NCERT Class 12 Mathematics Chapter 5 Continuity and Differentiability

NCERT Class 12 Mathematics Chapter 5 Continuity and Differentiability Solutions, NCERT Solutions For Class 12 Maths, NCERT Class 12 Mathematics Chapter 5 Continuity and Differentiability Notes to each chapter is provided in the list so that you can easily browse throughout different chapter NCERT Class 12 Mathematics Chapter 5 Continuity and Differentiability Question Answer and select needs one.

NCERT Class 12 Mathematics Chapter 5 Continuity and Differentiability

Join Telegram channel

Follow us:

Also, you can read the CBSE book online in these sections Solutions by Expert Teachers as per (CBSE) Book guidelines. NCERT Class 12 Mathematics Chapter 5 Continuity and Differentiability Question Answer. These solutions are part of NCERT All Subject Solutions. Here we have given NCERT Class 12 Mathematics Chapter 5 Continuity and Differentiability Solutions for All Subject, You can practice these here.

Continuity and Differentiability

Chapter – 5

1. Prove that the function f(x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5.

Ans: 

2. Examine the continuity of the function f(x) = 2x² – 1 at x = 3

Ans: 

3. Examine the following functions for continuity.

(a) f(x) = x – 5

Ans: 

Ans: 

Ans: 

Hence, f is continuous at every point in the domain of f and therefore, it is a continuous function.

(d) f(x) = |x – 5|

Ans: 

Therefore, f is continuous at all real numbers greater than 5

Hence, f is continuous at every real number and therefore, it is a continuous function.

4. Prove that the function f(x) = xⁿ is continuous at x = n, where n is a positive integer.

Ans: 

5. Is the function f defined by continuous at x = 0? At x = 1 ? At x = 2?

Ans: 

Find all points of discontinuity of f, where f is defined by 

Ans: 

It is observed that the left and right hand limit of f at x = 2 do not coincide.

Therefore, is not continuous at x = 2

Hence, x = 2 is the only point of discontinuity of f.

Ans: 

Therefore, f is continuous at all points x, such that x > 3

Hence, x = 3 is the only point of discontinuity of f.

Ans: 

It is observed that the left and right hand limit of f at x = 0 do not coincide.

Therefore, f is not continuous at x = 0

Ans: 

Therefore, the given function is a continuous function. 

Hence, the given function has no point of discontinuity.

Ans: 

Therefore, f is continuous at all points x, such that x > 1

Hence, the given function f has no point of discontinuity.

Ans: The given function is

The given function f is defined at all the points of the real line.

Let c be a point on the real line. 

Case I:

If c < 2 , then f(c) = c³ – 3

Therefore, f is continuous at all points x, such that x > 2

Thus, the given function f is continuous at every point on the real line.

Hence, f has no point of discontinuity.

Ans: The given function is

The given function f is defined at all the points of the real line. 

Let c be a point on the real line.

Case I:

If c < 1 , then f(c) = c¹⁰ – 1

Therefore, f is continuous at all points x, such that x < 1

Case II:

If c = 1, then the left hand limit of f at x = 1 is,

Therefore, f is continuous at all points x, such that x > 1

Thus from the above observation, it can be concluded that x = 1 is the only point of discontinuity of f.

13. Is the function defined by a continuous function?

Ans: The given function is 

The given function f is defined at all the points of the real line. 

Let c be a point on the real line.

Case I:

If c < 1 , then f(c) = c + 5

Therefore, f is continuous at all points x, such that x < 1

Case II:

If c = 1, then f(1) = 1 + 5 = 6

Therefore, f is continuous at all points x, such that x > 1

From the above observation it can be concluded that, x = 1 is the only point of discontinuity of f.

Discuss the continuity of the function f, where f is defined by

Ans: The given function is

The given function f is defined at all the points of the interval [0,10].

Let c be a point in the interval [0,10].

Case I:

If 0 ≤ c < 1 , then f(c) = 3

It is observed that the left and right hand limit of f at x = 1 do not coincide. Therefore, f is not continuous x = 1

Case III:

If 1 < c < 3, then f(c) = 4

It is observed that the left and right hand limit of f at x = 3 do not coincide. Therefore, f is discontinuous at x = 3

Case V:

If 3 < c <= 10 , then f(c) = 5

Therefore, f is continuous at all points of the interval (3,10]

Hence, f is discontinuous at x = 1 and x = 3

Ans: The given function is

The given function f is defined at all the points of the real line. 

Let c be a point on the real line.

