 𝑥 → 1.
Accordingly, lim𝑧→1
𝑧13−1
𝑧16−1
= lim𝑧→1
𝑥2−1
𝑥−1
= lim𝑧→1
𝑥2 − 12
𝑥 − 1
= 2.12−1 [lim𝑥→𝑎
𝑥𝑛−𝑎𝑛
𝑥−𝑎= 𝑛𝑎𝑛−1]
= 2
= lim𝑧→1
𝑧13 − 1
𝑧16 − 1
= 2
11. Evaluate the Given limit: lim𝑥→1
𝑎𝑥2+𝑏𝑥+𝑐
𝑐𝑥2+𝑏𝑥+𝑎, 𝑎 + 𝑏 + 𝑐 ≠ 0
Hint: At the time when we put variable value make sure function won’t take undefined form.
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Solution:
Solution step 1: lim𝑥→1
𝑎𝑥2+𝑏𝑥+𝑐
𝑐𝑥2+𝑏𝑥+𝑎=
𝑎(1)2+𝑏(1)+𝑐
𝑐(1)2+𝑏(1)+𝑎
=𝑎 + 𝑏 + 𝑐
𝑎 + 𝑏 + 𝑐
= 1 [𝑎 + 𝑏 + 𝑐 ≠ 0]
12. Evaluate the Given limit: lim𝑥→−2
1
𝑥+
1
2
𝑥+2
Hint: At the time when we put variable value make sure function won’t take undefined form.
Solution:
Solution step 1: lim𝑥→−2
1
𝑥+
1
2
𝑥+2
At 𝑥 = – 2, the value of the given function takes the form
Now, lim𝑥→−2
1
𝑥+
1
2
𝑥+2= lim
𝑥→−2
1
2𝑥
=1
2(−1)=
−1
4
13. Evaluate the Given limit: lim𝑥→0
sin𝑎𝑥
𝑏𝑥
Hint: At the time when we put variable value make sure function won’t take undefined form.
Solution:
Solution step 1: lim𝑥→0
sin𝑎𝑥
𝑏𝑥
At 𝑥 = 0, the value of the given function takes the form 0
0
Now, lim𝑥→0
sin𝑎𝑥
𝑏𝑥= lim
𝑥→0
sin𝑎𝑥
𝑎𝑥×
𝑎𝑥
𝑏𝑥
lim𝑥→0
(sin 𝑎𝑥
𝑏𝑥) × (
𝑎
𝑏)
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=𝑎
𝑏lim
𝑎𝑥→0(sin𝑎𝑥
𝑏𝑥) [𝑥 → 0 ⇒ 𝑎𝑥 → 0]
=𝑎
𝑏× 1 [lim
𝑦→0
sin𝑦
𝑦= 1]
=𝑎
𝑏
14. Evaluate the Given limit: lim𝑥→0
sin𝑎𝑥
sin𝑏𝑥, 𝑎, 𝑏 ≠ 0
Hint: At the time when we put variable value make sure function won’t take undefined form.
Solution:
Solution step 1: lim𝑥→0
sin𝑎𝑥
sin𝑏𝑥, 𝑎, 𝑏 ≠ 0
At 𝑥 = 0, the value of the given function takes the form 0
0
Now, lim𝑥→0
sin𝑎𝑥
sin𝑏𝑥= lim
𝑥→0
(sin𝑎𝑥
𝑎𝑥)×𝑎𝑥
(sin𝑏𝑥
𝑏𝑥)×𝑏𝑥
= (𝑎
𝑏) ×
lim𝑎𝑥→0
(sin 𝑎𝑥
𝑎𝑥)
lim𝑏𝑥→0
(sin 𝑏𝑥
𝑏𝑥) [
𝑥 → 0 ⇒ 𝑎𝑥 → 0and 𝑥 → 0 ⇒ 𝑏𝑥 → 0
]
= (𝑎
𝑏) ×
1
1 [lim
𝑦→0
sin𝑦
𝑦= 1]
=𝑎
𝑏
15. Evaluate the Given limit: lim𝑥→𝜋
sin(𝜋−𝑥)
𝜋(𝜋−𝑥)
Hint: At the time when we put variable value make sure function won’t take undefined form.
Solution:
Solution step 1: lim𝑥→𝜋
sin(𝜋−𝑥)
𝜋(𝜋−𝑥)
It is seen that 𝑥 → 𝜋 ⇒ (𝜋 – 𝑥) → 0
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∴ lim𝑥→𝑧
sin(𝜋 − 𝑥)
𝜋(𝜋 − 𝑥)=
1
𝜋lim
(𝜋−𝑥)→0
sin(𝜋 − 𝑥)
𝜋 − 𝑥
=1
𝜋× 1 [lim
𝑦→0
sin𝑦
𝑦= 1]
=1
𝜋
16. Evaluate the Given limit: lim𝑥→0
cos𝑥
𝜋−𝑥
Hint: At the time when we put variable value make sure function won’t take undefined form.
Solution:
Solution step 1: lim𝑥→0
cos𝑥
𝜋−𝑥=
cos0
𝜋−0=
1
𝜋
17. Evaluate the Given limit: lim𝑥→0
cos2𝑥−1
cos𝑥−1
Hint: At the time when we put variable value make sure function won’t take undefined form.
Solution:
Solution step 1: lim𝑥→0
cos2𝑥−1
cos𝑥−1
At 𝑥 = 0, the value of the given function takes the form 0
0
Now,
lim𝑥→0
cos2𝑥−1
cos𝑥−1= lim
𝑥→0
1−2 sin2 𝑥−1
1−2sin2𝑥
2−1
[cos 𝑥 = 1 − 2 sin2 𝑥
2]
= lim𝑥→0
sin2 𝑥
sin2 𝑥2
= lim𝑥→0
(sin2 𝑥
𝑥 ) × 𝑥2
(sin2 𝑥
2
(𝑥2)
2 ) ×𝑥2
4
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= 4lim𝑥→0
(sin2 𝑥
𝑥2 )
lim𝑥→0
(sin2 𝑥
2
(𝑥2)
2 )
= 4(lim𝑥→0
sin 𝑥
𝑥)2
(lim𝑥2→0
sin𝑥2
𝑥2
)
2 [𝑥 → 0 ⇒𝑥
2→ 0]
= 412
12 [lim
𝑦→0
sin𝑦
𝑦= 1]
= 4
18. Evaluate the Given limit: lim𝑥→0
𝑎𝑥+𝑥 cos𝑥
𝑏 sin𝑥
Hint: At the time when we put variable value make sure function won’t take undefined form.
Solution:
Solution step 1: lim𝑥→0
𝑎𝑥+𝑥 cos𝑥
𝑏 sin𝑥
At 𝑥 = 0, the value of the given function takes the form 0
0
Now,
lim𝑥→0
𝑎𝑥 + 𝑥 cos 𝑥
𝑏 sin 𝑥=
1
𝑏lim𝑥→0
(𝑎 + cos 𝑥)
sin 𝑥
=1
𝑏lim𝑥→0
(𝑥
sin 𝑥) × lim
𝑥→0(𝑎 + cos 𝑥)
=1
𝑏×
1
(lim𝑥→0
sin 𝑥𝑥 )
× lim𝑥→0
(𝑎 + cos 𝑥)
=1
𝑏× (𝑎 + cos 0) [lim
𝑥→0
sin𝑥
𝑥= 1]
=𝑎 + 1
𝑏
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19. Evaluate the Given limit: lim𝑥→0
𝑥 sec 𝑥
Hint: At the time when we put variable value make sure function won’t take undefined form.
Solution:
Solution step 1: lim𝑥→0
𝑥 sec 𝑥 = lim𝑥→0
𝑥
cos𝑥=
0
cos0=
0
1= 0
20. Evaluate the Given limit: lim𝑥→0
sin𝑎𝑥+𝑏𝑥
𝑎𝑥+sin𝑏𝑥 𝑎, 𝑏, 𝑎 + 𝑏 ≠ 0
Hint: At the time when we put variable value make sure function won’t take undefined form.
Solution:
Solution step 1: At 𝑥 = 0, the value of the given function takes the form 0
0
Now,
lim𝑥→0
sin 𝑎𝑥 + 𝑏𝑥
𝑎𝑥 + sin 𝑏𝑥
= lim𝑥→0
(sin 𝑎𝑥
𝑎𝑥 ) 𝑎𝑥 + 𝑏𝑥
𝑎𝑥 + 𝑏𝑥 (sin 𝑏𝑥
𝑏𝑥)
=( lim𝑎𝑥→0
sin𝑎𝑥
𝑎𝑥)×lim
𝑥→0(𝑎𝑥)+lim
𝑥→0𝑏𝑥
lim𝑥→0
𝑎𝑥+ lim𝑥→0
𝑏𝑥( lim𝑏𝑥→0
sin 𝑏𝑥
𝑏𝑥)
[As 𝑥 → 0 ⇒ 𝑎𝑥 → 0 and 𝑏𝑥 → 0]
=lim𝑥→0
(𝑎𝑥)+lim𝑥→0
𝑏𝑥
lim𝑥→0
𝑎𝑥+lim𝑥→0
𝑏𝑥 [lim
𝑥→0
sin𝑥
𝑥= 1]
=lim𝑥→0
(𝑎𝑥 + 𝑏𝑥)
lim𝑥→0
(𝑎𝑥 + 𝑏𝑥)
= lim𝑥→0
(1)
= 1
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21. Evaluate the Given limit: lim𝑥→0
(cosec 𝑥 − cot 𝑥)
Hint: At the time when we put variable value make sure function won’t take undefined form.
Solution:
Solution step 1: At 𝑥 = 0, the value of the given function takes the form ∞ − ∞
Now,
lim𝑥→0
(cosec 𝑥 − cot 𝑥)
= lim𝑥→0
(1
sin 𝑥−
cos 𝑥
sin 𝑥)
= lim𝑥→0
(1 − cos 𝑥
sin 𝑥)
= lim𝑥→0
(1 − cos 𝑥
sin 𝑥 )
(sin 𝑥
𝑥 )
=lim𝑥→0
1 − cos 𝑥𝑥
lim𝑥→0
sin 𝑥𝑥
=0
1 [lim
𝑥→0
1−cos𝑥
𝑥= 0 and lim
𝑥→0
sin𝑥
𝑥= 1]
= 0
22. lim𝑥→
𝜋
2
tan2𝑥
𝑥−𝜋
2
Hint: At the time when we put variable value make sure function won’t take undefined form.
Solution:
Solution step 1: lim𝑥→
𝜋
2
tan2𝑥
𝑥−𝜋
2
At 𝑥 =𝜋
2, the value of the given function takes the form
0
0
Now put 𝑥 −𝜋
2= 𝑦 so that 𝑥 →
𝜋
2, 𝑦 → 0
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∴ lim𝑥→
𝜋2
tan 2𝑥
𝑥 −𝜋2
= lim𝑥→0
tan 2 (𝑦 +𝜋2)
𝑦
= lim𝑦→0
tan(𝜋 + 2𝑦)
𝑦
= lim𝑦→0
tan2𝑦
𝑦 [tan(𝜋 + 2𝑦) = tan 2𝑦]
= lim𝑦→0
sin 2𝑦
𝑦 cos 2𝑦
= lim𝑦→0
(sin 2𝑦
2𝑦×
2
cos 2𝑦)
= ( lim2𝑦→0
sin2𝑦
2𝑦) × lim
𝑦→0(
2
cos2𝑦) [𝑦 → 0 ⇒ 2𝑦 → 0]
= 1 ×2
cos0 [lim
𝑥→0
sin𝑥
𝑥= 1]
= 1 ×2
1
= 2
23. Find lim𝑥→0
𝑓(𝑥) and lim𝑥→1
𝑓(𝑥) = {2𝑥 + 3, 𝑥 ≤ 0
3(𝑥 + 1), 𝑥 > 0
Hint: At the time when we put variable value make sure function won’t take undefined form.
Solution:
Solution step 1: The given function is 𝑓(𝑥) = {2𝑥 + 3, 𝑥 ≤ 0
3(𝑥 + 1), 𝑥 > 0
lim𝑥→0−
𝑓(𝑥) = lim𝑥→0
[2𝑥 + 3] = 2(0) + 3 = 3
lim𝑥→0−
𝑓(𝑥) = lim𝑥→0
3[𝑥 + 1] = 3(0 + 1) = 3
∴ lim𝑥→0−
𝑓(𝑥) = lim𝑥→0
𝑓(𝑥) = lim𝑥→0
𝑓(𝑥) = 3
lim𝑥→1−
𝑓(𝑥) = lim𝑥→1
3[𝑥 + 1] = 3(1 + 1) = 6
lim𝑥→1−
𝑓(𝑥) = lim𝑥→1
2(𝑥 + 1) = 3(1 + 1) = 6
∴ lim𝑥→1−
𝑓(𝑥) = lim𝑥→1+
𝑓(𝑥) = lim𝑥→1
𝑓(𝑥) = 6
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24. Find lim𝑥→1
𝑓(𝑥), where 𝑓(𝑥) = {𝑥2 − 1, 𝑥 ≤ 1
−𝑥2 − 1, 𝑥 > 1
Hint: At the time when we put variable value make sure function won’t take undefined form.
Solution:
Solution step 1: The given function is
𝑓(𝑥) = {𝑥2 − 1, 𝑥 ≤ 1
−𝑥2 − 1, 𝑥 > 1
lim𝑥→1−
𝑓(𝑥) = lim𝑥→1
[𝑥2 − 1] = 12 − 1 = 1 − 1 = 0
lim𝑥→1+
𝑓(𝑥) = lim𝑥→1
[−𝑥2 − 1] = −12 − 1 = −1 − 1 = −2
It is observed that that lim𝑥→1−
𝑓(𝑥) ≠ lim𝑥→1+
𝑓(𝑥).
Hence, lim𝑥→1
𝑓(𝑥) does not exist.
25. Evaluate lim𝑥→0
𝑓(𝑥), where 𝑓(𝑥) = {|𝑥|
𝑥, 𝑥 ≠ 0
0, 𝑥 = 0
Hint: At the time when we put variable value make sure function won’t take undefined form.
