Class 12 Math Ch-5 Continuity and Differentian MCQs Exam 2027

Chapter 5: Continuity and Differentiability (MCQ Bank)

I. Algebraic & Power Functions

1. $\frac{d}{dx} (\frac{x^4}{4})$ (BSEB 2022 A)

   (A) $4x^3$

   (B) $\frac{x^3}{4}$

   (C) $x^3$

   (D) $16x^3$

2. If $y = x^{20}$, then $\frac{d^2y}{dx^2} =$ (BSEB Previous Year)

   (A) $x^{18}$

   (B) $20x^{19}$

   (C) $380x^{18}$

   (D) $x^{10}$

3. If $y = x^5$, then $\frac{dy}{dx} =$ (BSEB Previous Year)

   (A) $5x$

   (B) $6x$

   (C) $5x^4$

   (D) $5x^2$

4. $\frac{d}{dx} (x^4) =$ (BSEB Previous Year)

   (A) $4x^3$

   (B) $12x^2$

   (C) $24x$

   (D) $24$

5. If $y = x^3$, then $\frac{d^2y}{dx^2} =$ (BSEB 2026)

   (A) $3x^2$

   (B) $6x$

   (C) $6$

   (D) $0$

6. $\frac{d}{dx} [ \frac{1}{3x-2} ] =$ (BSEB Previous Year)

   (A) $\frac{-1}{(3x-2)^2}$

   (B) $\frac{-3}{(3x-2)^2}$

   (C) $\frac{3}{(3x-2)^2}$

   (D) $\frac{3}{3x-2}$

7. $\frac{d}{dx} (\sqrt{x}) =$ (BSEB Previous Year)

   (A) $2\sqrt{x}$

   (B) $\frac{1}{2\sqrt{x}}$

   (C) $\frac{\sqrt{x}}{2}$

   (D) $\frac{1}{\sqrt{x}}$

8. If $y = x^5$, then $\frac{d^2y}{dx^2} =$ (2026)

   (A) $5x^4$

   (B) $20x^3$

   (C) $20x^4$

   (D) $x^5$

9. $\frac{d}{dx}(\sqrt{x}) =$ (2026)

   (A) $\frac{1}{2\sqrt{x}}$

   (B) $\frac{1}{\sqrt{x}}$

   (C) $2\sqrt{x}$

   (D) $\frac{2}{3}x^{3/2}$

10. If $y = x^3$, then $\frac{d^2y}{dx^2} =$ (2026)

    (A) $3x^2$

    (B) $6x$

    (C) 6

    (D) 0

11. $\frac{d^3}{dx^3}(x^4) =$ (2026)

    (A) $4x^3$

    (B) $12x^2$

    (C) $24x$

    (D) 24

12. $\frac{d}{dx}(\frac{1}{x+1}) =$ (2026)

    (A) $log(x+1)$

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

    (C) $\frac{1}{(x+1)^2}$

    (D) $-\frac{1}{x+1}$

II. Exponential & Logarithmic Functions

13. $\frac{d}{dx} [e^{x^2}] =$ (BSEB 2025 A)

    (A) $e^{x^2}$

    (B) $e^{2x}$

    (C) $2xe^{x^2}$

    (D) $2xe^{2x}$

14. $\frac{d}{dx} (5^x) =$ (BSEB Previous Year)

    (A) $5^x$

    (B) $x \cdot 5^{x-1}$

    (C) $\frac{5^x}{\log 5}$

    (D) $5^x \log_e 5$

15. $\frac{d}{dx} [\log x] =$ (BSEB 2019 A)

    (A) $\frac{1}{x}$

    (B) $-\frac{1}{x^2}$

    (C) $1$

    (D) $\frac{1}{x^2}$

16. $\frac{d}{dx} (\log 3^x) =$ (BSEB 2021 A)

    (A) $\frac{1}{3^x}$

    (B) $\log 3$

    (C) $x \log 3$

    (D) $1$

17. $\frac{d}{dx} (a^x) =$ (BSEB 2022, 2024 A)

    (A) $a^x$

    (B) $\frac{a^x}{\log a}$

    (C) $a^x \log a$

    (D) $x \log a$

18. $\frac{d}{dx} (\log \sqrt{x}) =$ (BSEB Previous Year)

