CBSE NCERT Solutions for Class 11 Mathematics Chapter 13 · 2019. 11. 29. ·...

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



Solution:


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


(𝑥+1)5−1

𝑥

Hint: At the time when we put variable value make sure function won’t take undefined form.

Solution:


(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


3𝑥2−𝑥−10

𝑥2−4




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


𝑥4−81

2𝑥2−5𝑥−3


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


𝑎𝑥+𝑏

𝑐𝑥+1




Solution:


𝑎𝑥+𝑏

𝑐𝑥+1=

𝑎(0)+𝑏

𝑐(0)+1= 𝑏

10. Evaluate the Given limit: lim𝑧→1

𝑧13−1

𝑧16−1


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


𝑎𝑥2+𝑏𝑥+𝑐

𝑐𝑥2+𝑏𝑥+𝑎, 𝑎 + 𝑏 + 𝑐 ≠ 0




Solution:



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


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


sin𝑎𝑥

𝑏𝑥


Solution:


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)



=𝑎

𝑏lim

𝑎𝑥→0(sin𝑎𝑥

𝑏𝑥) [𝑥 → 0 ⇒ 𝑎𝑥 → 0]

=𝑎

𝑏× 1 [lim

𝑦→0

sin𝑦

𝑦= 1]

=𝑎

𝑏


sin𝑎𝑥

sin𝑏𝑥, 𝑎, 𝑏 ≠ 0


Solution:


sin𝑎𝑥

sin𝑏𝑥, 𝑎, 𝑏 ≠ 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)


Solution:

Solution step 1: lim𝑥→𝜋

sin(𝜋−𝑥)

𝜋(𝜋−𝑥)

It is seen that 𝑥 → 𝜋 ⇒ (𝜋 – 𝑥) → 0



∴ 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


cos𝑥

𝜋−𝑥


Solution:


cos𝑥

𝜋−𝑥=

cos0

𝜋−0=

1

𝜋


cos2𝑥−1

cos𝑥−1


Solution:


cos2𝑥−1

cos𝑥−1


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



= 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


𝑎𝑥+𝑥 cos𝑥

𝑏 sin𝑥


Solution:


𝑎𝑥+𝑥 cos𝑥

𝑏 sin𝑥


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




𝑥 sec 𝑥


Solution:


𝑥 sec 𝑥 = lim𝑥→0

𝑥

cos𝑥=

0

cos0=

0

1= 0


sin𝑎𝑥+𝑏𝑥

𝑎𝑥+sin𝑏𝑥 𝑎, 𝑏, 𝑎 + 𝑏 ≠ 0


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




(cosec 𝑥 − cot 𝑥)


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


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



∴ 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


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−


3[𝑥 + 1] = 3(0 + 1) = 3

∴ lim𝑥→0−



𝑓(𝑥) = 3

lim𝑥→1−


3[𝑥 + 1] = 3(1 + 1) = 6

lim𝑥→1−


2(𝑥 + 1) = 3(1 + 1) = 6

∴ lim𝑥→1−

𝑓(𝑥) = lim𝑥→1+


𝑓(𝑥) = 6




𝑓(𝑥), where 𝑓(𝑥) = {𝑥2 − 1, 𝑥 ≤ 1

−𝑥2 − 1, 𝑥 > 1


Solution:

Solution step 1: The given function is

𝑓(𝑥) = {𝑥2 − 1, 𝑥 ≤ 1

−𝑥2 − 1, 𝑥 > 1

lim𝑥→1−


[𝑥2 − 1] = 12 − 1 = 1 − 1 = 0

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


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+


[|𝑥|

𝑥]

= lim𝑥→0

[𝑥

𝑥] [When 𝑥 is positive, |𝑥| = 𝑥]



= lim𝑥→0

(1)

= 1

It is observed that lim𝑥→0−


𝑓(𝑥).

Hence, lim𝑥→0



𝑓(𝑥), where 𝑓(𝑥) = {

𝑥

|𝑥|, 𝑥 ≠ 0

0, 𝑥 = 0


Solution:


𝑓(𝑥) = {

𝑥

|𝑥|, 𝑥 ≠ 0

0, 𝑥 = 0

lim𝑥→0−


[𝑥

|𝑥|]

= lim𝑥→0

[𝑥

−𝑥] [When 𝑥 < 0, |𝑥| = −𝑥]

= lim𝑥→0

(−1)

= −1

lim𝑥→0+


[𝑥

|𝑥|]

= lim𝑥→0

[𝑥

𝑥] [When 𝑥 > 0, |𝑥| = 𝑥]

It is observed that lim𝑥→0−


𝑓(𝑥).