Case I:

If c < 0, then f(c) = 2c

Therefore, f is continuous at x = 0

Case III:

If 0 < c < 1, then f(x) = 0

Therefore, f is continuous at all points x, such that x > 1

Hence, f is not continuous only at x = 1

Ans: The given function is

The given function f is defined at all the points. 

Let c be a point on the real line.

Case I:

If c < -1, then f (c) = -2

Therefore, f is continuous at all points x, such that x > 1

Thus, from the above observations, it can be concluded that f is continuous at all points of the real line.

17. Find the relationship between a and b so that the function f defined by

is continuous at x = 3.

Ans: 

18. For what value of λ is the function defined by

continuous at x = 0? What about continuity at x = 1?

Ans: 

19. Show that the function defined by g(x) = x-[x] is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x.

Ans: The given function is g(x) = x – [x]

It is evident that g is defined at all integral points. 

Let n be an integer.

Then, 

g(n) = n – [n] = n – n = 0 

The left hand limit of g at x = n is,

It is observed that the left and right hand limit of g at x = n do not coincide. 

Therefore, g is not continuous at x = n. 

Hence, g is discontinuous at all integral points.

20. Is the function defined by f(x) = x² – sin x + 5 continuous at x = π?

Ans: 

21. Discuss the continuity of the following functions:

(a) f(x) = sin x + cos x

(b) f(x) = sin x – cos X

(c) f(x) = sin x . cos x

Ans: It is known that if g and h are two continuous functions, then g + h, g-h and g,h are also continuous.

Let g(x) = sin x and h(x) = cos x are continuous functions.

It is evident that g(x) = sin x is defined for every real number.

Let c be a real number. Put x = c + h 

If x→c, then h → 0 

g(c) = sin c

Therefore, g(x) = sin x is a continuous function.

Let h(x) = cos x

It is evident that h(x) = cos x is defined for every real number.

Let c be a real number. Put x = c + h 

If x → c , then h → 0 

h(c) = cos c

Therefore, h(x) = cos x is a continuous function. 

Therefore, it can be concluded that,

(i) f(x) = g(x) + h(x) = sin x + cos x is a continuous function.

(ii) f(x) = g(x) – h(x) = sin x – cos x is a continuous function.

(iii) f(x) = g(x) × h(x) = sin x × cos x is a continuous function.

22. Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

Ans: It is known that if g and h are two continuous functions, then

Let g(x) = sin x and h(x) = cos x are continuous functions. 

It is evident that g(x) = sin x is defined for every real number. 

Let c be a real number. Put x = c + h

If x → c, then h → 0

Therefore, g(x) = sin x is a continuous function. 

Let h(x) = cos x

It is evident that h(x) = cos x is defined for every real number. 

Let c be a real number. Put x = c + h 

If x → c , then h → 0

h(c) = cos c

Therefore, h(x) = cos x is a continuous function. 

Therefore, it can be concluded that,

⇒ cot x,x ≠ n π(n ∈ Z) is continuous.

Therefore, cotangent is continuous except at x = n π(n ∈ Z).

23. Find all points of discontinuity of f, where

Ans: The given function is 

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

Therefore, f is continuous at x = 0

From the above observations, it can be concluded that f is continuous at all points of the real line.

Thus, f has no point of discontinuity.

24. Determine if f defined by 

is a continuous function?

Ans: The given function is 

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

Therefore, f is continuous at all points x, such that x ≠ 0

Case II:

If c = 0, then f(0) = 0

Therefore, f is continuous at x = 0

From the above observations, it can be concluded that f is continuous at every point of the real line.

Thus, f is a continuous function.

25. Examine the continuity of f, where f is defined by

Ans: The given function is

The given function f is defined at all the points of the real line. 

Let c be a point on the real line.

Case I:

If c ≠ 0 , then f(c) = sin c – cos c

Therefore, f is continuous at all points x, such that x ≠ 0

Therefore, f is continuous at x = 0

From the above observations, it can be concluded that f is continuous at every point of the real line.

Thus, f is a continuous function.

Find the values of k so that the function f is continuous at the indicated point in Exercises 26 to 29.

Ans: 

⇒ k = 6

Therefore, the value of k = 6

Ans: The given function is

The given function f is continuous at x = 2, if f is defined at x = 2 and if the value of the f at x = 2 equals the limit of f at x = 2

It is evident that f is defined at x = 2 and f (2) = k (2)² = 4k

Ans: The given function is

The given function f is continuous at x = π, if f is defined at x = π and if the value of the f at x = π equals the limit of f at x = π.