    (A) $\frac{1}{2\sqrt{x}}$

    (B) $\frac{1}{\sqrt{x}}$

    (C) $\frac{1}{2x}$

    (D) $\frac{\sqrt{x}}{2}$

19. $\frac{d}{dx} (e^{-3x}) =$ (BSEB Previous Year)

    (A) $\frac{e^{-3x}}{3}$

    (B) $\frac{e^{-3x}}{-3}$

    (C) $3e^{-3x}$

    (D) $-3e^{-3x}$

20. $\frac{d}{dx} (\log 5x) =$ (BSEB 2025 A)

    (A) $5x$

    (B) $\frac{1}{x}$

    (C) $\frac{5}{x}$

    (D) $\log 5 + \frac{1}{x}$

21. If $y = \log x^x$, then $\frac{dy}{dx} =$ (BSEB 2010)

    (A) 1

    (B) $\log x$

    (C) $1 + \log x$

    (D) $\log (ex)$

22. $\frac{d}{dx} (11^x) =$ (BSEB Previous Year)

    (A) $x11^{x-1}$

    (B) $11^x \log x$

    (C) $11^x \log 11$

    (D) $\frac{11^x}{\log 11}$

23. $\frac{d}{dx} [e^{\cos x}] =$ (BSEB 2020)

    (A) $(\sin x)e^{\cos x}$

    (B) $-(\sin x)e^{\cos x}$

    (C) $(\cos x)e^{\cos x}$

    (D) $-(\cos x)e^{\cos x}$

24. $\frac{d}{dx} (\log_{10} x) =$ (BSEB Previous Year)

    (A) $\frac{1}{10x}$

    (B) $\frac{1}{x}$

    (C) $10x$

    (D) $\frac{1}{x} \log_{10} e$

25. If $y = a^x$, then $\frac{d^2y}{dx^2} =$ (BSEB Previous Year)

    (A) $a^x \log a$

    (B) $a^x (\log a)^2$

    (C) $(a^x)^2 \log a$

    (D) None

26. $\frac{d^2}{dx^2} (e^{5x}) =$ (BSEB Previous Year)

    (A) $e^{5x}$

    (B) $10e^{5x}$

    (C) $5e^{5x}$

    (D) $25e^{5x}$

27. $\frac{d}{dx} (e^{x^3}) =$ (BSEB 2021 A)

    (A) $e^{x^3}$

    (B) $3x^2 e^x$

    (C) $3x^2 e^{x^2}$

    (D) $3x^2 e^{x^3}$

28. $\frac{d}{dx} (\log x^n) =$ (BSEB Previous Year)

    (A) $\frac{1}{x^n}$

    (B) $n$

    (C) $\frac{1}{x}$

    (D) $\frac{n}{x}$

29. $\frac{d}{dx} (e^x) =$ (BSEB Previous Year)

    (A) $e^x$

    (B) $e^{-x}$

    (C) $e^{2x}$

    (D) $e^{3x}$

30. $\frac{d}{dx} (2e^{2x}) =$ (BSEB Previous Year)

    (A) $2e^{2x}$

    (B) $e^{2x}$

    (C) $4e^{2x}$

    (D) $2e^x$

31. $\frac{d}{dx} (e^{13x}) =$ (BSEB Previous Year)

    (A) $e^{13x}$

    (B) $\frac{1}{13} e^{13x}$

    (C) $13e^{13x}$

    (D) $-13e^{13x}$

32. $\frac{d}{dx} (\log |x|) =$ (BSEB Previous Year)

    (A) $\frac{1}{|x|}$

    (B) $\frac{1}{x}$

    (C) $\frac{-1}{x}$

    (D) Undefined

33. $\frac{d}{dx} (e^{3x}) =$ (BSEB Previous Year)

    (A) $e^{3x}$

    (B) $\frac{e^{3x}}{3}$

    (C) $3e^{3x}$

    (D) $3e^{2x}$

34. $\frac{d}{dx} [e^{x^3}] =$ (BSEB 2024)

    (A) $3x^2 e^{x^3}$

    (B) $e^{x^3}$

    (C) $3x^2 e^x$

    (D) $x^3 e^{x^3-1}$

35. $\frac{d}{dx} [2^x] =$ (BSEB 2022)