Solution:
Solution step 1: The given function is 𝑓(𝑥) = {|𝑥|
𝑥, 𝑥 ≠ 0
0, 𝑥 = 0
lim𝑥→0−
𝑓(𝑥) = lim𝑥→0−
[|𝑥|
𝑥]
= lim𝑥→0
(−𝑥
𝑥) [When 𝑥 is negative, |𝑥| = −𝑥]
= lim𝑥→0
(−1)
= −1
lim𝑥→0+
𝑓(𝑥) = lim𝑥→0+
[|𝑥|
𝑥]
= lim𝑥→0
[𝑥
𝑥] [When 𝑥 is positive, |𝑥| = 𝑥]
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= lim𝑥→0
(1)
= 1
It is observed that lim𝑥→0−
𝑓(𝑥) ≠ lim𝑥→0+
𝑓(𝑥).
Hence, lim𝑥→0
𝑓(𝑥) does not exist.
26. Find lim𝑥→0
𝑓(𝑥), where 𝑓(𝑥) = {
𝑥
|𝑥|, 𝑥 ≠ 0
0, 𝑥 = 0
Hint: At the time when we put variable value make sure function won’t take undefined form.
Solution:
Solution step 1: The given function is
𝑓(𝑥) = {
𝑥
|𝑥|, 𝑥 ≠ 0
0, 𝑥 = 0
lim𝑥→0−
𝑓(𝑥) = lim𝑥→0−
[𝑥
|𝑥|]
= lim𝑥→0
[𝑥
−𝑥] [When 𝑥 < 0, |𝑥| = −𝑥]
= lim𝑥→0
(−1)
= −1
lim𝑥→0+
𝑓(𝑥) = lim𝑥→0+
[𝑥
|𝑥|]
= lim𝑥→0
[𝑥
𝑥] [When 𝑥 > 0, |𝑥| = 𝑥]
It is observed that lim𝑥→0−
𝑓(𝑥) ≠ lim𝑥→0+
𝑓(𝑥).
Hence, lim𝑥→0
𝑓(𝑥) does not exist.
27. Find lim𝑥→5
𝑓(𝑥), where 𝑓(𝑥) = |𝑥| − 5
Hint: At the time when we put variable value make sure function won’t take undefined form.
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Solution:
Solution step 1: The given function is 𝑓(𝑥) = |𝑥| − 5
lim𝑥→5−
𝑓(𝑥) = lim𝑥→5−
[|𝑥| − 5]
= lim𝑥→5
(𝑥 − 5) [When 𝑥 > 0, |𝑥| = 𝑥]
= 5 − 5
= 0
lim𝑥→5+
𝑓(𝑥) = lim𝑥→5+
(|𝑥| − 5)
lim𝑥→5
(𝑥 − 5) [When 𝑥 > 0, |𝑥| = 𝑥]
= 5 − 5
= 0
∴ lim𝑥→5−
𝑓(𝑥) = lim𝑥→5+
𝑓(𝑥) = 0
Hence, lim𝑥→5
𝑓(𝑥) = 0
28. Suppose 𝑓(𝑥) = {
𝑎 + 𝑏𝑥, 𝑖𝑓 𝑥 < 14, 𝑖𝑓 𝑥 = 0
𝑏 − 𝑎𝑥, 𝑖𝑓 𝑥 > 1
lim𝑥→1
𝑓(𝑥) = 𝑓(1) what are possible values of 𝑎 and 𝑏?
Hint: At the time when we put variable value make sure function won’t take undefined form.
Solution:
Solution step 1: The given function is
𝑓(𝑥) = {
𝑎 + 𝑏𝑥, 𝑖𝑓 𝑥 < 14, 𝑖𝑓 𝑥 = 0
𝑏 − 𝑎𝑥, 𝑖𝑓 𝑥 > 1
lim𝑥→1+
𝑓(𝑥) = lim𝑥→1
(𝑎 + 𝑏𝑥) = 𝑎 + 𝑏
lim𝑥→1+
𝑓(𝑥) = lim𝑥→1
(𝑏 − 𝑎𝑥) = 𝑏 − 𝑎
𝑓(1) = 4
It is given that lim𝑥→1
𝑓(𝑥) = 𝑓(1)
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∴ lim𝑥→1−
𝑓(𝑥) = lim𝑥→1+
𝑓(𝑥) = lim𝑥→1
𝑓(𝑥) = 𝑓(1)
⇒ 𝑎 + 𝑏 = 4 and 𝑏 − 𝑎 = 4
Thus, on solving these two equations, we obtain 𝑎 = 0 and 𝑏 = 4
Thus, the respective possible values of 𝑎 and 𝑏 are 0 and 4.
29. Let 𝑎1, 𝑎2, … , 𝑎𝑛be fixed real numbers and define a function
𝑓(𝑥) = (𝑥 − 𝑎1)(𝑥 − 𝑎2)… (𝑥 − 𝑎𝑛)
What is lim𝑥→𝑎1
𝑓(𝑥) ? For some 𝑎 ≠ 𝑎1, 𝑎2, … 𝑎𝑛compute lim𝑥→𝑎
𝑓(𝑥)
Hint: At the time when we put variable value make sure function won’t take undefined form.
Solution:
Solution step 1: The given function is 𝑓(𝑥) = (𝑥 − 𝑎1)(𝑥 − 𝑎2)… (𝑥 − 𝑎𝑛)
lim𝑥→𝑎1
𝑓(𝑥) = lim𝑥→𝑎1
[(𝑥 − 𝑎1)(𝑥 − 𝑎2)… (𝑥 − 𝑎𝑛)]
= [ lim𝑥→𝑎1
(𝑥 − 𝑎1)] [ lim𝑥→𝑎1
(𝑥 − 𝑎2)]… [ lim𝑥→𝑎1
(𝑥 − 𝑎𝑛)]
= (𝑎1 − 𝑎1)(𝑎1 − 𝑎2)… (𝑎1 − 𝑎𝑛) = 0
∴ lim𝑥→𝑎1
𝑓(𝑥) = 0
Now,lim𝑥→𝑎
𝑓(𝑥) = lim𝑥→𝑎
[(𝑥 − 𝑎1)(𝑥 − 𝑎2)… (𝑥 − 𝑎𝑛)]
= [lim𝑥→𝑎
(𝑥 − 𝑎1)] [lim𝑥→𝑎
(𝑥 − 𝑎2)]… [lim𝑥→𝑎
(𝑥 − 𝑎𝑛)]
= (𝑎 − 𝑎1)(𝑎 − 𝑎2)… (𝑎 − 𝑎𝑛)
∴ lim𝑥→𝑎
𝑓(𝑥) = (𝑎 − 𝑎1)(𝑎 − 𝑎2)… (𝑎 − 𝑎𝑛)
30. If 𝑓(𝑥) = {|𝑥| + 1, 𝑥 < 0
0, 𝑥 = 0|𝑥| − 1, 𝑥 > 0
For what value(s) of a does lim𝑥→𝑎
𝑓(𝑥) exists?
Hint: At the time when we put variable value make sure function won’t take undefined form.
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Solution:
Solution step 1: The given function is
𝑓(𝑥) = {|𝑥| + 1, 𝑥 < 0
0, 𝑥 = 0|𝑥| − 1, 𝑥 > 0
When 𝑎 = 0
lim𝑥→0
𝑓(𝑥) = lim𝑥→0−
(|𝑥| + 1)
= lim𝑥→0
(|𝑥| + 1)[If 𝑥 < 0, |𝑥| = −𝑥]
= −0 + 1
= 1
lim𝑥→0
𝑓(𝑥) = lim𝑥→0+
(|𝑥| − 1)
lim𝑥→0
(|𝑥| − 1)[ If 𝑥0, |𝑥| = 𝑥]
= 0 − 1
= −1
Here, it is observed that lim𝑥→0−
𝑓(𝑥) ≠ lim𝑥→0+
𝑓(𝑥).
∴ lim𝑥→0
𝑓(𝑥)does not exist.
When 𝑎 < 0
lim𝑥→𝑎−
𝑓(𝑥) = lim𝑥→𝑎−
(|𝑥| + 1)
= lim𝑥→𝑎
(−𝑥 + 1)[𝑥 < 𝑎 < 0 ⇒ |𝑥| = −𝑥]
= −𝑎 + 1
lim𝑥→𝑎+
𝑓(𝑥) = lim𝑥→𝑎+
(|𝑥| + 1)
= lim𝑥→𝑎
(−𝑥 + 1)[𝑎 < 𝑥 < 0 ⇒ |𝑥| = −𝑥]
= −𝑎 + 1
∴ lim𝑥→𝑎−
𝑓(𝑥) = lim𝑥→𝑎+
𝑓(𝑥) = −𝑎 + 1
Thus, limit of𝑓(𝑥) exists for all 𝑎 = 0, where 𝑎 < 0.
When 𝑎 > 0
lim𝑥→𝑎−
𝑓(𝑥) = lim𝑥→𝑎−
(|𝑥| − 1)
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= lim𝑥→𝑎
(−𝑥 − 1)[0 < 𝑥 < 𝑎 ⇒ |𝑥| = 𝑥]
= 𝑎 − 1
lim𝑥→𝑎+
𝑓(𝑥) = lim𝑥→𝑎+
(|𝑥| − 1)
= lim𝑥→𝑎
(𝑥 − 1)[0 < 𝑎 < 𝑥 ⇒ |𝑥| = 𝑥]
= 𝑎 − 1
∴ lim𝑥→𝑎−
𝑓(𝑥) = lim𝑥→𝑎+
𝑓(𝑥) = 𝑎 − 1
Thus, limit of 𝑓(𝑥) exist at 𝑥 = 𝑎, where 𝑎 > 0.
Thus, lim𝑥→𝑎
𝑓(𝑥) exists for all 𝑎 ≠ 0.
Marks for step 1: 3
Difficulty level 1: M
Page no.: 303
Question No.:
Question body: If the function 𝑓(𝑥) satisfies lim𝑥→1
𝑓(𝑥)−2
𝑥2−1= 𝜋,evaluate lim
𝑥→1𝑓(𝑥)
Full marks: 3
Overall difficulty Level: M
Hint: At the time when we put variable value make sure function won’t take undefined form.
Solution:
Solution step 1: lim𝑥→1
𝑓(𝑥)−2
𝑥2−1= 𝜋
⇒lim(𝑓(𝑥) − 2)
lim(𝑥2 − 1)= 𝜋
⇒ lim𝑥→1
𝑓(𝑥) − 2) = 𝜋lim𝑥→1
(𝑥2 − 1)
⇒ lim𝑥→1
(𝑓(𝑥) − 2) = 𝜋(12 − 1)
⇒ lim𝑥→1
𝑓(𝑥) − 2) = 0
⇒ lim𝑥→1
𝑓(𝑥) − lim𝑥→1
2 = 0
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⇒ lim𝑥→1
𝑓(𝑥) − 2 = 0
∴ lim𝑥→1
𝑓(𝑥) = 2
31. If
𝑓(𝑥) = {𝑚𝑥2 + 𝑛, 𝑥 < 0𝑛𝑥 + 𝑚, 0 ≤ 𝑥 ≤ 1
𝑛𝑥3 + 𝑚, 𝑥 > 1
For what integers 𝑚 and 𝑛 does lim𝑥→0
𝑓(𝑥) and lim𝑥→1
𝑓(𝑥) exist?
Hint: At the time when we put variable value make sure function won’t take undefined form.
Solution:
Solution step 1: The given function is
𝑓(𝑥) = {𝑚𝑥2 + 𝑛, 𝑥 < 0𝑛𝑥 + 𝑚, 0 ≤ 𝑥 ≤ 1
𝑛𝑥3 + 𝑚, 𝑥 > 1
lim𝑥→0−
𝑓(𝑥) = lim𝑥→0
(𝑚𝑥2 + 𝑛)
= 𝑚(0)2 + 𝑛
= 𝑛
lim𝑥→0+
𝑓(𝑥) = lim𝑥→0
(𝑛𝑥 + 𝑚)
= 𝑛(0) + 𝑚
= 𝑚.
Thus, lim𝑥→0
𝑓(𝑥) exists if 𝑚 = 𝑛.
lim𝑥→1−
𝑓(𝑥) = lim𝑥→1
(𝑛𝑥 + 𝑚)
= 𝑛(1) + 𝑚
= 𝑚 + 𝑛
lim𝑥→1+
𝑓(𝑥) = lim𝑥→1
(𝑛𝑥3 + 𝑚)
= 𝑛(1)3 + 𝑚
= 𝑚 + 𝑛
∴ lim𝑥→1−
𝑓(𝑥) = lim𝑥→1+
𝑓(𝑥) = lim𝑥→1
𝑓(𝑥)
Thus, lim𝑥→1
𝑓(𝑥)exists for any integral value of 𝑚 and 𝑛.
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Exercise 13.2
1. Find the derivative of (𝑥2) − 2 at 𝑥 = 10.
Hint: 𝑓′(𝑎) = limℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
Solution:
Solution step 1: Let 𝑓(𝑥) = 𝑥2 − 2. Accordingly,
𝑓′(10) = limℎ→0
𝑓(10 + ℎ) − 𝑓(10)
ℎ
= limℎ→0
[(10 + ℎ)2 − 2] − (102 − 2)
ℎ
= limℎ→0
102 + 2 ⋅ 10 ⋅ ℎ + ℎ2 − 2 − 102 + 2
ℎ
= limℎ→0
20ℎ + ℎ2
ℎ= lim
ℎ→0(20 + ℎ) = (20 + 0) = 20
Thus, the derivative of (𝑥2) − 2 at 𝑥 = 10 is 20.
2. Find the derivative of 99𝑥 at 𝑥 = 100.
Solution:
Let 𝑓(𝑥) = 99𝑥. Accordingly,
𝑓′(100) = limℎ→0
𝑓(100 + ℎ) − 𝑓(100)
ℎ
= limℎ→0
99(100 + ℎ) − 99(100)
ℎ
= limℎ→0
99 × 100 + 99ℎ − 99 × 100
ℎ
= limℎ→0
99ℎ
ℎ= lim
ℎ→0(99) = 99
Thus, the derivative of 99𝑥 at 𝑥 = 100 is 99.