It is evident that f is defined at x = π and f(π) = kπ + 1

Ans: The given function is

The given function f is continuous at x = 5, if f is defined at x = 5 and if the value of the f at x = 5 equals the limit of f at x = 5

It is evident that f is defined at x = 5 and f(5) = kx + 1 = 5k + 1

30. Find the values of a and b such that the function defined by

is a continuous function.

Ans: The given function is

It is evident that f is defined at all points of the real line. 

If f is a continuous function, then f is continuous at all real numbers. 

In particular, f is continuous at x = 2 and x = 10

Since f is continuous at x = 2 we obtain

On subtracting equation (1) from equation (2), we obtain

8a = 16

⇒ a = 2 

By putting a = 2 in equation (1), we obtain 

2(2) + b = 5 

⇒ 4 + b = 5 

⇒ b = 1 

Therefore, the values of a and b for which f is a continuous function are 2 and 1 respectively.

31. Show that the function defined by f(x) = cos(x²) is a continuous function.

Ans: The given function is f(x) = cos(x²) 

This function f is defined for every real number and f can be written as the composition of two functions as, 

f = goh, where g(x) = cos x and h(x) = x² 

[∵ (goh)(x) = g(h(x)) = g(x²) = cos(x²) = f(x)] 

It has to be proved first that g(x) = cos x and h(x) = x² are continuous functions. 

It is evident that g is defined for every real number.

Let c be a real number.

Let g(c) = cos c Put x = c + h

If x → c, then h → 0

Therefore, g(x) = cos x is a continuous function.

Let h(x) = x²

It is evident that h is defined for every real number. 

Let k be a real number, then h(k) = k²

Therefore, h is a continuous function.

It is known that for real valued functions g and h, such that (goh) is defined at c, if g is continuous at c and if f is continuous at g(c), then (fog) is continuous at c. 

Therefore, f(x) = (goh)(x) = cos(x²) is a continuous function.

32. Show that the function defined by f(x) = |cos x| is a continuous function.

Ans: The given function is f(x) = |cos x| 

This function f is defined for every real number and f can be written as the composition of two functions as, 

f = goh, where g(x) = |x| and h(x) = cos x 

[∵ (goh)(x) = g(h(x)) = g(cos x) = |cos x| = f(x)] 

It has to be proved first that g(x) = |x| and h(x) = cos x are continuous functions.

g(x) = |x| can be written as

It is evident that g is defined for every real number. 

Let c be a real number.

Case I:

If c < 0 , then g(c) = – c

Therefore, g is continuous at all points x, such that x < 0

Case II:

If c > 0, then g (c) = c

Therefore, g is continuous at all x = 0

From the above three observations, it can be concluded that g is continuous at all points. 

Let h(x) = cos x

It is evident that h(x) = cos x is defined for every real number.

Let c be a real number. Put x = c + h

If x → c, then h → 0

h(c) = cos c

Therefore, h(x) = cos x is a continuous function.

It is known that for real valued functions g and h, such that (goh) is defined at c, if g is continuous at c and if f is continuous at g(c), then (fog) is continuous at c.

Therefore, f (x) = (goh)(x) = g(h(x)) = g(cos x) = |cos x| is a continuous function.

33. Examine that sin |x| is a continuous function.

Ans: Let g(x) = |x| and h(x) = sin x

Now, g(x)=|x| is the absolute valued function, so it is continuous function for all x ∈ R

h(x) = sin x is the sine function, so it is continuous function for all x ∈ R

∴ (hog)(x) = h[g(x)] = h(|x|) = sin |x|

Since, g(x) and h(x) are both continuous functions for all x ∈R so, composition

of g(x) and h(x) is also a continuous function for all x ∈ R

Thus, f(x) = sin |x| is a continuous function.

34. Find all the points of discontinuity of f defined by f(x) = |x| – |x + 1|.

Ans: The given function is f (x) = |x| – |x + 1|

The two functions, g and h are defined as g(x) = |x| and h(x) = |x + 1| 

Then, f = g – h

The continuity of g and h are examined first.

g(x) = |x| can be written as

It is evident that g is defined for every real number. 

Let c be a real number.

Case I:

If c < 0 , then g(c) = – c

Therefore, g is continuous at all points x, such that x < 0

Case II:

If c > 0, then g(c) = c

Therefore, g is continuous at all points x, such that x > 0

Case III:

If c = 0 , then g(c) = g(0) = 0

Therefore, g is continuous at all x = 0

From the above three observations, it can be concluded that g is continuous at all points.

h(x) = |x + 1| can be written as

It is evident that h is defined for every real number.

Let c be a real number.