Hence, lim𝑥→0



𝑓(𝑥), where 𝑓(𝑥) = |𝑥| − 5




Solution:

Solution step 1: The given function is 𝑓(𝑥) = |𝑥| − 5

lim𝑥→5−


[|𝑥| − 5]

= lim𝑥→5

(𝑥 − 5) [When 𝑥 > 0, |𝑥| = 𝑥]

= 5 − 5

= 0

lim𝑥→5+


(|𝑥| − 5)

lim𝑥→5

(𝑥 − 5) [When 𝑥 > 0, |𝑥| = 𝑥]

= 5 − 5

= 0

∴ 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?


Solution:


𝑓(𝑥) = {

𝑎 + 𝑏𝑥, 𝑖𝑓 𝑥 < 14, 𝑖𝑓 𝑥 = 0

𝑏 − 𝑎𝑥, 𝑖𝑓 𝑥 > 1

lim𝑥→1+


(𝑎 + 𝑏𝑥) = 𝑎 + 𝑏

lim𝑥→1+


(𝑏 − 𝑎𝑥) = 𝑏 − 𝑎

𝑓(1) = 4

It is given that lim𝑥→1

𝑓(𝑥) = 𝑓(1)



∴ 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)


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?




Solution:


𝑓(𝑥) = {|𝑥| + 1, 𝑥 < 0

0, 𝑥 = 0|𝑥| − 1, 𝑥 > 0

When 𝑎 = 0

lim𝑥→0


(|𝑥| + 1)

= lim𝑥→0

(|𝑥| + 1)[If 𝑥 < 0, |𝑥| = −𝑥]

= −0 + 1

= 1

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)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) = −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)



= lim𝑥→𝑎

(−𝑥 − 1)[0 < 𝑥 < 𝑎 ⇒ |𝑥| = 𝑥]

= 𝑎 − 1

lim𝑥→𝑎+


(|𝑥| − 1)

= lim𝑥→𝑎

(𝑥 − 1)[0 < 𝑎 < 𝑥 ⇒ |𝑥| = 𝑥]

= 𝑎 − 1

∴ 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


Solution:


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



⇒ 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?


Solution:


𝑓(𝑥) = {𝑚𝑥2 + 𝑛, 𝑥 < 0𝑛𝑥 + 𝑚, 0 ≤ 𝑥 ≤ 1

𝑛𝑥3 + 𝑚, 𝑥 > 1

lim𝑥→0−


(𝑚𝑥2 + 𝑛)

= 𝑚(0)2 + 𝑛

= 𝑛

lim𝑥→0+


(𝑛𝑥 + 𝑚)

= 𝑛(0) + 𝑚

= 𝑚.

Thus, lim𝑥→0

𝑓(𝑥) exists if 𝑚 = 𝑛.

lim𝑥→1−


(𝑛𝑥 + 𝑚)

= 𝑛(1) + 𝑚

= 𝑚 + 𝑛

lim𝑥→1+


(𝑛𝑥3 + 𝑚)

= 𝑛(1)3 + 𝑚

= 𝑚 + 𝑛

∴ lim𝑥→1−



𝑓(𝑥)

Thus, lim𝑥→1

𝑓(𝑥)exists for any integral value of 𝑚 and 𝑛.



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.



3. Find the derivative of 𝑥 at 𝑥 = 1.



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



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



(ii)(𝑥 − 1)(𝑥 − 2)



ℎ

Solution:

Solution step 1: Let 𝑓(𝑥) = (𝑥– 1)(𝑥– 2). Accordingly, from the first principle,


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



ℎ

Solution:

Solution step 1: Let 𝑓(𝑥) =1

𝑥2 . Accordingly, from the first principle,


𝑓(𝑥 + ℎ) − 𝑓(𝑥)

ℎ

= limℎ→0

1(𝑥 + ℎ)2 −

1𝑥2

ℎ



= 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



ℎ

Solution:

Solution step 1: Let 𝑓(𝑥) =𝑥+1

𝑥−1. Accordingly, from the first principle,


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



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:


𝑓(𝑥) =𝑥100

100+

𝑥99

99+ ⋯ +

𝑥2

2+ 𝑥 + 1

𝑑

𝑑𝑥𝑓(𝑥) =

𝑑

𝑑𝑥[𝑥100

100+

𝑥99

99+ ⋯+

𝑥2

2+ 𝑥 + 1]

𝑑


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


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


Solution:

Solution step 1: Let 𝑓(𝑥) = 𝑥𝑛 + 𝑎𝑥𝑛−1 + 𝑎2𝑥𝑛−2 + ⋯+ 𝑎𝑛−1𝑥 + 𝑎𝑛



∴ 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)


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


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)


𝑑𝑥(𝑥𝑛) = 𝑛𝑥𝑛−1, text we obtain

𝑓′(𝑥) = 2𝑥 − (𝑎 + 𝑏) + 0 = 2𝑥 − 𝑎 − 𝑏

(ii)(𝑎𝑥2 + 𝑏)2

Hint: 𝑑


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)




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


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


Solution:

Solution step 1: Let 𝑓(𝑥) =𝑥𝑛−𝑎𝑛

𝑥−𝑎



⇒ 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


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


Solution:

Solution step 1: Let 𝑓(𝑥) = (5𝑥3 + 3𝑥 − 1)(𝑥 − 1)

By Leibnitz product rule,



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


Solution:

Solution step 1: Let 𝑓(𝑥) = 𝑥−3(5 + 3𝑥)


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


Solution:

Solution step 1: Let 𝑓(𝑥) = 𝑥5(3 − 6𝑥−9)




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


Solution:

Solution step 1: Let 𝑓(𝑥) = 𝑥−4(3 − 4𝑥−5)


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


Solution:

Solution step 1:Let 𝑓(𝑥) = 𝑐2

𝑥+1−

𝑥2

3𝑥−1



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.



ℎ

Solution:

Solution step 1: Let 𝑓(𝑥) = 𝑐𝑜𝑠 𝑥. Accordingly, from the first principle,


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)



= −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



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



ℎ



Solution:

Solution step 1: Let 𝑓(𝑥) = sec 𝑥. Accordingly, from the first principle,


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



ℎ

Solution:

Solution step 1: Let 𝑓(𝑥) = 5 𝑠𝑒𝑐𝑥 + 4 𝑐𝑜𝑠 𝑥. Accordingly, from the first principle,


𝑓(𝑥 + ℎ) − 𝑓(𝑥)

ℎ



= 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


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



ℎ

Solution:

Solution step 1: Let 𝑓(𝑥) = cosec 𝑥. Accordingly, from the first principle,


𝑓(𝑥 + ℎ) − 𝑓(𝑥)

ℎ


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]



= limℎ→0

1

ℎ[2 cos (

𝑥 + 𝑥 + ℎ2 ) ⋅ sin (

𝑥 − 𝑥 − ℎ2 )


= limℎ→0

1

ℎ[2 cos (

2𝑥 + ℎ2 ) sin (−

ℎ2)


= 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



ℎ

Solution:

Solution step 1: Let 𝑓(𝑥) = 3𝑐𝑜𝑡 𝑥 + 5𝑐𝑜𝑠𝑒𝑐 𝑥. Accordingly, from the first principle,


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)]



= limℎ→0

1

ℎ[sin(𝑥 − 𝑥 − ℎ)


= limℎ→0

1

ℎ[

sin(−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



= limℎ→0

1

ℎ[2 cos (

𝑥 + 𝑥 + ℎ2 ) ⋅ sin (

𝑥 − 𝑥 − ℎ2 )


= limℎ→0

1

ℎ[2 cos (

2𝑥 + ℎ2 ) sin (−

ℎ2)


= 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



ℎ



Solution:

Solution step 1: Let 𝑓(𝑥) = 5𝑠𝑖𝑛 𝑥– 6𝑐𝑜𝑠 𝑥 + 7. Accordingly, from the first principle,


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



ℎ

Solution:

Solution step 1: Let 𝑓(𝑥) = 2 tan 𝑥 – 7 sec 𝑥. Accordingly, from the first principle,


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





= 2limℎ→0

1

ℎ[sin(𝑥 + ℎ − 𝑥)


ℎ→0

1

ℎ[−2 sin (

𝑥 + 𝑥 + ℎ2 ) sin (

𝑥 − 𝑥 − ℎ2 )


= 2limℎ→0

[(sinℎ

ℎ)

1


ℎ→0

1

ℎ[−2 sin (

2𝑥 + ℎ2 ) sin (−

ℎ2)


= 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



ℎ

Solution:

Solution step 1: Let 𝑓(𝑥) =–𝑥. Accordingly, 𝑓(𝑥 + ℎ) = −(𝑥 + ℎ)

By first principle,


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



ℎ



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 + 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)



ℎ

Solution:

Solution step 1: Let 𝑓(𝑥) = sin(𝑥 + 1). Accordingly,𝑓(𝑥 + ℎ) = sin(𝑥 + ℎ + 1)