    (A) $x \cdot 2^{x-1}$

    (B) $\frac{2^x}{\log 2}$

    (C) $2^x \log 2$

    (D) $2^x$

36. $\frac{d}{dx} (\log \sqrt{x}) =$ (BSEB 2021)

    (A) $\frac{1}{2\sqrt{x}}$

    (B) $\frac{1}{\sqrt{x}}$

    (C) $\frac{1}{2x}$

    (D) $\frac{\sqrt{x}}{2}$

37. $\frac{d}{dx}(e^{3x}) =$ (2026)

    (A) $e^{3x}$

    (B) $3e^{3x}$

    (C) $\frac{1}{3}e^{3x}$

    (D) $3x e^{3x-1}$

38. $\frac{d}{dx} (2^x) =$ (2026)

    (A) $x \cdot 2^{x-1}$

    (B) $2^x \log 2$

    (C) $\frac{2^x}{\log 2}$

    (D) $2^x$

39. $\frac{d}{dx} (\log_a x) =$ (2026)

    (A) $\frac{1}{x \log_e a}$

    (B) $\frac{1}{x}$

    (C) $\frac{\log_e a}{x}$

    (D) $\frac{x}{\log_e a}$

40. $\frac{d}{dx}(e^{x-a}) =$ (2026)

    (A) $e^{x-a}$

    (B) $e^x$

    (C) $e^{-x}$

    (D) $(x-a)e^{x-a-1}$

41. $\frac{d}{dx}(e^{ax}) =$ (2026)

    (A) $e^{ax}$

    (B) $ae^{ax}$

    (C) $a^2 e^{ax}$

    (D) $a$

42. $\frac{d}{dx}(2^x) =$ (2026)

    (A) $2^x$

    (B) $x \cdot 2^{x-1}$

    (C) $\frac{2^x}{log 2}$

    (D) $2^x log 2$

III. Trigonometric Functions

43. $\frac{d}{dx} [2\sin^2 \theta + 2\cos^2 \theta]$ (BSEB Previous Year)

    (A) $4\sin \theta$

    (B) $4\cos \theta$

    (C) 0

    (D) 4

44. $\frac{d}{dx} [\tan ax]$ (BSEB 2021)

    (A) $a\tan ax$

    (B) $a\sec^2 ax$

    (C) $a\sec x$

    (D) $a\cot ax$

45. $\frac{d}{dx} (\sin^2 x)$ (BSEB 2019 C, 2022 A)

    (A) $\sin 2x$

    (B) $\cos 2x$

    (C) $\tan 2x$

    (D) $\cot 2x$

46. If $y = \tan^2 x$, then $\frac{dy}{dx} =$ (BSEB Previous Year)

    (A) $\sec^2 x$

    (B) $\sec^4 x$

    (C) $2 \tan x \sec x$

    (D) $2 \tan x \sec^2 x$

47. $\frac{d}{dx} (\sqrt{\cot x})$ (BSEB Previous Year)

    (A) $\frac{1}{2\sqrt{\cot x}}$

    (B) $\sqrt{\csc^2 x}$

    (C) $\frac{-\csc^2 x}{2\sqrt{\cot x}}$

    (D) $\frac{\csc^2 x}{2\sqrt{\cot x}}$

48. $\frac{d}{dx} (\tan \frac{x}{3})$ (BSEB 2022 A)

    (A) $\sec^2 \frac{x}{3}$

    (B) $\frac{1}{3} \sec^2 \frac{x}{3}$

    (C) $3 \sec^2 \frac{x}{3}$

    (D) $3 \cot \frac{x}{3}$

49. $\frac{d}{dx} [\csc x]$ (BSEB 2020)

    (A) $\csc x \cot x$

    (B) $-\csc x \cot x$

    (C) $\csc^2 x$

    (D) $-\csc^2 x$

50. $\frac{d}{dx} (\sin \frac{4x}{5})$ (BSEB Previous Year)

