NCERT Class 12 Mathematics Chapter 5 Continuity and Differentiability

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NCERT Class 12 Mathematics Chapter 5 Continuity and Differentiability

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Continuity and Differentiability

Chapter – 5

Exercise 5.1

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.

EXERCISE 5.2

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

EXERCISE 5.3

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

EXERCISE 5.4

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: 

EXERCISE 5.5

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

EXERCISE 5.6

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: 

EXERCISE 5.7

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.

Miscellaneous Exercise on Chapter 5

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.