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3. Find the derivative of 𝑥 at 𝑥 = 1.
Hint: 𝑓′(𝑎) = limℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
Solution:
Solution step 1: Let 𝑓(𝑥) = 𝑥. Accordingly,
𝑓′(1) = limℎ→0
𝑓(1 + ℎ) − 𝑓(1)
ℎ
= limℎ→0
(1 + ℎ) − 1
ℎ
= limℎ→0
ℎ
ℎ
= limℎ→0
(1)
= 1
Thus, the derivative of 𝑥 at 𝑥 = 1 is 1.
4. Find the derivative of the following functions from first principle.
(i) 𝑥3 − 27
Hint: 𝑓′(𝑎) = limℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
Solution:
Solution step 1: Let 𝑓(𝑥) = 𝑥3 − 27.Accordingly, from the first principle,
𝑓′(𝑥) = limℎ→0
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
= limℎ→0
[(𝑥 + ℎ)3 − 27] − (𝑥3 − 27)
ℎ
= limℎ→0
𝑥3 + ℎ3 + 3𝑥2ℎ + 3𝑥ℎ2 − 𝑥3
ℎ
= limℎ→0
ℎ3 + 3𝑥2ℎ + 3𝑥ℎ2
ℎ
= limℎ→0
(ℎ2 + 3𝑥2 + 3𝑥ℎ)
= 0 + 3𝑥2 + 0 = 3𝑥2
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(ii)(𝑥 − 1)(𝑥 − 2)
Hint: 𝑓′(𝑎) = limℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
Solution:
Solution step 1: Let 𝑓(𝑥) = (𝑥– 1)(𝑥– 2). Accordingly, from the first principle,
𝑓′(𝑥) = limℎ→0
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
= limℎ→0
(𝑥 + ℎ − 1)(𝑥 + ℎ − 2) − (𝑥 − 1)(𝑥 − 2)
ℎ
= limℎ→0
(𝑥2 + ℎ𝑥 − 2𝑥 + ℎ𝑥 + ℎ2 − 2ℎ − 𝑥 − ℎ + 2) − (𝑥2 − 2𝑥 − 𝑥 + 2)
ℎ
= limℎ→0
(ℎ𝑥 + ℎ𝑥 + ℎ2 − 2ℎ − ℎ)
ℎ
= limℎ→0
2ℎ𝑥 + ℎ2 − 3ℎ
ℎ
= limℎ→0
(2𝑥 + ℎ − 3)
= (2𝑥 + 0 − 3)
= 2𝑥 − 3
(iii)1
𝑥2
Hint: 𝑓′(𝑎) = limℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
Solution:
Solution step 1: Let 𝑓(𝑥) =1
𝑥2 . Accordingly, from the first principle,
𝑓′(𝑥) = limℎ→0
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
= limℎ→0
1(𝑥 + ℎ)2 −
1𝑥2
ℎ
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= limℎ→0
1
ℎ[𝑥2 − (𝑥 + ℎ)2
𝑥2(𝑥 + ℎ)2]
= limℎ→0
1
ℎ[𝑥2 − 𝑥2 − ℎ2 − 2ℎ𝑥
𝑥2(𝑥 + ℎ)2]
= limℎ→0
1
ℎ[−ℎ2 − 2ℎ𝑥
𝑥2(𝑥 + ℎ)2]
= limℎ→0
[−ℎ − 2𝑥
𝑥2(𝑥 + ℎ)2]
=0 − 2𝑥
𝑥2(𝑥 + 0)2
=−2
𝑥3
(iv) 𝑥+1
𝑥−1
Hint: 𝑓′(𝑎) = limℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
Solution:
Solution step 1: Let 𝑓(𝑥) =𝑥+1
𝑥−1. Accordingly, from the first principle,
𝑓′(𝑥) = limℎ→0
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
= limℎ→0
(𝑥 + ℎ + 1𝑥 + ℎ − 1
−𝑥 + 1𝑥 − 1
)
ℎ
= limℎ→0
1
ℎ[(𝑥 − 1)(𝑥 + ℎ + 1) − (𝑥 + 1)(𝑥 + ℎ − 1)
(𝑥 − 1)(𝑥 + ℎ − 1)]
= limℎ→0
1
ℎ[(𝑥2 + ℎ𝑥 + 𝑥 − 𝑥 − ℎ − 1) − (𝑥2 + ℎ𝑥 − 𝑥 + 𝑥 + ℎ − 1)
(𝑥 − 1)(𝑥 + ℎ − 1)]
= limℎ→0
1
ℎ[
−2ℎ
(𝑥 − 1)(𝑥 + ℎ − 1)]
= limℎ→0
[−2
(𝑥 − 1)(𝑥 + ℎ − 1)]
=−2
(𝑥 − 1)(𝑥 − 1)=
−2
(𝑥 − 1)2
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5. For the function:
𝑓(𝑥) =𝑥100
100+
𝑥99
99+ ⋯ +
𝑥2
2+ 𝑥 + 1
Prove That 𝑓′(1) = 100𝑓′(0)
Hint: 𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1
Solution:
Solution step 1: The given function is
𝑓(𝑥) =𝑥100
100+
𝑥99
99+ ⋯ +
𝑥2
2+ 𝑥 + 1
𝑑
𝑑𝑥𝑓(𝑥) =
𝑑
𝑑𝑥[𝑥100
100+
𝑥99
99+ ⋯+
𝑥2
2+ 𝑥 + 1]
𝑑
𝑑𝑥𝑓(𝑥) =
𝑑
𝑑𝑥(𝑥100
100) +
𝑑
𝑑𝑥(𝑥99
99) + ⋯+
𝑑
𝑑𝑥(𝑥2
2) +
𝑑
𝑑𝑥(𝑥) +
𝑑
𝑑𝑥(1)
On using theorem 𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1, we obtain
𝑑
𝑑𝑥𝑓(𝑥) =
100𝑥99
100+
99𝑥98
99+ ⋯+
2𝑥
2+ 1 + 0
= 𝑥99 + 𝑥98 + ⋯+ 𝑥 + 1
∴ 𝑓′(𝑥) = 𝑥99 + 𝑥98 + ⋯+ 𝑥 + 1
At 𝑥 = 0,
𝑓′(0) = 1
At 𝑥 = 1,
𝑓′(1) = 199 + 198 + ⋯+ 1 + 1 = [1 + 1 + ⋯+ 1 + 1]100 terms = 1 × 100 = 100
Thus, 𝑓′(1) = 100 × 𝑓1(0)
6. Find the derivative of (𝑥𝑛 + 𝑎𝑥𝑛−1 + 𝑎2𝑥𝑛−2 + ⋯ + 𝑎𝑛−1𝑥 + 𝑎𝑛) for some fixed real number 𝑎.
Hint: 𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1
Solution:
Solution step 1: Let 𝑓(𝑥) = 𝑥𝑛 + 𝑎𝑥𝑛−1 + 𝑎2𝑥𝑛−2 + ⋯+ 𝑎𝑛−1𝑥 + 𝑎𝑛
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∴ 𝑓′(𝑥) =𝑑
𝑑𝑥(𝑥𝑛 + 𝑎𝑥𝑛−1 + 𝑎2𝑥𝑛−2 + ⋯+ 𝑎𝑛−1𝑥 + 𝑎𝑛)
=𝑑
𝑑𝑥(𝑥𝑛) + 𝑎
𝑑
𝑑𝑥(𝑥𝑛−1) + 𝑎2
𝑑
𝑑𝑥(𝑥𝑛−2) + ⋯+ 𝑎𝑛−1
𝑑
𝑑𝑥(𝑥) + 𝑎𝑛
𝑑
𝑑𝑥(1)
On using theorem 𝑑
𝑑𝑥𝑥𝑛 = 𝑛𝑥𝑛−1, we obtain
𝑓′(𝑥) = 𝑛𝑥𝑛−1 + 𝑎(𝑛 − 1)𝑥𝑛−2 + 𝑎2(𝑛 − 2)𝑥𝑛−3 + ⋯+ 𝑎𝑛−1 + 𝑎𝑛(0)
= 𝑛𝑥𝑛−1 + 𝑎(𝑛 − 1)𝑥𝑛−2 + 𝑎2(𝑛 − 2)𝑥𝑛−3 + ⋯ + 𝑎𝑛−1
7. For some constants 𝑎 and 𝑏, find the derivative of:
(i) (𝑥 − 𝑎)(𝑥 − 𝑏)
Hint: 𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1
Solution:
Solution step 1: Let 𝑓(𝑥) = (𝑥– 𝑎)(𝑥– 𝑏)
⇒ 𝑓(𝑥) = 𝑥2 − (𝑎 + 𝑏)𝑥 + 𝑎𝑏
∴ 𝑓′(𝑥) =𝑑
𝑑𝑥(𝑥2 − (𝑎 + 𝑏)𝑥 + 𝑎𝑏)
=𝑑
𝑑𝑥(𝑥2) − (𝑎 + 𝑏)
𝑑
𝑑𝑥(𝑥) +
𝑑
𝑑𝑥(𝑎𝑏)
On using theorem 𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1, text we obtain
𝑓′(𝑥) = 2𝑥 − (𝑎 + 𝑏) + 0 = 2𝑥 − 𝑎 − 𝑏
(ii)(𝑎𝑥2 + 𝑏)2
Hint: 𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1
Solution:
Solution step 1: Let 𝑓(𝑥) = (𝑎𝑥2 + 𝑏)2
⇒ 𝑓(𝑥) = 𝑎2𝑥4 + 2𝑎𝑏𝑥2 + 𝑏2
∴ 𝑓′(𝑥) =𝑑
𝑑𝑥(𝑎2𝑥4 + 2𝑎𝑏𝑥2 + 𝑏2) = 𝑎2
𝑑
𝑑𝑥(𝑥4) + 2𝑎𝑏
𝑑
𝑑𝑥(𝑥2) +
𝑑
𝑑𝑥(𝑏2)
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On using theorem 𝑑
𝑑𝑥𝑥𝑛 = 𝑛𝑥𝑛−1, we obtain
𝑓′(𝑥) = 𝑎2(4𝑥3) + 2𝑎𝑏(2𝑥) + 𝑏2(0)
= 4𝑎2𝑥3 + 4𝑎𝑏𝑥
= 4𝑎𝑥(𝑎𝑥2 + 𝑏)
(iii)𝑥−𝑎
𝑥−𝑏
Hint: 𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1
Solution:
Solution step 1: Let 𝑓(𝑥) =(𝑥−𝑎)
(𝑥−𝑏)
⇒ 𝑓′(𝑥) =𝑑
𝑑𝑥(𝑥 − 𝑎
𝑥 − 𝑏)
By quotient rule,
𝑓′(𝑥) =(𝑥 − 𝑏)
𝑑𝑑𝑥
(𝑥 − 𝑎) − (𝑥 − 𝑎)𝑑𝑑𝑥
(𝑥 − 𝑏)
(𝑥 − 𝑏)2
=(𝑥 − 𝑏)(1) − (𝑥 − 𝑎)(1)
(𝑥 − 𝑏)2
=𝑥 − 𝑏 − 𝑥 + 𝑎
(𝑥 − 𝑏)2
=𝑎 − 𝑏
(𝑥 − 𝑏)2
8. Find the derivative of 𝑥𝑛−𝑎𝑛
𝑥−𝑎 for some constant 𝑎.