Case I:

If c < – 1, then h(c) = – (c + 1)

Therefore, h is continuous at x = -1

From the above three observations, it can be concluded that h is continuous at all points. 

It concludes that g and h are continuous functions. Therefore, f = g – h is also a continuous function.

Therefore, f has no point of discontinuity.

Differentiate the functions with respect to x in Exercises 1 to 8

1. sin(x² + 5)

Ans: Let f(x) = sin(x² + 5),  u(x) = x² + 5 and v(t) = sin t 

Then, (vou)(x) = v(u(x)) = v(x² + 5) = tan(x² + 5) = f(x) 

Thus, f is a composite of two functions. 

Put t = u(x) = x² + 5 

Then, we get

2. cos (sin x)

Ans: Let f(x) = cos(sin x) , u(x) = sin x and v(t) = cos t 

Then, (vou)(x) = v(u(x)) = v(sin x) = cos(sin x) = f(x) 

Here, f is a composite function of two functions. 

Put t = u(x) = sin x

3. sin (ax + b)

Ans: Let f(x) = sin(ax + b) , u(x) = ax + b and v(t) = sin t 

Then, (vou)(x) = v(u(x)) = v(ax + b) = sin(ax + b) = f(x) 

Here, f is a composite function of two functions u and v. 

Put, t = u(x) = ax + b 

Thus,

Alternate method:

4. sec(tan(√x))

Ans: Let f(x) = sec(tan(√x)), u(x)=√x,v(t)= tan t and w(s) = sec s 

Then, (wovou) (x) = w[v(u(x))] = w[v(√x)] = w(tan √x) = sec(tan √x) = f(x) 

Here, f is a composite function of three functions u, v and w. 

Put, s = v(t) = tan t and t = u(x) = √x 

Then,

= sec s tan s 

= sec(tan t) .tan(tan t)      [s = tan t]

= sec (tan √x) .tan(tan √x )     [t = √x ]

Now,

Hence, by chain rule, we get

Ans: 

Consider g(x) = sin(ax + b)

Let u(x) = ax + b, v(t) = sin t

Then (vou)(x) = v(u(x)) = v(ax + b) = sin(ax + b) = g(x)

∴ g is a composite function of two functions, u and v. 

Put, t = u(x) = ax + b

Consider h(x) = cos(cx + d) 

Let p(x) = cx + d, q(y) = cos y 

Then, (qop)(x) = q(p(x)) = q(cx + d) = cos(cx + d) = h(x) 

∴ h is a composite function of two functions, p and q. 

Put, y = p(x) = cx + d

6. cos x³ . sin² (x⁵)

Ans: Given,

cos x³ . sin² (x⁵).

= -3x² sin x³ . sin² (x⁵) + 2sin x⁵ cos x⁵ cos x³ 5x⁴

= 10x⁴ sin x⁵ cos x⁵ cos x³ – 3x² sin x³ sin²(x⁵)

7. 2√cot(x²)

Ans: 

8. cos (√x)

Ans: Let f(x) = cos(√x) 

Also, let u(x) = √x and, v(t) = cos t

Then, 

(vou)(x) = v(u(x)) 

= v(√x) 

= cos √x 

= f(x) 

Since, f is a composite function of u and v. 

t = u (x) = √x

Then,

9. Prove that the function f given by f(x) = |x – 1| , x ∈ R is not differentiable at x = 1.

Ans: Given, f(x) = |x – 1| , x ∈ R

It is known that a function f is differentiable at a point x = c in its domain if both

To check the differentiability of the given function at x = 1,

Consider LHD at x = 1

Since LHD and RHD at x = 1 are not equal, 

Therefore, f is not differentiable at x = 1

10. Prove that the greatest integer function defined by f(x) = [x],0 < x < 3  is not differentiable at x = 1 and x = 2. 

Ans: Given, f(x) = [x], 0 < x < 3 

It is known that a function f is differentiable at a point x = c in its domain if both

At x = 1,

Consider the LHD at x = 1

Since LHD and RHD at x = 1 are not equal, 

Hence, f is not differentiable at x = 1.

To check the differentiability of the given function at x = 2, 

Consider LHD at x = 2

Now, consider RHD at x = 2

Since, LHD and RHD at x = 2 are not equal.

Hence, f is not differentiable at x = 2

Find dy/dx  in the following:

1. 2x + 3y = sin x.

Ans: Given, 2x + 3y = sin x 

Differentiating with respect to x, we get

2. 2x + 3y = sin y.

Ans: Given, 2x + 3y = sin y 

Differentiating with respect to x, we get

3. ax + by² = cos y.

Ans: Given, ax + by² = cos y.