    (A) $\frac{4}{5} \cos \frac{4x}{5}$

    (B) $-\frac{4}{5} \cos \frac{4x}{5}$

    (C) $\frac{5}{4} \cos \frac{4x}{5}$

    (D) $-\frac{5}{4} \cos \frac{4x}{5}$

51. $\frac{d}{dx} (\sin 2x)$ (BSEB Previous Year)

    (A) $\cos 2x$

    (B) $\frac{\cos 2x}{2}$

    (C) $2 \sin 2x$

    (D) $2 \cos 2x$

52. $\frac{d}{dx} (\sin \sqrt{x})$ (BSEB 2021 A)

    (A) $\cos \sqrt{x}$

    (B) $\frac{\cos \sqrt{x}}{\sqrt{x}}$

    (C) $\frac{1}{\sqrt{x}} \cos \sqrt{x}$

    (D) $\frac{1}{2\sqrt{x}} \cos \sqrt{x}$

53. $\frac{d}{dx} (\tan x)$ (BSEB 2019 A)

    (A) $\sec^2 x$

    (B) $\sec x$

    (C) $\cot x$

    (D) $-\sec^2 x$

54. $\frac{d}{dx} [\cos(\pi x + \sin \pi)]$ (BSEB 2025 A)

    (A) $-\sin(\pi x + \sin \pi)$

    (B) $-\pi \sin(\pi x)$

    (C) $-\sin \pi x$

    (D) $\sin x$

55. If $y = \sin x^2$, then $\frac{dy}{dx} =$ (BSEB 2017 C)

    (A) $2x \sin x^2$

    (B) $x \sin x$

    (C) $x \cos x^2$

    (D) $2x \cos x^2$

56. $\frac{d^2}{dx^2} (\sin 2x) =$ (BSEB Previous Year)

    (A) $4 \sin 2x$

    (B) $4 \cos^2 2x$

    (C) $-4 \sin 2x$

    (D) $2 \sin 4x$

57. $\frac{d}{dx} (\tan x^2) =$ (BSEB Previous Year)

    (A) $\sec x^2$

    (B) $2x \sec^2 x^2$

    (C) $2x^2 \sec^2 x^2$

    (D) $\frac{\sec x^2}{2x}$

58. $\frac{d}{dx} (2\cos \frac{3x}{4}) =$ (BSEB Previous Year)

    (A) $-2\sin \frac{3x}{4}$

    (B) $-\frac{3}{8} \sin \frac{3x}{4}$

    (C) $-\frac{3}{2} \sin \frac{3x}{4}$

    (D) $-\frac{3}{4} \sin \frac{3x}{4}$

59. $\frac{d}{dx} (\sin x + \sin^2 x) =$ (BSEB Previous Year)

    (A) $\cos x + \sin 2x$

    (B) $\cos x + \cos 2x$

    (C) $\cos x + \sin 2x$

    (D) $\cos x – \sin 2x$

60. $\frac{d}{dx} (\tan kx) =$ (BSEB Previous Year)

    (A) $\sec^2 kx$

    (B) $k \sec^2 x$

    (C) $\frac{\sec^2 kx}{k}$

    (D) $k \sec^2 kx$

61. $\frac{d}{dx} (\sin x)$ (BSEB 2022 A)

    (A) $\cos x$

    (B) $-\sin x$

    (C) $-\cos x$

    (D) $\tan x$

62. $\frac{d}{dx} [ \frac{1}{4} \sec 4x ] =$ (BSEB Previous Year)

    (A) $\sec 4x \tan 4x$

    (B) $\sec^2 4x$

    (C) $\tan^2 4x$

    (D) $\frac{1}{16} \sec 4x \tan 4x$

63. $\frac{d}{dx} (\sec x) =$ (BSEB Previous Year)

    (A) $\sec^2 x$

    (B) $\tan^2 x$

    (C) $\sec x \tan x$

    (D) $0$

64. $\frac{d}{dx} (\cos 2x) =$ (BSEB Previous Year)

    (A) $-\sin 2x$

    (B) $2\sin 2x$

    (C) $-2\sin 2x$

    (D) $-\frac{1}{2} \sin 2x$

65. If $y = \sin^2 x$, then $\frac{dy}{dx} =$ (BSEB Previous Year)

    (A) $2\sin x$

    (B) $\cos^2 x$

    (C) $2\sin x \cos x$

    (D) $\sin x \cos x$

66. $\frac{d}{dx} (\cot x) =$ (BSEB Previous Year)