Hint: 𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1
Solution:
Solution step 1: Let 𝑓(𝑥) =𝑥𝑛−𝑎𝑛
𝑥−𝑎
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⇒ 𝑓′(𝑥) =𝑑
𝑑𝑥(𝑥𝑛 − 𝑎𝑛
𝑥 − 𝑎)
By quotient rule,
𝑓′(𝑥) =(𝑥 − 𝑎)
𝑑𝑑𝑥
(𝑥𝑛 − 𝑎𝑛) − (𝑥𝑛 − 𝑎𝑛)𝑑𝑑𝑥
(𝑥 − 𝑎)
(𝑥 − 𝑎)2
=(𝑥 − 𝑎)(𝑛𝑥𝑛−1 − 0) − (𝑥𝑛 − 𝑎𝑛)
(𝑥 − 𝑎)2
=𝑛𝑥𝑛 − 𝑎𝑛𝑥𝑛−1 − 𝑥𝑛 + 𝑎𝑛
(𝑥 − 𝑎)2
9. Find the derivative of
(i) 2𝑥 −3
4
Hint: 𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1
Solution:
Solution step 1:Let 𝑓(𝑥) = 2𝑥 −3
4
𝑓′(𝑥) =𝑑
𝑑𝑥(2𝑥 −
3
4)
= 2𝑑
𝑑𝑥(𝑥) −
𝑑
𝑑𝑥(3
4)
= 2 − 0
= 2
(ii)(5𝑥3 + 3𝑥 − 1)(𝑥 − 1)
Hint: 𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1
Solution:
Solution step 1: Let 𝑓(𝑥) = (5𝑥3 + 3𝑥 − 1)(𝑥 − 1)
By Leibnitz product rule,
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𝑓′(𝑥) = (5𝑥3 + 3𝑥 − 1)𝑑
𝑑𝑥(𝑥 − 1) + (𝑥 − 1)
𝑑
𝑑𝑥(5𝑥3 + 3𝑥 − 1)
= (5𝑥3 + 3𝑥 − 1)(1) + (𝑥 − 1)(5.3𝑥2 + 3 − 0)
= (5𝑥3 + 3𝑥 − 1) + (𝑥 − 1)(15𝑥2 + 3)
= 5𝑥3 + 3𝑥 − 1 + 15𝑥3 + 3𝑥 − 15𝑥2 − 3
= 20𝑥3 − 15𝑥2 + 6𝑥 − 4
(iii)𝑥−3(5 + 3𝑥)
Hint: 𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1
Solution:
Solution step 1: Let 𝑓(𝑥) = 𝑥−3(5 + 3𝑥)
By Leibnitz product rule,
𝑓′(𝑥) = 𝑥−3𝑑
𝑑𝑥(5 + 3𝑥) + (5 + 3𝑥)
𝑑
𝑑𝑥(𝑥−3)
= 𝑥−3(0 + 3) + (5 + 3𝑥)(−3𝑥−3−1)
= 𝑥−3(3) + (5 + 3𝑥)(−3𝑥−4)
= 3𝑥−3 − 15𝑥−4 − 9𝑥−3
= −6𝑥−3 − 15𝑥−4
= −3𝑥−3 (2 +5
𝑥)
=−3𝑥−3
𝑥(2𝑥 + 5)
=−3
𝑥4(5 + 2𝑥)
(iv)𝑥5(3 − 6𝑥−9)
Hint: 𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1
Solution:
Solution step 1: Let 𝑓(𝑥) = 𝑥5(3 − 6𝑥−9)
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By Leibnitz product rule,
𝑓′(𝑥) = 𝑥5𝑑
𝑑𝑥(3 − 6𝑥−9) + (3 − 6𝑥−9)
𝑑
𝑑𝑥(𝑥5)
= 𝑥5{0 − 6(−9)𝑥−9−1} + (3 − 6𝑥−9)(5𝑥4)
= 𝑥5(54𝑥−10) + 15𝑥4 − 30𝑥−5
= 54𝑥−5 + 15𝑥4 − 30𝑥−5
= 24𝑥−5 + 15𝑥4
= 15𝑥4 +24
𝑥5
(v)𝑥−4(3 − 4𝑥−5)
Hint: 𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1
Solution:
Solution step 1: Let 𝑓(𝑥) = 𝑥−4(3 − 4𝑥−5)
By Leibnitz product rule,
𝑓′(𝑥) = 𝑥−4𝑑
𝑑𝑥(3 − 4𝑥−5) + (3 − 4𝑥−5)
𝑑
𝑑𝑥(𝑥−4)
= 𝑥−4{0 − 4(−5)𝑥−5−1} + (3 − 4𝑥−5)(−4)𝑥−4−1
= 𝑥−4(20𝑥−6) + (3 − 4𝑥−5)(−4𝑥−5)
= 20𝑥−10 − 12𝑥−5 + 16𝑥−10
= 36𝑥−10 − 12𝑥−5
= −12
𝑥5+
36
𝑥10
(vi)2
𝑥+1−
𝑥2
3𝑥−1
Hint: 𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1
Solution:
Solution step 1:Let 𝑓(𝑥) = 𝑐2
𝑥+1−
𝑥2
3𝑥−1
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𝑓′(𝑥) =𝑑
𝑑𝑥(
2
𝑥 + 1) −
𝑑
𝑑𝑥(
𝑥2
3𝑥 − 1)
By quotient rule,
𝑓′(𝑥) = [(𝑥 + 1)
𝑑𝑑𝑥
(2) − 2𝑑𝑑𝑥
(𝑥 + 1)
𝑑𝑥] − [
(3𝑥 − 1)𝑑𝑑𝑥
(𝑥2) − 𝑥2 𝑑𝑑𝑥
(3𝑥 − 1)
𝑑𝑥]
= [(𝑥 + 1)(0) − 2(1)
(𝑥 + 1)2] − [
(3𝑥 − 1)(2𝑥) − (𝑥2)(3)
(3𝑥 − 1)2]
=−2
(𝑥 + 1)2− [
6𝑥2 − 2𝑥 − 3𝑥2
(3𝑥 − 1)2]
=−2
(𝑥 + 1)2− [
3𝑥2 − 2𝑥2
(3𝑥 − 1)2]
=−2
(𝑥 + 1)2−
𝑥(3𝑥 − 2)
(3𝑥 − 1)2
10. Find the derivative of cos 𝑥 from first principle.
Hint: 𝑓′(𝑎) = limℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
Solution:
Solution step 1: Let 𝑓(𝑥) = 𝑐𝑜𝑠 𝑥. Accordingly, from the first principle,
𝑓′(𝑥) = limℎ→0
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
= limℎ→0
cos(𝑥 + ℎ) − cos𝑥
ℎ
= limℎ→0
[cos𝑥cosℎ − sin𝑥sinℎ − cos𝑥
ℎ]
= limℎ→0
[−cos𝑥(1 − coshℎ) − sin𝑥sinℎ
ℎ]
= limℎ→0
[−cos𝑥(1 − cosℎ)
ℎ−
sin𝑥sinℎ
ℎ]
= −cos𝑥 (limℎ→0
1 − cosℎ
ℎ) − sin𝑥lim
ℎ→0(sinℎ
ℎ)
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= −cos𝑥(0) − sin𝑥(1) [limℎ→0
1 − cosℎ
ℎ= 0 and lim
ℎ→0
sinℎ
ℎ= 1]
∴ 𝑓′(𝑥) = −sin𝑥
11. Find the derivative of the following functions:
(i) sin 𝑥 cos 𝑥
Hint: 𝑓′(𝑎) = limℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
Solution:
Solution step 1: Let 𝑓(𝑥) = sin 𝑥 cos 𝑥. Accordingly, from the first principle,
𝑓′(𝑥) = &limℎ→0
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
= limℎ→0
sin(𝑥 + ℎ) cos(𝑥 + ℎ) − sin𝑥cos𝑥
ℎ
= limℎ→0
1
2ℎ[2 sin(𝑥 + ℎ) cos(𝑥 + ℎ) − 2sin𝑥 cos𝑥]
= limℎ→0
1
2ℎ[sin2(𝑥 + ℎ) − sin2𝑥]
= limℎ→0
1
2ℎ[2cos
2𝑥 + 2ℎ + 2𝑥
2⋅ sin
2𝑥 + 2ℎ − 2𝑥
2]
= limℎ→0
1
ℎ[cos
4𝑥 + 2ℎ
2sin
2ℎ
2]
= limℎ→0
1
ℎ[cos(2𝑥 + ℎ) sinℎ]
= limℎ→0
cos(2𝑥 + ℎ) ⋅ limℎ→0
sinℎ
ℎ
= cos(2𝑥 + 0) . 1
= cos2𝑥
(ii) sec 𝑥
Hint: 𝑓′(𝑎) = limℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
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Solution:
Solution step 1: Let 𝑓(𝑥) = sec 𝑥. Accordingly, from the first principle,
𝑓′(𝑥) = limℎ→0
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
= limℎ→0
sec(𝑥 + ℎ) − sec𝑥
ℎ
= limℎ→0
1
ℎ[
1
cos(𝑥 + ℎ)−
1
cos𝑥]
= limℎ→0
1
ℎ[cos𝑥 − cos(𝑥 + ℎ)
cos𝑥cos(𝑥 + ℎ)]
=1
cos𝑥limℎ→0
1
ℎ[−2 sin (
𝑥 + 𝑥 + ℎ2 ) sin (
𝑥 − 𝑥 − ℎ2 )
cos(𝑥 + ℎ)]
=1
cos𝑥⋅ lim
ℎ→0
1
ℎ[−2 sin (
2𝑥 + ℎ2 ) sin (−
ℎ2)
cos(𝑥 + ℎ)]
=1
cos𝑥limℎ→0
[sin (2𝑥 + ℎ
2 )sin (
ℎ2)
(ℎ2)
]
cos(𝑥 + ℎ)
=1
cos𝑥⋅ lim
ℎ2→0
sin (ℎ2)
(ℎ2)
⋅ limℎ→0
sin (2𝑥 + ℎ
2 )
cos(𝑥 + ℎ)
=1
cos𝑥⋅ 1 ⋅
sin𝑥
cos𝑥
= sec𝑥 tan𝑥
(iii)5sec 𝑥 + 4cos 𝑥
Hint: 𝑓′(𝑎) = limℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
Solution:
Solution step 1: Let 𝑓(𝑥) = 5 𝑠𝑒𝑐𝑥 + 4 𝑐𝑜𝑠 𝑥. Accordingly, from the first principle,
𝑓′(𝑥) = limℎ→0
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
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= limℎ→0
5 sec(𝑥 + ℎ) + 4 cos(𝑥 + ℎ) − [5sec𝑥 + 4cos𝑥]
ℎ
= 5limℎ→0
[sec(𝑥 + ℎ) − sec𝑥]
ℎ+ 4lim
ℎ→0
[cos(𝑥 + ℎ) − cos𝑥]
ℎ𝑙
= 5limℎ→0
1
ℎ[
1
cos(𝑥 + ℎ)−
1
cos𝑥] + 4lim
ℎ→0
1
ℎ[cos(𝑥 + ℎ) − cos𝑥]
= 5limℎ→0
1
ℎ[cos𝑥 − cos(𝑥 + ℎ)
cos𝑥cos(𝑥 + ℎ)] + 4lim
ℎ→0
1
ℎ[cos𝑥 cosℎ − sin𝑥 sinℎ − cos𝑥]
=5
cos𝑥limℎ→0
1
ℎ[−2 sin (
𝑥 + 𝑥 + ℎ2 ) sin (
𝑥 − 𝑥 − ℎ2 )
cos(𝑥 + ℎ)] + 4lim
ℎ→0
1
ℎ[−cos𝑥(1 − cosℎ) − sin𝑥sinℎ]
=5
cos𝑥⋅ lim
ℎ→0
1
ℎ[−2 sin (
2𝑥 + ℎ2 ) sin (−
ℎ2)
cos(𝑥 + ℎ)] + 4 [−cos𝑥lim
ℎ→0
(1 − cosℎ)
ℎ− sin𝑥lim
ℎ→0
sinℎ
ℎ]
=5
cos𝑥⋅ [lim
ℎ→0
sin (2𝑥 + ℎ
2 )
cos(𝑥 + ℎ)⋅ lim
ℎ→0
sin (ℎ2)
ℎ2
] − 4sin𝑥
=5
cos𝑥⋅sin𝑥
cos𝑥⋅ 1 − 4sin𝑥
= 5sec𝑥 tan𝑥 ⋅ −4sin𝑥
(iv) cosec 𝑥
Hint: 𝑓′(𝑎) = limℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
Solution:
Solution step 1: Let 𝑓(𝑥) = cosec 𝑥. Accordingly, from the first principle,
𝑓′(𝑥) = limℎ→0
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
𝑓′(𝑥) = limℎ→0
1
ℎ[csc(𝑥 + ℎ) − 𝑐𝑜𝑠𝑒𝑐𝑥]
= limℎ→0
1
ℎ[
1
sin(𝑥 + ℎ)−
1
sin𝑥]
= limℎ→0
1
ℎ[sin𝑥 − sin(𝑥 + ℎ)
sin(𝑥 + ℎ) sin𝑥]
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= limℎ→0
1
ℎ[2 cos (
𝑥 + 𝑥 + ℎ2 ) ⋅ sin (
𝑥 − 𝑥 − ℎ2 )
sin(𝑥 + ℎ) sin𝑥]
= limℎ→0
1
ℎ[2 cos (
2𝑥 + ℎ2 ) sin (−
ℎ2)
sin(𝑥 + ℎ) sin𝑥]
= limℎ→0
−cos (2𝑥 + ℎ
2 ) ⋅sin (
ℎ2)
(ℎ2)
sin(𝑥 + ℎ) sin𝑥
= limℎ→0
(−cos (
2𝑥 + ℎ2 )
sin(𝑥 + ℎ) sin𝑥) ⋅ lim
ℎ2→0
sin (ℎ2)
(ℎ2)
= (−cos𝑥
sin𝑥 sin𝑥) . 1 = −cosec𝑥 cot𝑥
(v)3cot + 5cosec 𝑥
Hint: 𝑓′(𝑎) = limℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
Solution:
Solution step 1: Let 𝑓(𝑥) = 3𝑐𝑜𝑡 𝑥 + 5𝑐𝑜𝑠𝑒𝑐 𝑥. Accordingly, from the first principle,
𝑓′(𝑥) = limℎ→0
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
= limℎ→0
3 cot(𝑥 + ℎ) + 5 𝑐𝑜𝑠𝑒𝑐(𝑥 + ℎ) − 3cot𝑥 − 5 𝑐𝑜𝑠𝑒𝑐𝑥
ℎ
= 3limℎ→0
1
ℎ[cot(𝑥 + ℎ) − cot𝑥] + 5lim
ℎ→0
1
ℎ[ 𝑐𝑜𝑠𝑒𝑐(𝑥 + ℎ) − 𝑐𝑜𝑠𝑒𝑐𝑥]…… . (1)
Now, limℎ→0
1
ℎ[cot(𝑥 + ℎ) − cot𝑥]
= limℎ→0
1
ℎ[cos(𝑥 + ℎ)
sin(𝑥 + ℎ)−
cos𝑥
sin𝑥]
= limℎ→0
1
ℎ[cos(𝑥 + ℎ) sin𝑥 − cos𝑥sin(𝑥 + ℎ)
sin𝑥sin(𝑥 + ℎ)]
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= limℎ→0
1
ℎ[sin(𝑥 − 𝑥 − ℎ)
sin𝑥sin(𝑥 + ℎ)]
= limℎ→0
1
ℎ[
sin(−ℎ)
sin𝑥sin(𝑥 + ℎ)]
= −(limℎ→0
sinℎ
ℎ) ⋅ (lim
ℎ→0
1
sin𝑥 ⋅ sin(𝑥 + ℎ))
= −1 ⋅1
sin𝑥 ⋅ sin(𝑥 + 0)=
−1
sin2𝑥= − 𝑐𝑜𝑠𝑒𝑐2𝑥 … (2)
limℎ→0
1
ℎ[ 𝑐𝑜𝑠𝑒𝑐(𝑥 + ℎ) − 𝑐𝑜𝑠𝑒𝑐𝑥]
= limℎ→0
1
ℎ[
1
sin(𝑥 + ℎ)−
1
sin𝑥]
= limℎ→0
1
ℎ[sin𝑥 − sin(𝑥 + ℎ)
sin(𝑥 + ℎ) sin𝑥]
= limℎ→0
1
ℎ[2 cos (
𝑥 + 𝑥 + ℎ2 ) ⋅ sin (
𝑥 − 𝑥 − ℎ2 )
sin(𝑥 + ℎ) sin𝑥]
= limℎ→0
1
ℎ[2 cos (
2𝑥 + ℎ2 ) sin (−
ℎ2)
sin(𝑥 + ℎ) sin𝑥]
= limℎ→0
−cos (2𝑥 + ℎ
2 ) ⋅sin (
ℎ2)
(ℎ2)
sin(𝑥 + ℎ) sin𝑥
= limℎ→0
(−cos (
2𝑥 + ℎ2 )
sin(𝑥 + ℎ) sin𝑥) ⋅ lim
ℎ→0
sin (ℎ2)
(ℎ2)
= (−cos𝑥
sin𝑥sin𝑥) ⋅ 1
= −cosec cot𝑥 …… (3)
From (1), (2) and (3) we obtain
𝑓′(𝑥) = −3 𝑐𝑜𝑠𝑒𝑐2𝑥 − 5 𝑐𝑜𝑠𝑒𝑐𝑥 cot 𝑥
(vi)5sin − 6cos 𝑥 + 7
Hint: 𝑓′(𝑎) = limℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
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Solution:
Solution step 1: Let 𝑓(𝑥) = 5𝑠𝑖𝑛 𝑥– 6𝑐𝑜𝑠 𝑥 + 7. Accordingly, from the first principle,
𝑓′(𝑥) = limℎ→0
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
= limℎ→0
1
ℎ[5 sin(𝑥 + ℎ) − 6 cos(𝑥 + ℎ) + 7 − 5sin𝑥 + 6cos𝑥 − 7]
= limℎ→0
1
ℎ[5{sin(𝑥 + ℎ) − sin𝑥} − 6{cos(𝑥 + ℎ) − cos𝑥}]
= 5limℎ→0
1
ℎ[2 cos (
𝑥 + ℎ + 𝑥
2) sin (
𝑥 + ℎ − 𝑥
2)] − 6lim
ℎ→0
cos𝑥cosℎ − sin𝑥sinℎ − cos𝑥
ℎ
= 5limℎ→0
1
ℎ[2 cos (
2𝑥 + ℎ
2) sin
ℎ
2] − 6lim
ℎ→0[−cos𝑥(1 − cosℎ) − sin𝑥sinℎ
ℎ]
= 5 [limℎ→0
cos (2𝑥 + ℎ
2)] [lim
ℎ
sinℎ2
ℎ2
] − 6 [(−cos𝑥) (limℎ→0
1 − cosℎ
ℎ) − sin𝑥lim
ℎ→0(sinℎ
ℎ)]
= 5cos𝑥 ⋅ 1 − 6[(−cos𝑥) ⋅ (0) − sin𝑥. 1]
= 5cos𝑥 + 6sin𝑥
(vii)2tan 𝑥 − 7sec 𝑥
Hint: 𝑓′(𝑎) = limℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
Solution:
Solution step 1: Let 𝑓(𝑥) = 2 tan 𝑥 – 7 sec 𝑥. Accordingly, from the first principle,
𝑓′(𝑥) = limℎ→0
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
= limℎ→0
1
ℎ[2 tan(𝑥 + ℎ) − 7 sec(𝑥 + ℎ) − 2tan𝑥 + 7sec𝑥]
= limℎ→0
1
ℎ[2{tan(𝑥 + ℎ) − tan𝑥} − 7{sec(𝑥 + ℎ) − sec𝑥}]
= 2limℎ→0
1
ℎ[sin(𝑥 + ℎ)
cos(𝑥 + ℎ)−
sin𝑥
cos𝑥] − 7lim
ℎ→0
1
ℎ[
1
cos(𝑥 + ℎ)−
1
cos𝑥]
= 2limℎ→0
1
ℎ[sin(𝑥 + ℎ) cos𝑥 − sin𝑥cos(𝑥 + ℎ)
cos𝑥cos(𝑥 + ℎ)] − 7lim
ℎ→0
1
ℎ[cos𝑥 − cos(𝑥 + ℎ)
cos𝑥cos(𝑥 + ℎ)]
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= 2limℎ→0
1
ℎ[sin(𝑥 + ℎ − 𝑥)
cos𝑥cos(𝑥 + ℎ)] − 7lim
ℎ→0
1
ℎ[−2 sin (
𝑥 + 𝑥 + ℎ2 ) sin (
𝑥 − 𝑥 − ℎ2 )
cos𝑥cos(𝑥 + ℎ)]
= 2limℎ→0
[(sinℎ
ℎ)
1
cos𝑥cos(𝑥 + ℎ)] − 7lim
ℎ→0
1
ℎ[−2 sin (
2𝑥 + ℎ2 ) sin (−
ℎ2)
cos𝑥cos(𝑥 + ℎ)]
= 2.1 ⋅1
cos𝑥cos𝑥− 7.1 (
sin𝑥
cos𝑥cos𝑥)
= 2sec2𝑥 − 7sec𝑥 tan𝑥
Miscellaneous Exercise
1. Find the derivative of the following functions from first principle:
(i) −𝑥
Hint: 𝑓′(𝑎) = limℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
Solution:
Solution step 1: Let 𝑓(𝑥) =–𝑥. Accordingly, 𝑓(𝑥 + ℎ) = −(𝑥 + ℎ)
By first principle,
𝑓′(𝑥) = limℎ→0
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
= limℎ→0
−(𝑥 + ℎ) − (−𝑥)
ℎ
= limℎ→0
−𝑥 − ℎ + 𝑥
ℎ
= limℎ→0
−ℎ
ℎ
= limℎ→0
(−1) = −1
(ii)(−𝑥)−1
Hint: 𝑓′(𝑎) = limℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
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Solution:
Solution step 1: Let 𝑓(𝑥) = (−𝑥)−1 =1
−𝑥=
−1
𝑥.
Accordingly, 𝑓(𝑥 + ℎ) =−1
(𝑥+ℎ)
By first principle,
𝑓′(𝑥) = limℎ→0
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
= limℎ→0
1
ℎ[
−1
𝑥 + ℎ− (
−1
𝑥)]
= limℎ→0
1
ℎ[−𝑥 + (𝑥 + ℎ)
𝑥 + ℎ+
1
𝑥]
= limℎ→0
1
ℎ[−𝑥 + (𝑥 + ℎ)
𝑥(𝑥 + ℎ)]
= limℎ→0
1
ℎ[−𝑥 + (𝑥 + ℎ)
𝑥(𝑥 + ℎ)]
= limℎ→0
1
ℎ[
ℎ
𝑥(𝑥 + ℎ)]
= limℎ→0
1
𝑥(𝑥 + ℎ)
=1
𝑥 ⋅ 𝑥=
1
𝑥2
(iii)sin(𝑥 + 1)
Hint: 𝑓′(𝑎) = limℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
Solution:
Solution step 1: Let 𝑓(𝑥) = sin(𝑥 + 1). Accordingly,𝑓(𝑥 + ℎ) = sin(𝑥 + ℎ + 1)
By first principle,
𝑓′(𝑥) = limℎ→0
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
= limℎ→0
1
ℎ[sin(𝑥 + ℎ + 1) − sin(𝑥 + 1)]
= limℎ→0
1
ℎ[2 cos (
𝑥 + ℎ + 1 + 𝑥 + 1
2) sin (
𝑥 + ℎ + 1 − 𝑥 − 1
2)]
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= limℎ→0
1
ℎ[2 cos (
2𝑥 + ℎ + 2
2) sin (
ℎ
2)]
= limℎ→0
[cos (2𝑥 + ℎ + 2
2) ⋅
sin (ℎ2)
(ℎ2)
]
= limℎ→0
cos (2𝑥 + ℎ + 2
2) ⋅ lim
ℎ2→0
sin (ℎ2)
𝑏2 + 0
sin (ℎ2)
ℎ2 + 0
[𝐴𝑠ℎ → 0 ⇒ℎ
2→ 0]
= cos (2𝑥 + 0 + 2
2) ⋅ 1 [lim
𝑥→0
sin𝑥
𝑥= 1]
= cos(𝑥 + 1)
(iv)cos (𝑥 −𝜋
8)
Hint: 𝑓′(𝑎) = limℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
Solution:
Solution step 1: Let 𝑓(𝑥) = cos (𝑥 −𝜋
8)
Accordingly, 𝑓(𝑥 + ℎ) = cos (𝑥 + ℎ −𝜋
8)
By first principle,
𝑓′(𝑥) = limℎ→0
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
= limℎ→0
1
ℎ[cos (𝑥 + ℎ −
𝜋
8) − cos (𝑥 −
𝜋
8)]
= limℎ→0
1
ℎ[−2sin
(𝑥 + ℎ −𝜋8 + 𝑥 −
𝜋8)
2sin(
𝑥 + ℎ −𝜋8 − 𝑥 +
𝜋8
2)]
= limℎ→0
1
ℎ[−2 sin(
2𝑥 + ℎ −𝜋4
2)sin
ℎ
2]
= limℎ→0
[− sin(2𝑥 + ℎ −
𝜋4
2)
sin (ℎ2)
2]
]
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= limℎ→0
[− sin(2𝑥 + ℎ −
𝜋4
2)] ∙ lim
ℎ2→0
sin (ℎ2)
(ℎ2)
[A𝑠 ℎ → 0 ⇒ℎ
2→ 0]
= −sin(2𝑥 + 0 −
𝜋4
2) . 1
= −sin (𝑥 −𝜋
8)
2. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): (𝑥 + 𝑎)
Hint: 𝑓′(𝑎) = limℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
Solution:
Solution step 1: Let 𝑓(𝑥) = 𝑥 + 𝑎. Accordingly, 𝑓(𝑥 + ℎ) = −(𝑥 + ℎ)
By first principle,
𝑓′(𝑥) = limℎ→0
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
= limℎ→0
𝑥 + ℎ + 𝑎 − 𝑥 − 𝑎
ℎ
= limℎ→0
(ℎ
ℎ)
= limℎ→0
(1)
= 1
3. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): (𝑝𝑥 + 𝑞) (𝑟
𝑥+ 𝑠)
Hint: 𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1
Solution:
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Solution step 1: Let 𝑓(𝑥) = (𝑝𝑥 + 𝑞) (𝑟
𝑥+ 𝑠)
𝑓′(𝑥) = (𝑝𝑥 + 𝑞) (𝑟
𝑥+ 𝑠) + (
𝑟
𝑥+ 𝑠) (𝑝𝑥 + 𝑞)′
= (𝑝𝑥 + 𝑞)(𝑟𝑥−1 + 𝑠)′ + (𝑟
𝑥+ 𝑠) (𝑝)
= (𝑝𝑥 + 𝑞)(−𝑟𝑥−2) + (𝑟
𝑥+ 𝑠) 𝑝
= (𝑝𝑥 + 𝑞) (−𝑟
𝑥2) + (
𝑟
𝑥+ 𝑠) 𝑝
=−𝑝𝑟
𝑥−
𝑞𝑟
𝑥2+
𝑝𝑟
𝑥+ 𝑝𝑠
= 𝑝𝑠 −𝑞𝑟
𝑥2
4. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): (𝑎𝑥 + 𝑏)(𝑐𝑥 + 𝑑)2
Hint: 𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1
Solution:
Solution step 1: Let 𝑓(𝑥) = (𝑎𝑥 + 𝑏)(𝑐𝑥 + 𝑑)2
By product rule,
𝑓′(𝑥) = (𝑎𝑥 + 𝑏)𝑑
𝑑𝑥(𝑐𝑥 + 𝑑)2 + (𝑐𝑥 + 𝑑)2
𝑑
𝑑𝑥(𝑎𝑥 + 𝑏)
= (𝑎𝑥 + 𝑏)𝑑
𝑑𝑥(𝑐2𝑥2 + 2𝑐𝑑𝑥 + 𝑑2) + (𝑐𝑥 + 𝑑)2
𝑑
𝑑𝑥(𝑎𝑥 + 𝑏)
= (𝑎𝑥 + 𝑏) [𝑑
𝑑𝑥(𝑐2𝑥2) +
𝑑
𝑑𝑥(2𝑐𝑑𝑥) +
𝑑
𝑑𝑥𝑑2] + (𝑐𝑥 + 𝑑)2 [
𝑑
𝑑𝑥𝑎𝑥 +
𝑑
𝑑𝑥𝑏]
