Question

NCERT Solutions Class 12 Maths Chapter 5 Continuity and Differentiability is suitable for studying for CBSE board exams and it is made by the experts in the subject matter. NCERT Solutions is one of the best study materials for students of class 12.

NCERT Solutions Class 12 is designed according to the latest syllabus proposed by the experts to provide complete detail of NCERT intext questions and also back of the chapter questions.

• Continuity – the concept of rigorous formulation and the intuitive concept of any function which have no breaks or jumps is termed continuity. Continuity can also be defined as the \(x\)-values when it is closer to the -values of the function. The concept of continuity can also be defined according to the saying of the function \(f(x)\) is continuous at any given point.
• Differentiability – a derivative that exists at each point in the domain of a function that has derivatives. A differential function when plotted in a graph has a non-vertical tangent line at the interior of the domain.
• Algebra of continuous functions – there are four basic arithmetic operations for carrying out operations on continuous algebraic functions. These four arithmetic functions are
  1. Addition operation of the continuous functions
  2. Subtraction of continuous functions
  3. Multiplication of continuous functions
  4. Division of continuous functions
• Derivatives of composite functions – we can define the derivative of composite functions as the product of the derivative of the inside function concerning the outside function. The composite functions of the derivatives are formed with the help of the chain rule formula.
• Derivatives of implicit functions – a dependent and the independent variable which is written in the form of \(y-3x^2+2\)
• Derivatives of inverse trigonometric functions \(-x+5 =0\) then the function is called an implicit function.

functions	\(( {dy \over dx})\)
arcsin x	\({1 \over \sqrt{1-x^2}}\)
arccos x	\(- {1 \over \sqrt{1-x^2}}\)
arctan x	\({1\over {1+x^2}}\)
account x	\(-{1\over {1+x^2}}\)
arcsec x	\({1 \over |x|\sqrt{x^2-1}}\)
arccsc x	\(-{1 \over |x|\sqrt{x^2-1}}\)

• Logarithmic Differentiation – the logarithmic differentiation is the method to calculate differentiated functions with the help of employing the logarithmic derivative of a given function \(f\).

NCERT Solutions Class 12 Maths is prepared by the mentors who have years of experience in the subject matter.

Answer 1

NCERT Solutions Class 12 Maths Chapter 5 Continuity and Differentiability

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

Answer:

Hence, the function f is continuous at x = 0.

Hence, the function f is continuous at x = -3.

Hence, the function f is continuous at x = -3.

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

Answer:

Hence, the function f is continuous at x = 3.

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

Answer:

Hence, the function f is continuous at x = n, where n is positive integer.

4. Is the function f defined by \(f(x) = \begin{cases}x,&x\le 1\\5,&x>1\end{cases}\) continuous at x = 0? At x = 1? At x = 2?

Answer:

Hence, the function f is continuous at x = 0.

Hence, the function f is continuous at x = 1.

Hence, the function f is continuous at x = 2.

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

\(f(x) = \begin{cases}2x + 3&\text{if} \;x\le2\\2x - 3& \text{if}\; x > 2\end{cases}\)

Answer:

\(f(x) = \begin{cases}2x + 3&\text{if} \;x\le2\\2x - 3& \text{if}\; x > 2\end{cases}\)

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

Let c be a point on the real line. Then, three cases arise. 

(i) c < 2 

(ii) c > 2 

(iii) c = 2 

Case (i) c < 2

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

Case (ii) c > 2

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

Case (iii) c = 2

Then, the left hand limit of f at x = 2 is,

The right hand limit of f at x = 2 is,

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

Therefore, f is not continuous at x = 2 

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

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

Answer:

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 < −3 

Case II:

Therefore, f is continuous at x = −3 

Case III:

Therefore, f is continuous in (−3, 3). 

Case IV: 

If c = 3, then the left hand limit of f at x = 3 is,

The right hand limit of f at x = 3 is,

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

Therefore, f is not continuous at x = 3 

Case V:

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

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

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

Answer:

It is known that, x < 0 

⇒ |x| = -x and x > 0

⇒ |x| = x

Therefore, the given function can be rewritten as

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 < 0 

Case II:

If c = 0, then the left hand limit of f at x = 0 is,

The right hand limit of f at x = 0 is,

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 

Case III:

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

Hence, x = 0 is the only point of discontinuity of f.