Differentiating with respect to x, we get

4. xy + y² = tan x + y.

Ans: Given, xy + y² = tan x + y.

Differentiating with respect to x, we get

5. x² + xy + y² = 100

Ans: Given, x² + xy + y² = 100

Differentiating with respect to x, we get

6. x³ + x²y + xy² + y³ = 81

Ans: Given, x³ + x²y + xy² + y³ = 81

Differentiating with respect to x, we get

7. sin² y + cos xy = k.

Ans: 

8. sin² x + cos² y = 1

Ans: Given, sin² x + cos² y = 1

Differentiating with respect to x, we get

Ans: Given,

Ans: 

Ans: Given,

Ans: 

Ans: 

Ans: Given, 

y = sin⁻¹ (2x √1 – x²) 

y = sin⁻¹ (2x √1 – x²) 

⇒ sin y = (2x √1 – x²)

Ans: Given,

Differentiating with respect to x, we get

Differentiate the following w.r.t. x:

Ans: 

Ans: 

3. eˣ³.

Ans: Let y = eˣ³ 

By using the quotient rule, we get

4. sin(tan⁻¹ e⁻ˣ).

Ans: Let y = sin(tan⁻¹ e⁻ˣ)

By using the chain rule, we get

5. log (cos eˣ).

Ans: Let y = log (cos eˣ). 

By using the chain rule, we get

6. eˣ + eˣ² +….+ eˣ⁵.

Ans: 

Ans: 

8. log (log x), x> 1

Ans: 

Ans: Let

By using the quotient rule, we get

10. cos (log x + eˣ), x > 0.

Ans: 

Differentiate the functions given in Exercises 1 to 11 w.r.t. x.

1. cos x.cos 2x. cos 3x.

Ans: Let y = cos x. cos 2x.cos 3x

Taking logarithm on both the sides, we obtain 

log y = log(cos x.cos 2x.cos 3x) 

⇒ logy = log(cos x) + log (cos2x) + log(cos3x)

Differentiating both sides with respect to x, we obtain

Ans: 

Differentiating both sides with respect to x, we obtain

3. (log x)ᶜᵒˢ ˣ.

Ans: Let y = (log x)ᶜᵒˢ ˣ

Taking logarithm on both the sides, we obtain

log y = cos x. log (log x)

Differentiating both sides with respect to x, we obtain

4. xˣ – 2 ˢⁱⁿ ˣ.

Ans: Let y = xˣ – 2 ˢⁱⁿ ˣ 

Also, let xˣ = u and 2 ˢⁱⁿ ˣ = v 

∴ y = u – v

u = xˣ

Taking logarithm on both the sides, we obtain log(u) = x log x

Differentiating both sides with respect to x, we obtain

5. (x + 3)².(x+4)³.(x+5)⁴.

Ans: Let y = (x + 3)².(x+4)³.(x+5)⁴.

Taking logarithm on both the sides, we obtain 

log y = log(x + 3)² + log(x + 4)³ + log(x + 5)⁴

⇒ log y = 2 log(x + 3) + 3log(x + 4) + 4log(x + 5)

Differentiating both sides with respect to x, we obtain

Ans: 

Differentiating both sides with respect to x, we obtain

7. (log x)ˣ + xˡᵒᵍ ˣ.

Ans: Let y = (log x)ˣ + xˡᵒᵍ ˣ 

Also, let u = (log x)ˣ and v = xˡᵒᵍ ˣ

∴ y = u + v

Then, u = (log x)ˣ

Taking logarithm on both the sides, we obtain 

⇒ log u = log[(log x)ˣ] 

⇒ log u = x log(log(x)

Differentiating both sides with respect to x, we obtain

8. (sin x)ˣ + sin⁻¹ √x.

Ans: Let y = (sin x)ˣ + sin⁻¹ √x.

Also, let u = (sin x)ˣ and v = sin⁻¹ √x

∴ y = u + v

Then, u = (sin x)ˣ

Taking logarithm on both the sides, we obtain 

⇒ log(u) = log(sin x)ˣ

⇒ log(u) = x log(sin x)

Differentiating both sides with respect to x, we obtain

9. xˢⁱⁿ ˣ + (sin x)ᶜᵒˢ ˣ.