= (𝑎𝑥 + 𝑏)(2𝑐2𝑥 + 2𝑐𝑑) + (𝑐𝑥 + 𝑑2)𝑎
= 2𝑐(𝑎𝑥 + 𝑏)(𝑐𝑥 + 𝑑) + 𝑎(𝑐𝑥 + 𝑑)2
5. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): 𝑎𝑥+𝑏
𝑐𝑥+𝑑
Hint: 𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1
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Solution:
Solution step 1: Let 𝑓(𝑥) =𝑎𝑥+𝑏
𝑐𝑥+𝑑
By quotient rule,
𝑓′(𝑥) =(𝑐𝑥 + 𝑑)
𝑑𝑑𝑥
(𝑎𝑥 + 𝑏) − (𝑎𝑥 + 𝑏)𝑑𝑑𝑥
(𝑐𝑥 + 𝑑)
(𝑐𝑥 + 𝑑)2(𝑐𝑥 + 𝑑)
=(𝑐𝑥 + 𝑑)(𝑎) − (𝑎𝑥 + 𝑏)(𝑐)
(𝑐𝑥 + 𝑑)2
=𝑎𝑐𝑥 + 𝑎𝑑 − 𝑎𝑐𝑥 − 𝑏𝑐
(𝑐𝑥 + 𝑑)2
=𝑎𝑑 − 𝑏𝑐
(𝑐𝑥 + 𝑑)2
6. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): 1+
1
𝑥
1−1
𝑥
Hint: 𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1
Solution:
Solution step 1: 𝑓(𝑥) =1+
1
𝑥
1−1
𝑥
=𝑥+1
𝑥𝑥−1
𝑥
=𝑥+1
𝑥−1, where 𝑥 ≠ 0
By quotient rule,
𝑓′(𝑥) =(𝑥 − 1) − (𝑥 − 1) − (𝑥 + 1) − (𝑥 + 1)
(𝑥 − 1)2, 𝑥 ≠ 0,1
=(𝑥 − 1)(1) − (𝑥 + 1)(1)
(𝑥 − 1)2, 𝑥 ≠ 0,1
=𝑥 − 1 − 𝑥 − 1
(𝑥 − 1)2, 𝑥 ≠ 0,1
=−2
(𝑥 − 1)2, 𝑥 ≠ 0,1
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7. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): 1
𝑎𝑥2+𝑏𝑥+𝑐
Hint: 𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1
Solution:
Solution step 1: Let 𝑓(𝑥) =1
𝑎𝑥2+𝑏𝑥+𝑐
By quotient rule,
𝑓′(𝑥) =(𝑎𝑥2 + 𝑏𝑥 + 𝑐)
𝑑𝑑𝑥
(1) −𝑑𝑑𝑥
(𝑎𝑥2 + 𝑏𝑥 + 𝑐)
𝑑𝑥
=(𝑎𝑥2 + 𝑏𝑥 + 𝑐)(0) − (2𝑎𝑥 + 𝑏)
(𝑎𝑥2 + 𝑏𝑥 + 𝑐)2
=−(2𝑎𝑥 + 𝑏)
(𝑎𝑥2 + 𝑏𝑥 + 𝑐)2
8. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): 𝑎𝑥+𝑏
𝑝𝑥2+𝑞𝑥+𝑟
Hint: 𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1
Solution:
Solution step 1: Let 𝑓(𝑥) =𝑎𝑥+𝑏
𝑝𝑥2+𝑞𝑥+𝑟
By quotient rule,
𝑓′(𝑥) =(𝑝𝑥2 + 𝑞𝑥 + 𝑟)
𝑑𝑑𝑥
(𝑎𝑥 + 𝑏) − (𝑎𝑥 + 𝑏)𝑑𝑑𝑥
(𝑝𝑥2 + 𝑞𝑥 + 𝑟)
(𝑝𝑥2 + 𝑞𝑥 + 𝑟)2
=(𝑝𝑥2 + 𝑞𝑥 + 𝑟)(𝑎) − (𝑎𝑥 + 𝑏)(2𝑝𝑥 + 𝑞)
(𝑝𝑥2 + 𝑞𝑥 + 𝑟)2
=𝑎𝑝𝑥2 + 𝑎𝑞𝑥 + 𝑎𝑟 − 2𝑎𝑝𝑥2 − 𝑎𝑞𝑥 − 2𝑏𝑝𝑥 − 𝑏𝑞
(𝑝𝑥2 + 𝑞𝑥 + 𝑟)2
=−𝑎𝑝𝑥2 − 2𝑏𝑝𝑥 + 𝑎𝑟 − 𝑏𝑞
(𝑝𝑥2 + 𝑞𝑥 + 𝑟)2
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9. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): 𝑝𝑥2+𝑞𝑥+𝑟
𝑎𝑥+𝑏
Hint: 𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1
Solution:
Solution step 1: Let 𝑓(𝑥) =𝑝𝑥2+𝑞𝑥+𝑟
𝑎𝑥+𝑏
By quotient rule,
𝑓′(𝑥) =(𝑎𝑥 + 𝑏)
𝑑𝑑𝑥
(𝑝𝑥2 + 𝑞𝑥 + 𝑟) − (𝑝𝑥2 + 𝑞𝑥 + 𝑟)𝑑𝑑𝑥
(𝑎𝑥 + 𝑏)
(𝑎𝑥 + 𝑏)2
=(𝑎𝑥 + 𝑏)(2𝑝𝑥 + 𝑞) − (𝑝𝑥2 + 𝑞𝑥 + 𝑟)(𝑎)
(𝑎𝑥 + 𝑏)2
=2𝑎𝑝𝑥2 + 𝑎𝑞𝑥 + 2𝑏𝑝𝑥 + 𝑏𝑞 − 𝑎𝑝𝑥2 − 𝑎𝑞𝑥 − 𝑎𝑟
(𝑎𝑥 + 𝑏)2
=𝑎𝑝𝑥2 + 2𝑏𝑝𝑥 + 𝑏𝑞 − 𝑎𝑟
(𝑎𝑥 + 𝑏)2
10. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): 𝑎
𝑥4 −𝑏
𝑥2 + cos
Hint: 𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1
Solution:
Solution step 1:
Let 𝑓(𝑥) = &𝑎
𝑥4 −𝑏
𝑥2 + cos𝑥
𝑓′(𝑥) =𝑑
𝑑𝑥(
𝑎
𝑥4) −
𝑑
𝑑𝑥(
𝑏
𝑥2) +
𝑑
𝑑𝑥(cos𝑥)
= 𝑎𝑑
𝑑𝑥(𝑥−4) − 𝑏
𝑑
𝑑𝑥(𝑥−2) +
𝑑
𝑑𝑥(cos𝑥)
(= 𝑎(−4𝑥−5) − 𝑏(−2𝑥−3) + (−sin𝑥) [𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1 and
𝑑
𝑑𝑥(cos𝑥) = −sin𝑥]
=−4𝑎
𝑥5+
2𝑏
𝑥3− sin𝑥
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11. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): 4√𝑥 − 2
Hint: 𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1
Solution:
Solution step 1: Let 𝑓(𝑥) = 4√𝑥 − 2
𝑓′(𝑥) =𝑑
𝑑𝑥(4√𝑥 − 2) =
𝑑
𝑑𝑥(4√𝑥) −
𝑑
𝑑𝑥(2)
= 4𝑑
𝑑𝑥(𝑥
12) − 0 = 4 (
1
2𝑥
12−1)
= (2𝑥−12) =
2
√𝑥
12. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): (𝑎𝑥 + 𝑏)𝑛
Hint: 𝑓′(𝑎) = limℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
Solution:
Solution step 1: 𝑓(𝑥) = (𝑎𝑥 + 𝑏)𝑛
Accordingly, 𝑓(𝑥 + ℎ) = {𝑎(𝑥 + ℎ) + 𝑏}𝑛 = (𝑎𝑥 + 𝑎ℎ + 𝑏)𝑛
By first principle,
𝑓′(𝑥) = limℎ→0
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
= limℎ→0
(𝑎𝑥 + 𝑎ℎ + 𝑏)𝑛 − (𝑎𝑥 + 𝑏)𝑛
ℎ
= limℎ→0
(𝑎𝑥 + 𝑏)𝑛 (1 +𝑎ℎ
𝑎𝑥 + 𝑏)𝑛
− (𝑎𝑥 + 𝑏)𝑛
ℎ
= (𝑎𝑥 + 𝑏)𝑛 limℎ→0
(1 +𝑎ℎ
𝑎𝑥 + 𝑏)𝑛
− 1
ℎ
= (𝑎𝑥 + 𝑏)𝑛 limℎ→0
1
𝑛[{1 + 𝑛 (
𝑎ℎ
𝑎𝑥 + 𝑏) +
𝑛(𝑛 − 1)
∟2(
𝑎ℎ
𝑎𝑥 + 𝑏)
2
+ ⋯} − 1] (Using binomial theorem)
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= (𝑎𝑥 + 𝑏)𝑛 limℎ→0
1
ℎ[𝑛(
𝑎ℎ
𝑎𝑥 + 𝑏) +
𝑛(𝑛 − 1)𝑎2ℎ2
∟2(𝑎𝑥 + 𝑏)2+ ⋯ (Terms containing higher degrees ofℎ)]
= (𝑎𝑥 + 𝑏)𝑛lim𝑏→0
[𝑛𝑎
(𝑎𝑥 + 𝑏)+
𝑛(𝑛 − 1)𝑎2ℎ
∟2(𝑎𝑥 + 𝑏)2+ ⋯]
= (𝑎𝑥 + 𝑏)𝑛 [𝑛𝑎
(𝑎𝑥 + 𝑏)+ 0]
= 𝑛𝑎(𝑎𝑥 + 𝑏)𝑛
(𝑎𝑥 + 𝑏)
= 𝑛𝑎(𝑎𝑥 + 𝑏)𝑛−1
13. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): (𝑎𝑥 + 𝑏)𝑛(𝑐𝑥 + 𝑑)𝑚
Hint: 𝑓′(𝑎) = limℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
Solution:
Solution step 1: Let 𝑓(𝑥) = (𝑎𝑥 + 𝑏)𝑛(𝑐𝑥 + 𝑑)𝑚
𝑓′(𝑥) = (𝑎𝑥 + 𝑏)𝑛𝑑
𝑑𝑥(𝑐𝑥 + 𝑑)𝑚 + (𝑐𝑥 + 𝑑)𝑚
𝑑
𝑑𝑥(𝑎𝑥 + 𝑏)𝑛 … (1)
Now, let 𝑓1(𝑥) = (𝑐𝑥 + 𝑑)𝑚
𝑓1(𝑥 + ℎ) = (𝑐𝑥 + 𝑐ℎ + 𝑑)𝑚
𝑓1′(𝑥) = lim
ℎ→0
𝑓1(𝑥 + ℎ) − 𝑓1(𝑥)
ℎ
= limℎ→0
(𝑐𝑥 + 𝑐ℎ + 𝑑)𝑚 − (𝑐𝑥 + 𝑑)𝑚
ℎ
= (𝑐𝑥 + 𝑑)𝑚 limℎ→0
1
ℎ[(1 +
𝑐ℎ
𝑐𝑥 + 𝑑)𝑚
− 1]
= (𝑐𝑥 + 𝑑)𝑚 limℎ→0
1
ℎ[(1 +
𝑚𝑐ℎ
(𝑐𝑥 + 𝑑)+
𝑚(𝑚 − 1)
2
(𝑐2ℎ2)
(𝑐𝑥 + 𝑑)2+ ⋯) − 1]
= (𝑐𝑥 + 𝑑)𝑚 limℎ→0
1
ℎ[
𝑚𝑐ℎ
(𝑐𝑥 + 𝑑)+
𝑚(𝑚 − 1)𝑐2ℎ2
2(𝑐𝑥 + 𝑑)2+ ⋯ (Terms containing higher degrees of ℎ)]
= (𝑐𝑥 + 𝑑)𝑚 limℎ→0
[𝑚𝑐
(𝑐𝑥 + 𝑑)+
𝑚(𝑚 − 1)𝑐2ℎ
2(𝑐𝑥 + 𝑑)2+ ⋯ ]
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= (𝑐𝑥 + 𝑑)𝑚 [𝑚𝑐
𝑐𝑥 + 𝑑+ 0]
=𝑚𝑐(𝑐𝑥 + 𝑑)𝑚
(𝑐𝑥 + 𝑑)
= 𝑚𝑐(𝑐𝑥 + 𝑑)𝑚−1
𝑑
𝑑𝑥(𝑐𝑥 + 𝑑)𝑚 = 𝑚𝑐(𝑐𝑥 + 𝑑)𝑚−1 …(2)
Similarly, 𝑑
𝑑𝑥(𝑎𝑥 + 𝑏)𝑛 = 𝑛𝑎(𝑎𝑥 + 𝑏)𝑛−1 …(3)
Therefore, from (1), (2) and (3) we obtain
𝑓′(𝑥) = (𝑎𝑥 + 𝑏)𝑛{𝑚𝑐(𝑐𝑥 + 𝑑)𝑚−1} + (𝑐𝑥 + 𝑑)𝑚{𝑛𝑎(𝑎𝑥 + 𝑏)𝑛−1}
= (𝑎𝑥 + 𝑏)𝑛−1(𝑐𝑥 + 𝑑)𝑚−1[𝑚𝑐(𝑎𝑥 + 𝑏) + 𝑛𝑎(𝑐𝑥 + 𝑑)]
14. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): sin(𝑥 + 𝑎)
Hint: 𝑓′(𝑎) = limℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
Solution:
Solution step 1: Let 𝑓(𝑥) = sin(𝑥 + 𝑎), therefore 𝑓(𝑥 + ℎ) = sin(𝑥 + ℎ + 𝑎)
By first principle,
𝑓′(𝑥) = limℎ→0
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
= limℎ→0
sin(𝑥 + ℎ + 𝑎) − sin(𝑥 + 𝑎)
ℎ
= limℎ→0
1
ℎ[2 cos (
𝑥 + ℎ + 𝑎 + 𝑥 + 𝑎
2) sin (
𝑥 + ℎ + 𝑎 − 𝑥 − 𝑎
2)]
= limℎ→0
1
ℎ[2 cos (
2𝑥 + 2𝑎 + ℎ
2) sin (
ℎ
2)]
= limℎ→0
[cos (2𝑥 + 2𝑎 + ℎ
2){
sin (ℎ2)
(ℎ2)
}]
= limℎ→0
cos (2𝑥 + 2𝑎 + ℎ
2) lim
𝑏2→0
{sin (
𝑛2)
(ℎ2)
} [ As ℎ → 0 ⇒ℎ
2→ 0]
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= cos (2𝑥 + 2𝑎
2) × 1 [lim
𝑥→0
sin𝑥
𝑥= 1]
= cos(𝑥 + 𝑎)
15. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): cosec 𝑥 cot 𝑥
Hint: 𝑓′(𝑎) = limℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
Solution:
Solution step 1: Let 𝑓(𝑥) = cosec𝑥 cot𝑥
By product rule,
𝑓′(𝑥) = cosec𝑥(cot𝑥)′ + cot𝑥(cosec𝑥)′…(1)
Let 𝑓1(𝑥) = cot𝑥. Accordingly,𝑓1(𝑥 + ℎ) = cot(𝑥 + ℎ)
By first principle,
𝑓1′(𝑥) = lim
ℎ→0
𝑓1(𝑥 + ℎ) − 𝑓1(𝑥)
ℎ
= limℎ→0
cot(𝑥 + ℎ) − cot𝑥
ℎ
= limℎ→0
1
ℎ(cos(𝑥 + ℎ)
sin(𝑥 + ℎ)−
cos𝑥
sin𝑥)
= limℎ→0
1
ℎ[sin𝑥cos(𝑥 + ℎ) − cos𝑥sin(𝑥 + ℎ)
sin𝑥sin(𝑥 + ℎ)]
= limℎ→0
1
ℎ[sin(𝑥 − 𝑥 − ℎ)
sin𝑥sin(𝑥 + ℎ)]
=1
sin𝑥limℎ
1
ℎ[
sin(−ℎ)
sin(𝑥 + ℎ)]
=−1
sin𝑥⋅ (lim
ℎ→0
sinℎ
ℎ) (lim
ℎ→0
1
sin(𝑥 + ℎ))
=−1
sin𝑥⋅ 1 ⋅ (
1
sin(𝑥 + 0))
=−1
sin2𝑥
= − 𝑐𝑜𝑠𝑒𝑐2𝑥
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∴ (cot𝑥)′ = −cosec2𝑥 … (2)
Now, let 𝑓2(𝑥) = 𝑐𝑜𝑠𝑒𝑐 𝑥.