Answer 2

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

Answer:

It is known that, x < 0 ⇒ |x| = -x

Therefore, the given function can be rewritten as

Let c be any real number. Then,

Therefore, the given function is a continuous function. 

Hence, the given function has no point of discontinuity.

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

Answer:

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 < 1 

Case II:

Therefore, f is continuous at x = 1 

Case III:

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

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

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

Answer:

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 < 2 

Case II:

Therefore, f is continuous at x = 2 

Case III:

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.

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

Answer:

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 < 1 

Case II: 

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

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

Therefore, f is not continuous at x = 1

Case III:

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.

12. Is the function defined by

a continuous function?

Answer:

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 < 1 

Case II:

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

Therefore, f is not continuous at x = 1 

Case III:

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

Answer:

The given function is

The given function is defined at all 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:

Therefore, f is continuous at x = 0

Case III:

Therefore, f is continuous at all points of the interval (0, 1). 

Case IV:

It is observed that the left and right hand limits of f at x = 1 do not coincide. 

Therefore, f is not continuous at x = 1 

Case V:

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

Hence, f is not continuous only at x = 1

Answer 3

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

Answer:

The given function is defined at all 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 < −1 

Case II:

Therefore, f is continuous at x = −1 

Case III:

Therefore, f is continuous at all points of the interval (−1, 1). 

Case IV:

Therefore, f is continuous at x = 2 

Case V:

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.

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

is continuous at x = 3.

Answer:

If f is continuous at x = 3, then

Therefore, from (1), we obtain

Therefore, the required relationship is given by, \(a = b + \frac23\)

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

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

Answer:

The given function is 

If f is continuous at x = 0, then

Therefore, there is no value of λ for which f is continuous at x = 0

At x = 1,

f(1) = 4x + 1 = 4 × 1 + 1 = 5

Therefore, for any values of λ, f is continuous at x = 1

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

Answer:

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,

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

Therefore, f is not continuous at x = n 

Hence, g is discontinuous at all integral points.

18. Is the function defined by f(x) = x² - sin x + 5 continuous at x = p?

Answer:

The given function is f(x) = x² - sin x + 5

It is evident that f is defined at x = p

Therefore, the given function f is continuous at x = π

19. 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

Answer:

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

g + h, g - h and g.h are also continuous. 

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

Let g (x) = sin x 

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 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 is a continuous function. 

Therefore, it can be concluded that 

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

(b) f (x) = g (x) − h (x) = sin x − cos x is a continuous function 

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

20. Find the points of discontinuity of f, where

Answer:

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

Let c be a real number. 

Case I:

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

Case II:

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

Case III:

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.

Answer 4

21. Determine if f defined by

is a continuous function?

Answer:

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

Let c be a real number. 

Case I:

Therefore, f is continuous at all points 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.

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

Answer:

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

Let c be a real number. 

Case I:

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

Case II:

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.

23. Find the values of k so that the function f is continuous at the indicated point.

Answer:

The given function is

The given function f is continuous at \(x = \frac\pi2\), if f is defined at \(x = \frac\pi2\) and if the value of the f at \(x = \frac\pi2\) equals the limit of f at \(x = \frac\pi2\).

It is evident that f is defined at \(x = \frac\pi2\) and \(f(\frac\pi2)=3\)

Therefore, the required value of k is 6.

24. Find the values of k so that the function f is continuous at the indicated point.

at x = 2.

Answer:

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 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

Therefore, the required value of k is \(\frac34\).

25. Find the values of k so that the function f is continuous at the indicated point.

at x = π.

Answer:

The given function is 

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

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

Therefore, the required value of k is \(- \frac2\pi\).

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

at x = 5.

Answer:

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 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

Therefore, the required value of k is \(\frac95\).

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

is a continuous function.

Answer:

It is evident that the given function 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

Since f is continuous at x = 10, 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.

Answer 5

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

Answer:

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 = g o h, where g (x) = cos x and h (x) = x

It has to be first proved 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. 

Then, g (c) = cos c

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

h (x) = x² Clearly, 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 (g o h) is defined at c, if g is continuous at c and if f is continuous at g (c), then (f o g) is continuous at c. 