    (A) $\tan x$

    (B) $\csc^2 x$

    (C) $-\csc^2 x$

    (D) $\csc x \cot x$

67. $\frac{d}{dx} (\cos x) =$ (BSEB Previous Year)

    (A) $\cos x$

    (B) $\sin x$

    (C) $-\cos x$

    (D) $-\sin x$

68. $\frac{d}{dx} (\sin \frac{x}{5}) =$ (BSEB Previous Year)

    (A) $\cos \frac{x}{5}$

    (B) $\frac{1}{5} \cos \frac{x}{5}$

    (C) $5 \cos \frac{x}{5}$

    (D) $-\frac{1}{5} \sin \frac{x}{5}$

69. $\frac{d}{d\theta} (\cos^3 \theta) =$ (BSEB Previous Year)

    (A) $-3\sin^2 \theta$

    (B) $3\sin^2 \theta \cos \theta$

    (C) $-3\cos^2 \theta \sin \theta$

    (D) $3\cos^2 \theta \sin \theta$

70. $\frac{d}{dx} (\sqrt{\tan x}) =$ (BSEB Previous Year)

    (A) $2\sqrt{\tan x}$

    (B) $\frac{\sec^2 x}{2\sqrt{\tan x}}$

    (C) $2\tan x \sec x$

    (D) $\frac{\sec x}{2\sqrt{\tan x}}$

71. $\frac{d}{dx} [\sin 4x] =$ (BSEB 2020)

    (A) $4 \sin 4x$

    (B) $4 \cos 4x$

    (C) $4x \sin 4x$

    (D) $4x \cos 4x$

72. $\frac{d}{dx} [\sqrt{\sin x}] =$ (BSEB 2021)

    (A) $\frac{\cos x}{2\sqrt{\sin x}}$

    (B) $\frac{1}{2\sqrt{\sin x}}$

    (C) $\frac{\sin x}{2\sqrt{\cos x}}$

    (D) $\cos \sqrt{x}$

73. If $y = \sin(x^3)$, then $\frac{dy}{dx} =$ (BSEB 2015)

    (A) $x^3 \cos(x^3)$

    (B) $3x^2 \sin(x^3)$

    (C) $3x^2 \cos(x^3)$

    (D) $\cos(x^3)$

74. $\frac{d}{dx} [3 \sin x – 4 \sin^3 x] =$ (BSEB Previous Year)

    (A) $3 \cos 3x$

    (B) $\cos 3x$

    (C) $3 \sin 3x$

    (D) $-3 \cos 3x$

75. $\frac{d}{dx} [\sec x] =$ (BSEB 2020)

    (A) $\sec x \cot x$

    (B) $\sec x \tan x$

    (C) $\tan x$

    (D) $\sec^2 x$

76. $\frac{d}{dx} [\tan x] =$ (BSEB 2019 A)

    (A) $\sec^2 x$

    (B) $\sec x$

    (C) $\cot x$

    (D) $-\sec^2 x$

77. $\frac{d}{dx} (\sin^2 x) =$ (BSEB 2019 C)

    (A) $\sin 2x$

    (B) $\cos 2x$

    (C) $\tan 2x$

    (D) $\cot 2x$

78. $\frac{d}{dx} (\tan \frac{x}{3}) =$ (BSEB 2022 A)

    (A) $\sec^2 \frac{x}{3}$

    (B) $\frac{1}{3} \sec^2 \frac{x}{3}$

    (C) $3 \sec^2 \frac{x}{3}$

    (D) $3 \cot \frac{x}{3}$

79. $\frac{d}{dx} (\sin \sqrt{x}) =$ (BSEB 2021 A)

    (A) $\cos \sqrt{x}$

    (B) $\frac{\cos \sqrt{x}}{\sqrt{x}}$

    (C) $\frac{1}{\sqrt{x}} \cos \sqrt{x}$

    (D) $\frac{1}{2\sqrt{x}} \cos \sqrt{x}$

80. $\frac{d}{dx}(\cos x^3) =$ (2026)