Accordingly, 𝑓2(𝑥 + ℎ) = 𝑐𝑜𝑠𝑒𝑐 (𝑥 + ℎ)
By first principle,
𝑓2′(𝑥) = lim
ℎ→0
𝑓2(𝑥 + ℎ) − 𝑓2(𝑥)
ℎ
= limℎ→0
1
ℎ[ 𝑐𝑜𝑠𝑒𝑐(𝑥 + ℎ) − 𝑐𝑜𝑠𝑒𝑐𝑥] = lim
ℎ→0
1
ℎ[
1
sin(𝑥 + ℎ)−
1
sin𝑥]
= limℎ→0
1
ℎ[sin𝑥 − sin(𝑥 + ℎ)
sin𝑥sin(𝑥 + ℎ)]
=1
sin𝑥∙ limℎ→0
1
ℎ[2 cos (
𝑥 + 𝑥 + ℎ2 ) sin (
𝑥 − 𝑥 − ℎ2 )
sin(𝑥 + ℎ)]
=1
sin𝑥∙ limℎ→0
1
ℎ= [
2 cos (2𝑥 + ℎ
2 ) sin (−ℎ2 )
sin(𝑥 + ℎ)]
=1
sin𝑥∙ limℎ→0
[− sin (
ℎ2)
(ℎ2)
⋅cos (
2𝑥 + ℎ2 )
sin(𝑥 + ℎ)]
=−1
sin𝑥∙ limℎ→0
sin (ℎ2)
(ℎ2)
⋅ limℎ→0
cos (2𝑥 + ℎ
2 )
sin(𝑥 + ℎ)
=−1
sin𝑥. 1.
cos (2𝑥 + 0
2 )
sin(𝑥 + 0)
=−1
sin𝑥⋅cos𝑥
sin𝑥
= −cosec𝑥 ⋅ cot𝑥
∴ ( 𝑐𝑜𝑠𝑒𝑐𝑥)′ = − 𝑐𝑜𝑠𝑒𝑐𝑥. cot𝑥 …(3)
From (1), (2) and (3) we obtain
𝑓′(𝑥) = 𝑐𝑜𝑠𝑒𝑐𝑥(− 𝑐𝑜𝑠𝑒𝑐2𝑥) + cot𝑥(− 𝑐𝑜𝑠𝑒𝑐𝑥cot𝑥)
= − 𝑐𝑜𝑠𝑒𝑐3𝑥 − cot2𝑥 𝑐𝑜𝑠𝑒𝑐𝑥
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16. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): cos𝑥
1+sin𝑥
Hint: 𝑓′(𝑎) = limℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
Solution:
Solution step 1: Let
𝑓(𝑥) =cos𝑥
1 + sin𝑥
By quotient rule, 𝑓′(𝑥) =(1+sin𝑥)
𝑑
𝑑𝑥(cos𝑥)−(cos𝑥)
𝑑
𝑑𝑥(1+sin𝑥)
(1+sin𝑥)2
=(1 + sin𝑥)(−sin𝑥) − (cos𝑥)(cos𝑥)
(1 + sin𝑥)2
=−sin𝑥 − sin2𝑥 − cos2𝑥
(1 + sin𝑥)2
=−sin𝑥 − (sin2𝑥 + cos2𝑥)
(1 + sin𝑥)2
=−sin𝑥 − 1
(1 + sin𝑥)2
=−(1 + sin𝑥)
(1 + sin𝑥)2
=−1
(1 + sin𝑥)
17. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): sin𝑥+cos𝑥
sin𝑥−cos𝑥
Hint: 𝑓′(𝑎) = limℎ→0
𝑓(𝑎+ℎ)−𝑓(𝑎)
ℎ
Solution:
Solution step 1: Let 𝑓(𝑥)sin𝑥+cos𝑥
sin𝑥−cos𝑥
By quotient rule,
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𝑓′(𝑥) =(sin𝑥 − cos𝑥)
𝑑𝑑𝑥
(sin𝑥 + cos𝑥) − (sin𝑥 + cos𝑥)𝑑𝑑𝑥
(sin𝑥 − cos𝑥)
(sin𝑥 − cos𝑥)2
=(sin𝑥 − cos𝑥)(cos𝑥 − sin𝑥) − (sin𝑥 + cos𝑥)(cos𝑥 + sin𝑥)
(sin𝑥 − cos𝑥)2
=−(sin𝑥 − cos𝑥)2 − (sin𝑥 + cos𝑥)2
(sin𝑥 − cos𝑥)2
=−[sin2𝑥 + cos2𝑥 − 2sin𝑥cos𝑥 + sin2𝑥 + cos2𝑥 + 2sin𝑥cos𝑥]
(sin𝑥 − cos𝑥)2
=−[1 + 1]
(sin𝑥 − cos𝑥)2
=−2
(sin𝑥 − cos𝑥)2
18. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): sec𝑥−1
sec𝑥+1
Hint: 𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1
Solution:
Solution step 1: Let
𝑓(𝑥) =sec𝑥 − 1
sec𝑥 + 1 𝑓(𝑥) =
1cos𝑥 − 1
1cos𝑥
+ 1=
1 − cos𝑥
1 + cos𝑥
By quotient rule,
𝑓′(𝑥) =(1 + cos𝑥)
𝑑𝑑𝑥
(1 − cos𝑥) − (1 − cos𝑥)𝑑𝑑𝑥
(1 + cos𝑥)
(1 + cos𝑥)2
=(1 + cos𝑥)(sin𝑥) − (1 − cos𝑥)(− sin 𝑥)
(1 + cos𝑥)2
=sin𝑥 + cos𝑥sin𝑥 + sin𝑥 − sin𝑥cos𝑥
(1 + cos𝑥)2
=2sin𝑥
(1 + cos𝑥)2
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=2sin𝑥
(1 +1
sec𝑥)2 =
2sin𝑥
(sec𝑥 + 1)2
sec2𝑥
=2sin𝑥sec2𝑥
(sec𝑥 + 1)2
=
2sin𝑥cos 𝑥 sec 𝑥
(sec𝑥 + 1)2
=2sec𝑥 + tan𝑥
(sec𝑥 + 1)2
19. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): sin𝑛 𝑥
Full marks: 3
Overall difficulty Level: M
Hint: 𝑑
𝑑𝑥(sin𝑛𝑥) = 𝑛sin(𝑛−1)𝑥cos𝑥
Solution:
Solution step 1: Let 𝑦 = sin𝑛 𝑥
Accordingly, for 𝑛 = 1, 𝑦 = sin 𝑥
∴𝑑𝑦
𝑑𝑥= cos𝑥, i.e.,
𝑑
𝑑𝑥sin𝑥 = cos𝑥
For 𝑛 = 1, 𝑦 = sin2 𝑥
∴𝑑𝑦
𝑑𝑥=
𝑑
𝑑𝑥(sin𝑥sin𝑥)
= (sin𝑥)′sin𝑥 + sin𝑥(sin𝑥)′ [By Leibnitz product rule]
= cos𝑥sin𝑥 + sin𝑥cos𝑥
= 2 sin 𝑥 cos 𝑥 … (1)
For 𝑛 = 1, 𝑦 = sin3 𝑥
∴𝑑𝑦
𝑑𝑥=
𝑑
𝑑𝑥(sin𝑥sin2𝑥)
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= (sin𝑥)′sin2𝑥 + sin𝑥(sin2𝑥)′ [By Leibnitz product rule]
= cos𝑥sin2𝑥 + sin𝑥(2sin𝑥cos𝑥)[Using(1)]
=cosxsin2x+2sin2xcosx
=3sin2xcosx
We assert that 𝑑
𝑑𝑥(sin𝑛𝑥) = 𝑛sin(𝑛−1)𝑥cos𝑥
Let our assertion be true for 𝑛 = 𝑘.
i.e., 𝑑
𝑑𝑥(sin𝑘𝑥) = 𝑘sin(𝑘−1)𝑥cos𝑥 … (2)
Consider 𝑑
𝑑𝑥(sin𝑘+1𝑥) =
𝑑
𝑑𝑥(sin𝑥sin𝑘𝑥)
= (sin𝑥)′sin𝑘𝑥 + sin𝑥(sin𝑘𝑥)′ [By Leibnitz product rule]
= cos𝑥 sin𝑘𝑥 + sin𝑥(𝑘sin(𝑘−1)𝑥cos𝑥)[Using(2)]
= cos 𝑥 sinkx+ksinkx cos 𝑥
= (k+1)sinkx cos 𝑥
Thus, our assertion is true for 𝑛 = 𝑘 + 1
Hence, by mathematical induction, 𝑑
𝑑𝑥(sin𝑛𝑥) = 𝑛sin(𝑛−1)𝑥cos𝑥
20. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): 𝑎+𝑏sin𝑥
𝑐+𝑑cos𝑥
Hint: 𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1
Solution:
Solution step 1: Let 𝑓(𝑥) =𝑎+𝑏sin𝑥
𝑐+𝑑cos𝑥
By quotient rule,
𝑓′(𝑥) =(𝑐 + 𝑑cos𝑥)
𝑑𝑑𝑥
(𝑎 + 𝑏sin𝑥) − (𝑎 + 𝑏sin𝑥)𝑑𝑑𝑥
(𝑐 + 𝑑cos𝑥)
(𝑐 + 𝑑cos𝑥)2
=(𝑐 + 𝑑cos𝑥)(𝑏cos𝑥) − (𝑎 + 𝑏cos𝑥)(−𝑑sin𝑥)
(𝑐 + 𝑑cos𝑥)2
=𝑐𝑏cos𝑥 + 𝑏𝑑cos2𝑥 + 𝑎𝑑sin𝑥 + 𝑏𝑑sin2𝑥
(𝑐 + 𝑑cos𝑥)2
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=𝑏𝑐cos𝑥 + 𝑎𝑑sin𝑥 + 𝑏𝑑(cos2𝑥 + sin2𝑥)
(𝑐 + 𝑑cos𝑥)2
=𝑏𝑐cos𝑥 + 𝑎𝑑sin𝑥 + 𝑏𝑑
(𝑐 + 𝑑cos𝑥)2
21. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): sin(𝑥+𝑎)
cos𝑥
Hint: 𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1
Solution:
Solution step 1:
Let 𝑓(𝑥) =sin(𝑥+𝑎)
cos𝑥
By quotient rule,
𝑓′(𝑥) =cos𝑥
𝑑𝑑𝑥
[sin(𝑥 + 𝑎)] − sin(𝑥 + 𝑎)𝑑𝑑𝑥
cos𝑥
cos2𝑥
𝑓′(𝑥) =cos𝑥
𝑑𝑑𝑥
[sin(𝑥 + 𝑎)] − sin(𝑥 + 𝑎) (−sin𝑥)
cos2𝑥… (𝑖)
Let 𝑔(𝑥) = sin(𝑥 + 𝑎) Accordingly, 𝑔(𝑥 + ℎ) = sin(𝑥 + ℎ + 𝑎)
By first principle,
𝑔′(𝑥) = limℎ→0
𝑔(𝑥 + ℎ) − 𝑔(𝑥)
ℎ
= limℎ→0
1
ℎ[sin(𝑥 + ℎ + 𝑎) − sin(𝑥 + 𝑎)]
= limℎ→0
1
ℎ[2 cos (
𝑥 + ℎ + 𝑎 + 𝑥 + 𝑎
2) sin (
𝑥 + ℎ + 𝑎 − 𝑥 − 𝑎
2)]
= limℎ→0
1
ℎ[2 cos (
2𝑥 + ℎ + 𝑎 + ℎ
2) sin (
ℎ
2)]
= limℎ→0
[cos (2𝑥 + 2𝑎 + ℎ
2){
sin (ℎ2)
(ℎ2)
}]
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= limℎ→0
cos (2𝑥 + 2𝑎 + ℎ
2) lim
ℎ2,0
{sin (
ℎ2)
(ℎ2)
} [ As ℎ → 0 ⇒ℎ
2→ 0]
= cos(𝑥 + 𝑎) … (𝑖𝑖)
From (𝑖) and (𝑖𝑖), we obtain
𝑓′(𝑥) =cos𝑥 ⋅ cos(𝑥 + 𝑎) + sin𝑥sin(𝑥 + 𝑎)
cos2𝑥
=cos(𝑥 + 𝑎 − 𝑥)
cos2𝑥
=cos𝑎
cos2𝑥
22. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): (𝑥4)(5sin𝑥 − 3cos𝑥)
Hint: 𝑑
𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1
Solution:
Solution step 1: Let 𝑓(𝑥)=(𝑥4)(5sin𝑥 − 3cos𝑥)
By product rule,
𝑓′(𝑥) = 𝑥4𝑑
𝑑𝑥(5sin𝑥 − 3cos𝑥) + (5sin𝑥 − 3cos𝑥)
𝑑
𝑑𝑥(𝑥4)
= 𝑥4 [5𝑑
𝑑𝑥(sin𝑥) − 3
𝑑
𝑑𝑥(cos𝑥)] + (5sin𝑥 − 3cos𝑥)
𝑑
𝑑𝑥(𝑥4)
= 𝑥4[5cos𝑥 − 3(−sin𝑥)] + (5sin𝑥 − 3cos𝑥)(4𝑥3)
= 𝑥3[5𝑥cos𝑥 + 3𝑥sin𝑥 + 20sin𝑥 − 12cos𝑥]
23. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): (𝑥2 + 1)cos𝑥
Hint: Apply product rule.
Solution:
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Solution step 1:
Let 𝑓(𝑥) = (𝑥2 + 1)cos𝑥
By product rule,
𝑓′(𝑥) = (𝑥2 + 1)𝑑
𝑑𝑥(cos𝑥) + cos𝑥
𝑑
𝑑𝑥(𝑥2 + 1)
= (𝑥2 + 1)(−sin𝑥) + cos𝑥(2𝑥)
= −𝑥2sin𝑥 − sin𝑥 + 2𝑥cos𝑥
24. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): (𝑎𝑥2 + sin𝑥)(𝑝 + 𝑞cos𝑥)
Hint: Apply product rule.