By first principle,


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)]



= 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)



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 + 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]

]



= 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)



ℎ

Solution:

Solution step 1: Let 𝑓(𝑥) = 𝑥 + 𝑎. Accordingly, 𝑓(𝑥 + ℎ) = −(𝑥 + ℎ)

By first principle,


𝑓(𝑥 + ℎ) − 𝑓(𝑥)

ℎ

= limℎ→0

𝑥 + ℎ + 𝑎 − 𝑥 − 𝑎

ℎ

= limℎ→0

(ℎ

ℎ)

= limℎ→0

(1)

= 1


fixed non-zero constants and m and n are integers): (𝑝𝑥 + 𝑞) (𝑟

𝑥+ 𝑠)

Hint: 𝑑


Solution:



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


fixed non-zero constants and m and n are integers): (𝑎𝑥 + 𝑏)(𝑐𝑥 + 𝑑)2

Hint: 𝑑


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


fixed non-zero constants and m and n are integers): 𝑎𝑥+𝑏

𝑐𝑥+𝑑

Hint: 𝑑




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


fixed non-zero constants and m and n are integers): 1+

1

𝑥

1−1

𝑥

Hint: 𝑑


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




fixed non-zero constants and m and n are integers): 1


Hint: 𝑑


Solution:

Solution step 1: Let 𝑓(𝑥) =1


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


fixed non-zero constants and m and n are integers): 𝑎𝑥+𝑏

𝑝𝑥2+𝑞𝑥+𝑟

Hint: 𝑑


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




fixed non-zero constants and m and n are integers): 𝑝𝑥2+𝑞𝑥+𝑟

𝑎𝑥+𝑏

Hint: 𝑑


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


fixed non-zero constants and m and n are integers): 𝑎

𝑥4 −𝑏

𝑥2 + cos

Hint: 𝑑


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




fixed non-zero constants and m and n are integers): 4√𝑥 − 2

Hint: 𝑑


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


fixed non-zero constants and m and n are integers): (𝑎𝑥 + 𝑏)𝑛



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 + 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


1

𝑛[{1 + 𝑛 (

𝑎ℎ

𝑎𝑥 + 𝑏) +

𝑛(𝑛 − 1)

∟2(

𝑎ℎ

𝑎𝑥 + 𝑏)

2

+ ⋯} − 1] (Using binomial theorem)




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


fixed non-zero constants and m and n are integers): (𝑎𝑥 + 𝑏)𝑛(𝑐𝑥 + 𝑑)𝑚



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]


1

ℎ[(1 +

𝑚𝑐ℎ

(𝑐𝑥 + 𝑑)+

𝑚(𝑚 − 1)

2

(𝑐2ℎ2)

(𝑐𝑥 + 𝑑)2+ ⋯) − 1]


1

ℎ[

𝑚𝑐ℎ

(𝑐𝑥 + 𝑑)+

𝑚(𝑚 − 1)𝑐2ℎ2

2(𝑐𝑥 + 𝑑)2+ ⋯ (Terms containing higher degrees of ℎ)]


[𝑚𝑐

(𝑐𝑥 + 𝑑)+

𝑚(𝑚 − 1)𝑐2ℎ

2(𝑐𝑥 + 𝑑)2+ ⋯ ]



= (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)]


fixed non-zero constants and m and n are integers): sin(𝑥 + 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 + 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]



= cos (2𝑥 + 2𝑎

2) × 1 [lim

𝑥→0

sin𝑥

𝑥= 1]

= cos(𝑥 + 𝑎)


fixed non-zero constants and m and n are integers): cosec 𝑥 cot 𝑥



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,


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)


= limℎ→0

1

ℎ[sin(𝑥 − 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



∴ (cot𝑥)′ = −cosec2𝑥 … (2)

Now, let 𝑓2(𝑥) = 𝑐𝑜𝑠𝑒𝑐 𝑥.

Accordingly, 𝑓2(𝑥 + ℎ) = 𝑐𝑜𝑠𝑒𝑐 (𝑥 + ℎ)

By first principle,


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



=1

sin𝑥∙ limℎ→0

1

ℎ[2 cos (

𝑥 + 𝑥 + ℎ2 ) sin (

𝑥 − 𝑥 − ℎ2 )

sin(𝑥 + ℎ)]

=1


1

ℎ= [

2 cos (2𝑥 + ℎ

2 ) sin (−ℎ2 )

sin(𝑥 + ℎ)]

=1


[− sin (

ℎ2)

(ℎ2)

⋅cos (

2𝑥 + ℎ2 )

sin(𝑥 + ℎ)]