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

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

Answer:

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, 

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

Clearly, g is defined for all real numbers. 

Let c be a real number. 

Case I:

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

Case II:

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

Case III:

Therefore, g is continuous at x = 0 

From the above three observations, it can be concluded that g is continuous at all points. 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 (g o h) is defined at c, if g is continuous at c and if f is continuous at g (c), then (f o g) is continuous at c.

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

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

Answer:

Let f(x) = sin|x|

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

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

Clearly, g is defined for all real numbers. 

Let c be a real number. 

Case I:

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

Case II:

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

Case III:

Therefore, g is continuous at x = 0 

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

h (x) = sin x 

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

Let c be a real number. 

Put x = c + k 

If x → c, then k → 0 

h (c) = sin c

Therefore, h is a continuous function. 

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

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

31. Differentiate the functions with respect to x.

sin(x² + 5)

Answer:

Thus, f is a composite of two functions.

Alternate method

32. Differentiate the functions with respect to x.

cos(sin x)

Answer:

Thus, f is a composite function of two functions. 

Put t = u (x) = sin x

Alternate method

33. Differentiate the functions with respect to x.

sin(ax + b)

Answer:

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

Put t = u (x) = ax + b

Therefore,

Alternate method

Answer 6

34. Differentiate the functions with respect to x.

sec(tan(√x))

Answer:

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

Hence, by chain rule, we obtain

Alternate method

35. Differentiate the functions with respect to x.

cos x³.sin²(x⁵)

Answer:

The given function is cos x³.sin²(x⁵)

 36. Differentiate the functions with respect to x.

\(2\sqrt{cot(x^2)}\)

Answer:

37. Differentiate the functions with respect to x.

cos(√x)

Answer:

Clearly, f is a composite function of two functions, u and v, such that

t = u(x) = √x

By using chain rule, we obtain

Alternate method

38. Find \(\frac{dy}{dx}\):

2x + 3y = sin x

Answer:

The given relationship is 2x + 3y = sin x

Differentiating this relationship with respect to x, we obtain

39. Find \(\frac{dy}{dx}\):

2x + 3y = sin y

Answer:

The given relationship is 2x + 3y = sin y

Differentiating this relationship with respect to x, we obtain

40. Find \(\frac{dy}{dx}\):

ax + by² = cos y

Answer:

The given relationship is ax + by² = cos y

Differentiating this relationship with respect to x, we obtain

Using chain rule, we obtain

From (1) and (2), we obtain

41. Find \(\frac{dy}{dx}\):

xy + y² = tan x + y

Answer:

The given relationship is xy + y² = tan x + y

Differentiating this relationship with respect to x, we obtain

42. Find \(\frac{dy}{dx}\):

x² + xy + y² = 100

Answer:

The given relationship is x² + xy + y² = 100

Differentiating this relationship with respect to x, we obtain

43. Find \(\frac{dy}{dx}\):

x³ + x²y + xy² + y³ = 81

Answer:

The given relationship is x³ + x²y + xy² + y³ = 81

Differentiating this relationship with respect to x, we obtain

44. Find \(\frac{dy}{dx}\):

sin²y + cos xy = π

Answer:

The given relationship is sin²y + cos xy = π

Differentiating this relationship with respect to x, we obtain

Using chain rule, we obtain

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

45. Find \(\frac{dy}{dx}\):

sin²x + cos²y = 1

Answer:

The given relationship is sin²x + cos²y = 1

Differentiating this relationship with respect to x, we obtain

46. Find \(\frac{dy}{dx}\):

\(y = tan^{-1} \left(\frac{3x - x^3}{1 - 3x^2}\right), -\frac1{\sqrt3} < x< \frac1{\sqrt3}\)

Answer:

The given relationship is \(y = tan^{-1} \left(\frac{3x - x^3}{1 - 3x^2}\right)\)

Comparing equations (1) and (2), we obtain

\(x = tan \frac y3\)

Differentiating this relationship with respect to x, we obtain

Answer 7

47. Find \(\frac{dy}{dx}\):

\(y = cos^{-1} \left(\frac{1 - x^2}{1 + x^2}\right) ,0< x< 1\)