Solution:
Solution step 1:
Let 𝑓(𝑥) = (𝑎𝑥2 + sin𝑥)(𝑝 + 𝑞cos𝑥)
By product rule,
𝑓′(𝑥) = (𝑎𝑥2 + sin𝑥)𝑑
𝑑𝑥(𝑝 + 𝑞cos𝑥) + (𝑝 + 𝑞cos𝑥)
𝑑
𝑑𝑥(𝑎𝑥2 + sin𝑥)
= (𝑎𝑥2 + sin𝑥)(−𝑞sin𝑥) + (𝑝 + 𝑞cos𝑥)(2𝑎𝑥 + cos𝑥)
= −𝑞sin𝑥(𝑎𝑥2 + sin𝑥) + (𝑝 + 𝑞cos𝑥)(2𝑎𝑥 + cos𝑥)
25. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): (𝑥 + cos𝑥)(𝑥 − tan𝑥)
Hint: Apply product rule.
Solution:
Solution step 1: Let 𝑓(𝑥) = (𝑥 + cos𝑥)(𝑥 − tan𝑥)
By product rule,
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𝑓′(𝑥) = (𝑥 + cos𝑥)𝑑
𝑑𝑥(𝑥 − tan𝑥) + (𝑥 − tan𝑥)
𝑑
𝑑𝑥(𝑥 + cos𝑥)
= (𝑥 + cos𝑥) [𝑑
𝑑𝑥(𝑥) −
𝑑
𝑑𝑥(tan𝑥)] + (𝑥 − tan𝑥)(1 − sin𝑥)
= (𝑥 + cos𝑥) [1 −𝑑
𝑑𝑥tan𝑥] + (𝑥 − tan𝑥)(1 − sin𝑥)… (𝑖)
Let 𝑔(𝑥) = tan 𝑥 Accordingly, 𝑔(𝑥 + ℎ) = tan(𝑥 + ℎ)
By first principle,
𝑔′(𝑥) = limℎ→0
𝑔(𝑥 + ℎ) − 𝑔(𝑥)
ℎ= lim
ℎ→0(tan(𝑥 + ℎ) − tan𝑥
ℎ)
= limℎ→0
1
ℎ[sin(𝑥 + ℎ)
cos(𝑥 + ℎ)−
sin𝑥
cos𝑥]
= limℎ→0
1
ℎ[sin(𝑥 + ℎ) cos𝑥 − sin𝑥cos(𝑥 + ℎ)
cos(𝑥 + ℎ) cos𝑥]
=1
cos𝑥⋅ lim
ℎ→1ℎ
[sin(𝑥 + ℎ − 𝑥)
cos(𝑥 + ℎ)]
=1
cos𝑥⋅ lim
ℎ→0
1
ℎ[
sinℎ
cos(𝑥 + ℎ)]
=1
cos𝑥⋅ (lim
ℎ→0
sinℎ
ℎ) ⋅ ( lim
ℎ→∞
1
cos(𝑥 + ℎ))
=1
cos𝑥⋅ 1 ⋅
1
cos(𝑥 + 0)
=1
cos2𝑥
= sec2𝑥 … (𝑖𝑖)
Therefore, from (𝑖) and (𝑖𝑖) we obtain
𝑓′(𝑥) = (𝑥 + cos𝑥)(1 − sec2𝑥) + (𝑥 − tan𝑥)(1 − sin𝑥)
= (𝑥 + cos𝑥)(−tan2𝑥) + (𝑥 − tan𝑥)(1 − sin𝑥)
= −tan2𝑥(𝑥 + cos𝑥) + (𝑥 − tan𝑥)(1 − sin𝑥)
26. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): 4𝑥+5sin𝑥
3𝑥+7cos𝑥
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Hint: Apply quotient rule .
Solution:
Solution step 1: Let 𝑓(𝑥) =4𝑥+5sin𝑥
3𝑥+7cos𝑥
By quotient rule,
𝑓′(𝑥) =(3𝑥 + 7cos𝑥)
𝑑𝑑𝑥
(4𝑥 + 5sin𝑥) − (4𝑥 + 5sin𝑥)𝑑𝑑𝑥
(3𝑥 + 7cos𝑥)
(3𝑥 + 7cos𝑥)2
=(3𝑥 + 7cos𝑥) [4
𝑑𝑑𝑥
(𝑥) + 5𝑑𝑑𝑥
(sin𝑥)] − (4𝑥 + 5sin𝑥) [3𝑑𝑑𝑥
𝑥 + 7𝑑𝑑𝑥
cos𝑥]
(3𝑥 + 7cos𝑥)2
=(3𝑥 + 7cos𝑥)(4 + 5cos𝑥) − (4𝑥 + 5sin𝑥)(3 − 7sin𝑥)
(3𝑥 + 7cos𝑥)2
=12𝑥 + 15𝑥cos𝑥 + 28cos𝑥 + 35cos2𝑥 − 12𝑥 + 28𝑥sin𝑥 − 15sin𝑥 + 35sin2𝑥
(3𝑥 + 7cos𝑥)2
=15𝑥cos𝑥 + 28cos𝑥 + 28𝑥sin𝑥 − 15sin𝑥 + 35(cos2𝑥 + sin2𝑥)
(3𝑥 + 7cos𝑥)2
=35 + 15𝑥cos𝑥 + 28cos𝑥 + 28𝑥sin𝑥 − 15sin𝑥
(3𝑥 + 7cos𝑥)2
27. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): 𝑥2 cos(
𝜋
4)
sin𝑥
Hint: Apply quotient rule .
Solution:
Solution step 1: Let 𝑓(𝑥) =𝑥2 cos(
𝜋
4)
sin𝑥
By quotient rule,
𝑓′(𝑥) = cos𝜋
4⋅ [
sin𝑥𝑑𝑑𝑥
(𝑥2) − 𝑥2 𝑑𝑑𝑥
(sin𝑥)
sin2𝑥]
= cos𝜋
4⋅ [
sin𝑥 ⋅ 2𝑥 − 𝑥2cos𝑥
sin2𝑥]
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=𝑥cos
𝜋4
[2sin𝑥 − 𝑥cos𝑥]
sin2𝑥
28. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): 𝑥
1+tan𝑥
Hint: Apply quotient rule .
Solution:
Solution step 1: Let 𝑓(𝑥) =𝑥
1+tan𝑥
𝑓′(𝑥) =(1 + tan𝑥)
𝑑𝑑𝑥
(𝑥) − 𝑥𝑑𝑑𝑥
(1 + tan𝑥)
(1 + tan𝑥)2
𝑓′(𝑥) =(1 + tan𝑥) − 𝑥 ⋅
𝑑𝑑𝑥
(1 + tan𝑥)
(1 + tan𝑥)2…(𝑖)
Let 𝑔(𝑥) = 1 + tan𝑥. Accordingly 𝑔(𝑥 + ℎ) = 1 + tan(𝑥 + ℎ).
By first principle,
𝑔′(𝑥) = limℎ→0
𝑔(𝑥 + ℎ) − 𝑔(𝑥)
ℎ
= limℎ→0
[1 + tan(𝑥 + ℎ) − 1 − tan𝑥
ℎ]
= limℎ→0
1
ℎ[sin(𝑥 + ℎ)
cos(𝑥 + ℎ)−
sin𝑥
cos𝑥]
= limℎ→0
1
ℎ[sin(𝑥 + ℎ) cos𝑥 − sin𝑥cos(𝑥 + ℎ)
cos(𝑥 + ℎ) cos𝑥]
= limℎ→0
1
ℎ[sin(𝑥 + ℎ − 𝑥)
cos(𝑥 + ℎ) cos𝑥]
= limℎ→0
1
ℎ[
sinℎ
cos(𝑥 + ℎ) cos𝑥]
= (limℎ→0
sinℎ
ℎ) ⋅ (lim
ℎ→0
1
cos(𝑥 + ℎ) cos𝑥)
= 1 ×1
cos2𝑥= sec2𝑥
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⇒𝑑
𝑑𝑥(1 + tan𝑥) = sec2𝑥 …(𝑖𝑖)
From (𝑖) and (𝑖𝑖), we obtain
𝑓′(𝑥) =1 + tan𝑥 − 𝑥sec2𝑥
(1 + tan𝑥)2
29. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): (𝑥 + sec 𝑥)(𝑥 − tan 𝑥)
Hint: Apply product rule.
Solution:
Solution step 1: Let 𝑓(𝑥) = (𝑥 + sec𝑥)(𝑥 − tan𝑥)
By product rule,
𝑓′(𝑥) = (𝑥 + sec𝑥)𝑑
𝑑𝑥(𝑥 − tan𝑥) + (𝑥 − tan𝑥)
𝑑
𝑑𝑥(𝑥 + sec𝑥)
= (𝑥 + sec𝑥) [𝑑
𝑑𝑥(𝑥) −
𝑑
𝑑𝑥tan𝑥] + (𝑥 − tan𝑥) [
𝑑
𝑑𝑥(𝑥) +
𝑑
𝑑𝑥sec𝑥]
= (𝑥 + sec𝑥) [1 −𝑑
𝑑𝑥tan𝑥] + (𝑥 − tan𝑥) [1 +
𝑑
𝑑𝑥sec𝑥]… (𝑖)
Let 𝑓1(𝑥) = tan 𝑥 , 𝑓2(𝑥) = sec 𝑥
Accordingly, 𝑓1(𝑥 + ℎ) = tan(𝑥 + ℎ) and 𝑓2(𝑥 + ℎ) = sec(𝑥 + ℎ)
𝑓1′(𝑥) = lim
ℎ→0(𝑓1(𝑥 + ℎ) − 𝑓1(𝑥)
ℎ)
= limℎ→0
(tan(𝑥 + ℎ) − tan𝑥
ℎ)
= limℎ→0
[tan(𝑥 + ℎ) − tan𝑥
ℎ]
= limℎ→0
1
ℎ[sin(𝑥 + ℎ)
cos(𝑥 + ℎ)−
sin𝑥
cos𝑥]
= limℎ→0
1
ℎ[sin(𝑥 + ℎ) cos𝑥 − sin𝑥cos(𝑥 + ℎ)
cos(𝑥 + ℎ) cos𝑥]
= limℎ→0
1
ℎ[sin(𝑥 + ℎ − 𝑥)
cos(𝑥 + ℎ) cos𝑥]
Class–XI–CBSE-Mathematics Limits and Derivatives
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= limℎ→0
1
ℎ[
sinℎ
cos(𝑥 + ℎ) cos𝑥]
= (limℎ→0
sinℎ
ℎ) ⋅ (lim
ℎ→0
1
cos(𝑥 + ℎ) cos𝑥)
= 1 ×1
cos2𝑥= sec2𝑥
⇒𝑑
𝑑𝑥tan𝑥 = sec2𝑥 … (𝑖𝑖)
𝑓2′(𝑥) = lim
ℎ→0(𝑓2(𝑥 + ℎ) − 𝑓2(𝑥)
ℎ)
= limℎ→0
(sec(𝑥 + ℎ) − sec𝑥
ℎ)
= limℎ→0
1
ℎ[
1
cos(𝑥 + ℎ)−
1
cos𝑥]
= limℎ→0
1
ℎ[cos𝑥 − cos(𝑥 + ℎ)
cos(𝑥 + ℎ) cos𝑥]
=1
cos𝑥⋅ lim
ℎ→0
1
ℎ[−2 sin (
𝑥 + 𝑥 + ℎ2
) ⋅ sin (𝑥 − 𝑥 − ℎ
2)
cos(𝑥 + ℎ)]
=1
cos𝑥⋅ lim
ℎ→0
1
ℎ[−2 sin (
2𝑥 + ℎ2 ) ⋅ sin (
−ℎ2 )
cos(𝑥 + ℎ)]
=1
cos𝑥⋅ lim
ℎ→0
[ sin (
2𝑥 + ℎ2 ){
sin (ℎ2)
ℎ2
}
cos(𝑥 + ℎ)
]
= sec𝑥 ⋅
{limℎ→0
sin (2𝑥 + ℎ
2 )} {limℎ2→0
sin (ℎ2)
ℎ2 → 0
ℎ2}
limℎ→0
cos(𝑥 + ℎ)
= sec𝑥 ⋅sin𝑥 ⋅ 1
cos𝑥
⇒𝑑
𝑑𝑥sec𝑥 = sec𝑥 tan𝑥 … (𝑖𝑖𝑖)
From (𝑖), (𝑖𝑖) and (𝑖𝑖𝑖), we obtain
Class–XI–CBSE-Mathematics Limits and Derivatives
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𝑓′(𝑥) = (𝑥 + sec𝑥)(1 − sec2𝑥) + (𝑥 − tan𝑥)(1 + sec𝑥 tan𝑥)
30. Find the derivative of the following functions (it is to be understood that 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞, 𝑟 and 𝑠 are
fixed non-zero constants and m and n are integers): 𝑥
sin𝑛𝑥
Hint: Apply quotient rule.
Solution:
Solution step 1: Let 𝑓(𝑥) =𝑥
sin𝑛𝑥
By quotient rule,
𝑓′(𝑥) =sin𝑛𝑥
𝑑𝑑𝑥
𝑥 − 𝑥𝑑𝑑𝑥
sin𝑛𝑥
sin2𝑛𝑥
It can be easily shown that 𝑑
𝑑𝑥sin𝑛𝑥 = 𝑛sin𝑛−1𝑥cos𝑥
Therefore,
𝑓′(𝑥) =sin𝑛𝑥
𝑑𝑑𝑥
𝑥 − 𝑥𝑑𝑑𝑥
sin𝑛𝑥
sin2𝑛𝑥
=sin𝑛𝑥 ⋅ 1 − 𝑥(𝑛sin𝑛−1𝑥cos𝑥)
sin2𝑛𝑥
=sin𝑛−1𝑥(sin𝑥 − 𝑛𝑥cos𝑥)
sin2𝑛𝑥
=sin𝑥 − 𝑛𝑥cos𝑥
sin𝑛+1𝑥