    (A) $-3x^2 \sin x^3$

    (B) $\sin x^3$

    (C) $3x^2 \sin x^3$

    (D) $3x^2$

81. $\frac{d}{dx}(\tan 5x) =$ (2026)

    (A) $5 \sec^2 5x$

    (B) $\sec^2 5x$

    (C) $5 \sec 5x$

    (D) $-5 \sec^2 5x$

82. $\frac{d}{dx}(\tan x^2) =$ (2026)

    (A) $2x \sec^2 x^2$

    (B) $\sec^2 x^2$

    (C) $2x \tan x^2$

    (D) $2 \sec^2 x^2$

83. $\frac{d}{dx}(\sqrt{\sin x}) =$ (2026)

    (A) $\frac{\cos x}{2\sqrt{\sin x}}$

    (B) $\frac{\sin x}{2\sqrt{\cos x}}$

    (C) $\frac{\cos x}{\sin x}$

    (D) $\frac{1}{2\sqrt{sin x}}$

84. $\frac{d}{dx}(\sin 2x + \cos 2x) =$ (2026)

    (A) $2 \cos 2x – 2 \sin 2x$

    (B) $2 \cos 2x + 2 \sin 2x$

    (C) $\cos 2x – \sin 2x$

    (D) 0

85. $\frac{d^2}{dx^2}(\sin 2x) =$ (2026)

    (A) $4 \sin 2x$

    (B) $-4 \sin 2x$

    (C) $2 \cos 2x$

    (D) $-2 \cos 2x$

86. $\frac{d}{dx}(\sin 2x) =$ (2026)

    (A) $\cos 2x$

    (B) $2 \cos 2x$

    (C) $-2 \cos 2x$

    (D) $\frac{cos 2x}{2}$

87. $\frac{d}{dx}(\tan kx) =$ (2026)

    (A) $k \sec^2 kx$

    (B) $\sec^2 kx$

    (C) $\frac{sec^2 kx}{k}$

    (D) $\tan k$

IV. Chain Rule & Combined Functions

88. If $y = \log \cos x^2$, then the value of $\frac{d}{dx}$ at $x = \sqrt{\pi}$ is: (BSEB 2018 A)

    (A) $1$

    (B) $\frac{\pi}{4}$

    (C) $0$

    (D) $\sqrt{\pi}$

89. If $y = \sin(\log x)$, then $\frac{dy}{dx} =$ (BSEB Previous Year)

    (A) $\frac{1}{x} \cos(\log x)$

    (B) $\frac{1}{x} \sin(\log x)$

    (C) $0$

    (D) $1$

90. $\frac{d}{dx} [\log (\tan x)] =$ (BSEB Previous Year)

    (A) $\sec^2 x$

    (B) $\frac{1}{\sin x \cos x}$

    (C) $2\csc 2x$

    (D) Both (B) and (C)

91. $\frac{d}{dx} (e^x + \cos 5x) =$ (BSEB Previous Year)

    (A) $e^x + \cos 5x$

    (B) $e^x + 5\sin 5x$

    (C) $e^x – 5\sin 5x$

    (D) $e^x – 5\cos 5x$

92. $\frac{d}{dx} [\log(\sec x + \tan x)] =$ (BSEB 2018 A, 2024 A)

    (A) $\sec x + \tan x$

    (B) $\sec x$

    (C) $\tan x$

    (D) $\sec x + \tan x$

93. $\frac{d}{dx} (\sqrt{x^2 + ax + 1}) =$ (BSEB Previous Year)

    (A) $\frac{x + a}{2\sqrt{x^2 + ax + 1}}$

    (B) $\frac{2x + a}{2\sqrt{x^2 + ax + 1}}$

    (C) $\frac{2x + a}{\sqrt{x^2 + ax + 1}}$

    (D) $\frac{1}{2\sqrt{x^2 + ax + 1}}$

94. $\frac{d}{dx} (\frac{1+e^x}{\sin x}) =$ (BSEB 2021 A)

    (A) $-\frac{1+e^x}{\sin^2 x}$

    (B) $\csc x + e^x$

    (C) $-\csc x \cot x + e^x$

    (D) $\csc x \cot x + e^x$

95. $\frac{d}{dx} (x^3 + e^x) =$ (BSEB Previous Year)