Ans: Let y = x ˢⁱⁿ ˣ + (sin x)ᶜᵒˢ ˣ 

Also, let u = xˢⁱⁿ ˣ and v = (sin x)ᶜᵒˢ ˣ

∴ y = u + v

Then, u = xˢⁱⁿ ˣ          

Taking logarithm on both the sides, we obtain 

⇒ log u = log(xˢⁱⁿ ˣ) 

⇒ log u = sin x log x 

Differentiating both sides with respect to x, we obtain

Taking logarithm on both the sides, we obtain 

⇒ log v = log(sin x)ᶜᵒˢ ˣ

⇒ log v = cos x log(sin x)

Differentiating both sides with respect to x, we obtain

Ans: 

Then, u = xˣ ᶜᵒˢ ˣ

Taking logarithm on both the sides, we obtain 

⇒ log(u) = log(xˣ ᶜᵒˢ ˣ) 

⇒ log(u) = x cos x log x

Differentiating both sides with respect to x, we obtain

Taking logarithm on both the sides, we obtain 

⇒ log(v) = log(x² + 1) – log(x² – 1)

Differentiating both sides with respect to x, we obtain

Ans: 

∴ y = u + v

Then, u = (x cos x)ˣ

Taking logarithm on both the sides, we obtain

⇒ log u = (x cos x)ˣ

⇒ log u = x log(x cos x)

⇒ log u = x[log x + log cos x]

⇒ log u = x log x + x log cos x

Differentiating both sides with respect to x, we obtain

12. xʸ + yˣ = 1

Ans: The given function is xʸ + yˣ = 1

Let, xʸ = u and yˣ = v 

∴ u + v = 1

Then, u = xʸ 

Taking logarithm on both the sides, we obtain 

⇒ log u = log(xʸ) 

⇒ log(u) = y log x

Differentiating both sides with respect to x, we obtain

Therefore, from (1), (2) and (3);

13. yˣ = xʸ.

Ans: The given function is yˣ = xʸ

Taking logarithm on both the sides, we obtain 

x log y = y log x

Differentiating both sides with respect to x, we obtain

14. (cos x)ʸ = (cos y)ˣ.

Ans: The given function is (cos x)ʸ = (cos y)ˣ

Taking logarithm on both the sides, we obtain

y log cos x = x log cos y

Differentiating both sides with respect to x, we obtain

Ans: 

Taking logarithm on both the sides, we obtain

⇒ log x + log y = (x – y) log e

⇒ log x + log y = (x – y) × 1

⇒ log x + log y = (x – y)

Differentiating both sides with respect to x, we obtain

16. Find the derivative of the function given by f(x) = (1 + x)(1 + x²)(1 + x⁴)(1 + x⁸) and hence find f’ (1) .

Ans: The given function is f(x) = (1 + x)(1 + x²)(1 + x⁴)(1 + x⁸)

Taking logarithm on both the sides, we obtain 

log(f(x)) = log(1 + x) + log(1 + x²) + log(1 + x⁴) + log(1 + x⁸)

Differentiating both sides with respect to x, we obtain

17. Differentiate (x² – 5x + 8)(x³ + 7x + 9) in three ways mentioned below:

Do they all give the same answer?

(i) by using product rule.

Ans: By using product rule

(ii) by expanding the product to obtain a single polynomial.

Ans: By expanding the product to obtain a single polynomial.

(iii) by logarithmic differentiation.

Ans: By logarithmic differentiation. 

y = (x² – 5x + 8)(x³ + 7x + 9)

Taking logarithm on both the sides, we obtain 

log(y) = log(x² – 5x + 8) + log(x³ + 7x + 9)

Differentiating both sides with respect to x, we obtain

From the above three observations, it can be concluded that all the results of dy/dx are same.

18. If u, v and w are functions of x, then show that

in two ways – first by repeated application of product rule, second by logarithmic differentiation.

Ans: Let y = u.v.w = u.(v.w) 

By applying product rule, we get

Taking logarithm on both the sides of the equation y = u.v.w, we obtain log y = logu + log v + log w

Differentiating both sides with respect to x, we obtain

If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx

1. x = 2at², y = at⁴.

Ans: 

2. x = a cos θ, y = b cos θ.

Ans: Given, x = a cos θ, y = b cos θ.

Then,

3. x = sin t, y = cos 2t.

Ans: Given, x = sin t, y = cos 2t

Ans: 

5. x = cos θ – cos 2 θ, y = sin θ – sin 2 θ.

Ans: Given, x = cos θ – cos 2 θ, y = sin θ – sin 2 θ.

Then,

6. x = a(θ – sin θ) y = a(1 + cos θ).

Ans: Given, x = a(θ – sin θ), y = a(1 + cos θ).

Ans: 

Ans: Given,

Then,

Therefore,

9. x = a sec θ, y = b tan θ.

Ans: Given, x = a sec θ, y = b tan θ

Then, 

10. x = a(cos θ + θ  sin θ), y = a(sin θ – θ cos θ).