Answer:

The given relationship is,

On comparing L.H.S. and R.H.S. of the above relationship, we obtain 

\(tan \frac y2=x\)

Differentiating this relationship with respect to x, we obtain

48. Find \(\frac{dy}{dx}\):

\(y = sin^{-1} \left(\frac{1 - x^2}{1 + x^2}\right) ,0< x< 1\)

Answer:

The given relationship is \(y = sin^{-1} \left(\frac{1 - x^2}{1 + x^2}\right) \) 

Differentiating this relationship with respect to x, we obtain

Using chain rule, we obtain

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

49. Find \(\frac{dy}{dx}\):

\(y = cos^{-1} \left(\frac{2x}{1 + x^2}\right) ,-1< x< 1\) 

Answer:

The given relationship is \(y = cos^{-1} \left(\frac{2x}{1 + x^2}\right)\) 

Differentiating this relationship with respect to x, we obtain

Answer 8

50. Find \(\frac{dy}{dx}\):

\(y = sin^{-1}(2x \sqrt{1 -x^2}), -\frac1{\sqrt 2} < x < \frac1{\sqrt 2}\)

Answer:

The given relationship is \(y = sin^{-1}(2x \sqrt{1 -x^2})\) 

Differentiating this relationship with respect to x, we obtain

51. Find \(\frac{dy}{dx}\):

\(y = sec^{-1}\left(\frac{1}{2x^2 - 1}\right),0< x < \frac1{\sqrt2}\)

Answer:

The given relationship is \(y = sec^{-1}\left(\frac{1}{2x^2 - 1}\right)\) 

\(y = sec^{-1}\left(\frac{1}{2x^2 - 1}\right)\) 

Differentiating this relationship with respect to x, we obtain

52. Differentiate the following w.r.t. x:

\(\frac{e^x}{sin x}\)

Answer:

Let \(y = \frac{e^x}{sin x}\) 

By using the quotient rule, we obtain

53. Differentiate the following w.r.t. x:

\(e^{sin^{-1}x}\)

Answer:

Let y = \(e^{sin^{-1}x}\) 

By using the chain rule, we obtain

54. Show that the function given by f(x) = e²x is strictly increasing on R.

Answer:

Let x₁ and x₂ be any two numbers in R. 

Then, we have:

Hence, f is strictly increasing on R.

55. Differentiate the following w.r.t. x:

\(e^{x^3}\)

Answer:

Let y = \(e^{x^3}\)

By using the chain rule, we obtain

56. Differentiate the following w.r.t. x:

sin(tan^-1 e^-x)

Answer:

Let y = sin(tan^-1 e^-x)

By using the chain rule, we obtain

57. Differentiate the following w.r.t. x:

log(cos e^x)

Answer:

Let y = log(cos e^x)

By using the chain rule, we obtain

58. Differentiate the following w.r.t. x:

\(e^x + e^{x^2} + .... +e^{x^5}\)

Answer:

59. Differentiate the following w.r.t. x:

\(\sqrt{e^{\sqrt x}}, x > 0\)

Answer:

By differentiating this relationship with respect to x, we obtain

60. Differentiate the following w.r.t. x:

log(log x), x > 1

Answer:

Let y = log(log x)

By using the chain rule, we obtain

Answer 9

61. Differentiate the following w.r.t. x:

\(\frac{cos\,x}{log\, x}, x> 0\)

Answer:

Let \(y = \frac{cos\,x}{log\, x}\) 

By using the quotient rule, we obtain

62. Differentiate the following w.r.t. x:

cos(log x + e^x), x > 0

Answer:

Let y = cos(log x + e^x), x > 0

By using the chain rule, we obtain

63. Differentiate the function with respect to x.

cos x. cos 2x. cos 3x

Answer:

Let y = cos x. cos 2x. cos 3x

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

64. Differentiate the function with respect to x.

\(\sqrt{\frac{(x -1)(x -2)}{(x - 3)(x - 4)(x - 5)}}\)

Answer:

Let \(y = \sqrt{\frac{(x -1)(x -2)}{(x - 3)(x - 4)(x - 5)}}\) 

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

65. Differentiate the function with respect to x.

(log x)^{cos x}

Answer:

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

66. Differentiate the function with respect to x.

x^x - 2^{sin x}

Answer:

u = x^x

Taking logarithm on both the sides, we obtain

log u = x log x

Differentiating both sides with respect to x, we obtain

v = 2^{sin x}

Taking logarithm on both the sides with respect to x, we obtain

log v = sin x.log 2

Differentiating both sides with respect to x, we obtain

67. Differentiate the function with respect to x.

(x + 3)².(x + 4)³.(x +5)⁴

Answer:

Let y = (x + 3)².(x + 4)³.(x +5)⁴

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

68. Differentiate the function with respect to x.

\(\left(x + \frac1x\right)^x + x^{(1 + \frac1x)}\)

Answer:

Differentiating both sides with respect to x, we obtain

Differentiating both sides with respect to x, we obtain

Therefore, from (1), (2), and (3), we obtain

69. Differentiate the function with respect to x.

(log x)^x + x^{log x}

Answer:

Differentiating both sides with respect to x, we obtain

Differentiating both sides with respect to x, we obtain

Therefore, from (1), (2), and (3), we obtain

70. Differentiate the function with respect to x.

(sin x)^x + sin^-1 √x

Answer:

Differentiating both sides with respect to x, we obtain

Differentiating both sides with respect to x, we obtain

Therefore, from (1), (2), and (3), we obtain

71. Differentiate the function with respect to x.

x^{sin x} + (sin x)^{cos x}

Answer:

Differentiating both sides with respect to x, we obtain

Differentiating both sides with respect to x, we obtain

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

Answer 10

72. Differentiate the function with respect to x.

\(x^{x cos x} + \frac{x^2 + 1}{x^2-1}\)

Answer:

Differentiating both sides with respect to x, we obtain

Differentiating both sides with respect to x, we obtain

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

73. Find \(\frac{dy}{dx}\) of function.

x^y + y^x = 1

Answer:

The given function is x^y + y^x = 1

Let x y = u and y x = v 

Then, the function becomes u + v = 1

Differentiating both sides with respect to x, we obtain

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

74. Find \(\frac{dy}{dx}\) of function.

y^x = x^y

Answer:

The given function is y^x = x^y

Taking logarithm on both the sides, we obtain 

x log y = y log x

Differentiating both sides with respect to x, we obtain

75. Find \(\frac{dy}{dx}\) of function.

(cos x)^y = (cos y)^x

Answer:

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, we obtain

76. Find \(\frac{dy}{dx}\) of function.

xy = e^(x-y) 

Answer:

The given function is xy = e^(x-y) 

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

77. 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.

Answer:

Let y = u.v.w = u.(v.w)

By applying product rule, we obtain

By taking logarithm on both sides of the equation y = u.v.w, we obtain 

log y = log u + log v + log w

Differentiating both sides with respect to x, we obtain

78. If x and y are connected parametrically by the equation, without eliminating the parameter, find \(\frac{dy}{dx}\).

x = 2at², y = at⁴

Answer:

The given equations are x = 2at², y = at⁴

79. If x and y are connected parametrically by the equation, without eliminating the parameter, find \(\frac{dy}{dx}\).

x = a cos θ, y = b cos θ

Answer:

The given equations are x = a cos θ and y = b cos θ

80. If x and y are connected parametrically by the equation, without eliminating the parameter, find \(\frac{dy}{dx}\).

x = sin t, y = cos 2t

Answer:

The given equations are x = sin t and y = cos 2t

81. If x and y are connected parametrically by the equation, without eliminating the parameter, find \(\frac{dy}{dx}\).

x = 4t, y = \(\frac4t\) 

Answer:

The given equations are x = 4t and y = \(\frac4t\) 

82. If x and y are connected parametrically by the equation, without eliminating the parameter, find \(\frac{dy}{dx}\).

x = cosθ - cos2θ, y = sinθ - sin2θ

Answer:

The given equations are x = cosθ - cos2θ, y = sinθ - sin2θ

83. If x and y are connected parametrically by the equation, without eliminating the parameter, find \(\frac{dy}{dx}\).

x = a(θ - sinθ), y = a(1 + cosθ)

Answer:

The given equations are x = a(θ - sinθ), y = a(1 + cosθ)