=−1


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




fixed non-zero constants and m and n are integers): cos𝑥

1+sin𝑥



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)


fixed non-zero constants and m and n are integers): sin𝑥+cos𝑥

sin𝑥−cos𝑥



ℎ

Solution:

Solution step 1: Let 𝑓(𝑥)sin𝑥+cos𝑥

sin𝑥−cos𝑥

By quotient rule,



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


=−[sin2𝑥 + cos2𝑥 − 2sin𝑥cos𝑥 + sin2𝑥 + cos2𝑥 + 2sin𝑥cos𝑥]


=−[1 + 1]


=−2



fixed non-zero constants and m and n are integers): sec𝑥−1

sec𝑥+1

Hint: 𝑑


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



=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


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)



= (sin𝑥)′sin2𝑥 + sin𝑥(sin2𝑥)′ [By Leibnitz product rule]

= cos𝑥sin2𝑥 + sin𝑥(2sin𝑥cos𝑥)[Using(1)]

=cosxsin2x+2sin2xcosx

=3sin2xcosx

We assert that 𝑑


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



fixed non-zero constants and m and n are integers): 𝑎+𝑏sin𝑥

𝑐+𝑑cos𝑥

Hint: 𝑑


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)


=𝑐𝑏cos𝑥 + 𝑏𝑑cos2𝑥 + 𝑎𝑑sin𝑥 + 𝑏𝑑sin2𝑥




=𝑏𝑐cos𝑥 + 𝑎𝑑sin𝑥 + 𝑏𝑑(cos2𝑥 + sin2𝑥)


=𝑏𝑐cos𝑥 + 𝑎𝑑sin𝑥 + 𝑏𝑑



fixed non-zero constants and m and n are integers): sin(𝑥+𝑎)

cos𝑥

Hint: 𝑑


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)

}]



= 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


fixed non-zero constants and m and n are integers): (𝑥4)(5sin𝑥 − 3cos𝑥)

Hint: 𝑑


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]


fixed non-zero constants and m and n are integers): (𝑥2 + 1)cos𝑥

Hint: Apply product rule.

Solution:



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


fixed non-zero constants and m and n are integers): (𝑎𝑥2 + sin𝑥)(𝑝 + 𝑞cos𝑥)


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)


fixed non-zero constants and m and n are integers): (𝑥 + cos𝑥)(𝑥 − tan𝑥)


Solution:

Solution step 1: Let 𝑓(𝑥) = (𝑥 + cos𝑥)(𝑥 − tan𝑥)

By product rule,



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 + 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


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)


fixed non-zero constants and m and n are integers): 4𝑥+5sin𝑥

3𝑥+7cos𝑥



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


fixed non-zero constants and m and n are integers): 𝑥2 cos(

𝜋

4)

sin𝑥


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]



=𝑥cos

𝜋4

[2sin𝑥 − 𝑥cos𝑥]

sin2𝑥


fixed non-zero constants and m and n are integers): 𝑥

1+tan𝑥


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 + 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



= limℎ→0

1



= limℎ→0

1

ℎ[

sinℎ


= (limℎ→0

sinℎ

ℎ) ⋅ (lim

ℎ→0

1

cos(𝑥 + ℎ) cos𝑥)

= 1 ×1

cos2𝑥= sec2𝑥



⇒𝑑

𝑑𝑥(1 + tan𝑥) = sec2𝑥 …(𝑖𝑖)

From (𝑖) and (𝑖𝑖), we obtain

𝑓′(𝑥) =1 + tan𝑥 − 𝑥sec2𝑥

(1 + tan𝑥)2


fixed non-zero constants and m and n are integers): (𝑥 + sec 𝑥)(𝑥 − tan 𝑥)


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)


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



= limℎ→0

1





= limℎ→0

1

ℎ[

sinℎ


= (limℎ→0

sinℎ

ℎ) ⋅ (lim

ℎ→0

1

cos(𝑥 + ℎ) cos𝑥)

= 1 ×1

cos2𝑥= sec2𝑥

⇒𝑑

𝑑𝑥tan𝑥 = sec2𝑥 … (𝑖𝑖)


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



=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



𝑓′(𝑥) = (𝑥 + sec𝑥)(1 − sec2𝑥) + (𝑥 − tan𝑥)(1 + sec𝑥 tan𝑥)


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

CBSE NCERT Solutions for Class 11 Mathematics Chapter 13 · 2019. 11. 29. ·...

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Transcript of CBSE NCERT Solutions for Class 11 Mathematics Chapter 13 · 2019. 11. 29. ·...