Ans: Given, 

x = a(cos θ + θ  sin θ), y = a(sin θ – θ cos θ).

Then,

Therefore, 

Ans: 

Find the second order derivatives of the functions given in Exercises 1 to 10.

1. x² + 3x + 2

Ans: 

2. x²⁰.

Ans: 

3. x. cos x.

Ans: Consider, y = x cos x

Then,

4. log x.

Ans: Let y = log x

Then,

5. x³ log x.

Ans: Let y = x³ log x

Then,

= 2x + 6 log x + 3x 

= 5x + 6x log x 

= x(5 + 6 log x)

6. eˣ sin 5x.

Ans: Let y = eˣ sin 5x

Then,

= eˣ (sin 5x + 5 cos 5x) + eˣ (5cos 5x – 25sin 5x) 

= eˣ (10cos 5x – 24 sin 5x)

= 2eˣ (5 cos 5x – 12 sin 5x)

7. e⁶ˣ cos 3x.

Ans: Let y = e⁶ˣ cos 3x

Then, 

= 6e⁶ˣ cos 3x – 3e⁶ˣ sin 3x     …(1)

Therefore,

= 36e⁶ˣ cos 3x – 18e⁶ˣ sin 3x -3[sin 3x.e⁶ˣ .6+e⁶ˣ .cos 3x.3] 

= 36e⁶ˣ cos 3x – 18e⁶ˣ sin 3x – 18e⁶ˣ sin 3x – 9e⁶ˣ cos 3x 

= 27e⁶ˣ cos 3x – 36e⁶ˣ sin 3x 

= 9e⁶ˣ (3cos 3x – 4sin 3x)

8. tan⁻¹ x.

Ans: 

9. Log(log x).

Ans: Consider, y = log(log x) 

Then,

10. sin (log x).

Ans: Let y = sin (log x)

Then, 

Ans: Given, y = 5cos x – 3sin x

Then, 

Ans: Given, y = cos⁻¹ x

Then,

⇒ y = cos⁻¹ x 

⇒ x = cos y

Putting x = cos y in equation (1), we get

13. If y = 3 cos(log x) + 4sin(log x), show that x² y₂ + xy₁ + y = 0

Ans: Given, y = 3 cos(log x) + 4sin(log x)

Then,

Therefore,

Hence proved.

Ans: Given, y = Aeᵐˣ⁺ + Beⁿˣ

Then,

Ans:

Ans: Given, eʸ (x+1) = 1

⇒ eʸ (x+1) = 1

Hence proved.

17. If y = (tan⁻¹ x)², show that (x² + 1)² y₂ + 2x(x² + 1) y₁ = 2

Ans: Given, y = (tan⁻¹ x)²

Then,

Hence proved.

Differentiate w.r.t. x the function in Exercises 1 to 11.

1. (3x² – 9x + 5)⁹

Ans: Let y = (3x² – 9x + 5)⁹

using chain rule, we get

= 9 (3x² – 9x + 5)⁸ .(6x-9) 

= 9 (3x² – 9x + 5)⁸ .3(2x – 3) 

= 27 (3x² – 9x + 5)⁸ (2x – 3)

2. sin³ x + cos⁶ x.

Ans: Let y = sin³ x + cos⁶ x

using chain rule, we get

= 3sin² x .cos x+6 cos⁵ x.(-sin x) 

= 3sin x cos x (sin x – 2cos⁴ x)

3. (5x)³ ᶜᵒˢ²ˣ.

Ans: Let y = (5x) ³ ᶜᵒˢ²ˣ 

Taking logarithm on both the sides, we obtain 

log y = 3 cos 2x log 5x 

Differentiating both sides with respect to x, we get

4. sin⁻¹ (x √x), 0 ≤ x ≤ 1

Ans: Let y = sin⁻¹ (x √x)

using chain rule, we get

Ans: Let

using quotient rule, we get

Ans: Let

Then,

7. (log x)ˡᵒᵍ ˣ, x > 1

Ans: Let y = (log x)ˡᵒᵍ ˣ

Taking logarithm on both the sides, we obtain

log y = log x. log(log x)

Differentiating both sides with respect to x, we obtain

8. cos(a cos x + b sin x), for some constant a and b.

Ans: Let y = cos(a cos x + b sin x)

Using chain rule, we get

= – sin(a cos x + b sin x) .[a(-sin x) + b cos x] 

= (a sin x – b cos x) .sin(a cos x+b sin x)

Ans: Ley y = (sin x – cos x) ˢⁱⁿ ˣ ⁻ ᶜᵒˢ ˣ

Taking log on both the sides, we obtain

10. xˣ + xᵃ + aˣ + aᵃ, for some fixed a > 0 and x > 0.

Ans: Let y = xˣ + xᵃ + aˣ + aᵃ

Also, let xˣ = u , xᵃ = v, aˣ = w and aᵃ = s 

Therefore, 

⇒ y = u + v + w + s

Now, u = xˣ

Taking logarithm on both the sides, we obtain 

⇒ log u = log xˣ

⇒ log u = x log x

Differentiating both sides with respect to x, we obtain

Now, w = aˣ 

Taking logarithm on both the sides, we obtain 

⇒ log w = log aˣ

⇒ log w = x log a

Differentiating both sides with respect to x, we obtain

11. x ˣ² ⁻ ³ + (x – 3)ˣ², for x > 3.

Ans: Let y = x ˣ² ⁻ ³ + (x – 3)ˣ²

Also, let u = x ˣ² ⁻ ³ and v = (x – 3)ˣ²

Therefore, 

y = u + v

Now, u = x ˣ² ⁻ ³

Taking logarithm on both the sides, we obtain 

log u = log (x ˣ² ⁻ ³) 

= (x² – 3) log x

Differentiating both sides with respect to x, we obtain

Now, v = (x – 3)ˣ² 

Taking logarithm on both the sides, we obtain 

log v = log(x – 3)ˣ² 

= x² log(x – 3)

Differentiating both sides with respect to x, we obtain

From (1), (2), and (3), we obtain

Ans: The given function is y = 12(1 – cos t) , x = 10(t – sin t) 

Hence,

Ans: The given function is y = sin⁻¹ x + sin⁻¹ √1 – x²

Hence, 

14. If x √1 + y + y √1 + x = 0 , for, – 1 < x < 1 prove that

Ans: The given function is x √1 + y + y √1 + x = 0 

⇒ x √1 + y = – y √1 + x

Squaring both sides, we obtain 

x² (1 + y) = y² (1 + x) 

⇒ x² + x² y = y² + xy²

⇒ x² – y² = xy² – x² y 

⇒ x² – y² = xy(y – x) 

⇒ (x + y)(x – y) = xy(y – x) 

⇒ x + y = – xy 

⇒ (1 + x) y = – x

Differentiating both sides with respect to x, we obtain

Hence proved.

15. If (x – a)² + (y – b)² = c² for some c > 0, prove that is a constant independent of a and b.

Ans: The given function is (x – a)² + (y – b)² = c²

Differentiating both sides with respect to x, we obtain

Therefore, 

Hence, 

= – c

-c is a constant and is independent of a and b. 

Hence proved.

Ans: The given function is cos y = x cos(a + y)

Therefore,

Hence proved.

Ans: The given function is x = a(cos t + t sin t) and y = a(sin t – t cos t) 

Therefore,

18. If f(x) = |x|³, show that f” (x) exists for all real x, and find it.

Ans: It is known that

Therefore, when x ≥ 0, f(x) = |x|³ = x³

In this case, f’ (x) = 3x² and hence, f” (x) = 6x

When x < 0, f(x) = |x|³ = (- x)³ = – x³ 

In this case, f’ (x) = – 3x² and hence, f” (x) = – 6x 

Thus, for f(x) = |x|³, f” (x) exists for all real x and is given by,

19. Using the fact that sin(A + B) = sin A  cos B + cos A  sin B and the differentiation, obtain the sum formula for cosines.

Ans: Given, sin(A + B) = sin A cos B + cos A sin B 

Differentiating both sides with respect to x, we obtain

20. Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.

Ans: Consider,

It can be seen from the above graph that the given function is continuous everywhere but not differentiable at exactly two points which are 0 and 1.

Ans: 

Ans: The given function is y = eᵃ ᶜᵒˢ ⁻¹ˣ 

Taking logarithm on both the sides, we obtain 

⇒ log y = a cos⁻¹ x log e

⇒ log y = a cos⁻¹ x

Differentiating both sides with respect to x, we obtain

Hence proved.

Parimol

Hi! my Name is Parimal Roy. I have completed my Bachelor’s degree in Philosophy (B.A.) from Silapathar General College. Currently, I am working as an HR Manager at Dev Library. It is a website that provides study materials for students from Class 3 to 12, including SCERT and NCERT notes. It also offers resources for BA, B.Com, B.Sc, and Computer Science, along with postgraduate notes. Besides study materials, the website has novels, eBooks, health and finance articles, biographies, quotes, and more.