    (A) $3x^2 + e^x$

    (B) $3x^2 + 3e^x$

    (C) $3x^2 + e^x$

    (D) $3x^2 + e^x$

96. $\frac{d}{dx} (\log \cos x) =$ (BSEB Previous Year)

    (A) $\tan x$

    (B) $-\tan x$

    (C) $\cot x$

    (D) $-\cot x$

97. $\frac{d}{dx} (\sin 2x + e^x – \cos x) =$ (BSEB Previous Year)

    (A) $\cos 2x + e^x – \sin x$

    (B) $2\cos 2x + e^x + \sin x$

    (C) $2\cos 2x + e^x – \sin x$

    (D) $-2\cos 2x + e^x + \sin x$

98. $\frac{d}{dx} [x^2 \cdot e^x] =$ (BSEB 2023)

    (A) $e^x(x^2 + 2x)$

    (B) $e^x(x^2 – 2x)$

    (C) $2xe^x$

    (D) $x^2e^x$

99. If $y = \cos(\log x)$, then $\frac{dy}{dx} =$ (BSEB 2016)

    (A) $-\sin(\log x)$

    (B) $\frac{-\sin(\log x)}{x}$

    (C) $\frac{\cos(\log x)}{x}$

    (D) $-\sin(\log x) \cdot \log x$

100. $\frac{d}{dx} [e^x \sin x] =$ (BSEB Previous Year)

     (A) $e^x (\sin x + \cos x)$

     (B) $e^x (\sin x – \cos x)$

     (C) $e^x \cos x$

     (D) $e^x \sin x$

101. $\frac{d}{dx} [x^2 \sin \frac{1}{x}] =$ (BSEB Previous Year)

     (A) $2x \sin \frac{1}{x} – \cos \frac{1}{x}$

     (B) $2x \sin \frac{1}{x} + \cos \frac{1}{x}$

     (C) $x \sin \frac{1}{x} + \cos \frac{1}{x}$

     (D) $2x \cos \frac{1}{x}$

102. $\frac{d}{dx} (\log \cos x) =$ (BSEB Previous Year)

     (A) $\tan x$

     (B) $-\tan x$

     (C) $\cot x$

     (D) $-\cot x$

103. $\frac{d}{dx} [ \sqrt{x^2+ax+1} ] =$ (BSEB 2023)

     (A) $\frac{2x+a}{2\sqrt{x^2+ax+1}}$

     (B) $\frac{x+a}{2\sqrt{x^2+ax+1}}$

     (C) $\frac{2x+a}{\sqrt{x^2+ax+1}}$

     (D) $\frac{1}{2\sqrt{x^2+ax+1}}$

104. If $y = \log(\sin x)$, then $\frac{dy}{dx} =$ (2026)

     (A) $\cot x$

     (B) $\tan x$

     (C) $\frac{1}{\sin x}$

     (D) $\cos x$

105. If $y = \sin(\log x)$, then $\frac{dy}{dx} =$ (2026)

     (A) $\frac{\cos(\log x)}{x}$

     (B) $\cos(\log x)$

     (C) $\frac{\sin(\log x)}{x}$

     (D) $-\frac{\cos(\log x)}{x}$

106. If $y = \cos( \sin x)$, then $\frac{dy}{dx} =$ (2026)

     (A) $-\sin(\sin x) \cdot \cos x$

     (B) $\sin(\sin x) \cdot \cos x$

     (C) $-\sin(\sin x)$

     (D) $\cos(\cos x)$

107. $\frac{d}{dx}(\log \cos x) =$ (2026)

     (A) $\tan x$

     (B) $-\tan x$

     (C) $\cot x$

     (D) $-\cot x$

108. $\frac{d}{dx}(e^{\cot x}) =$ (2026)

     (A) $-\csc^2 x \cdot e^{\cot x}$

     (B) $\sin x \cdot e^{\cot x}$

     (C) $\cos x \cdot e^{\cot x}$

     (D) $e^{\cot x}$

109. If $y = \cos(\log x)$, then $\frac{dy}{dx} =$ (2026)

     (A) $-sin(log x)$

     (B) $-\frac{sin(log x)}{x}$

     (C) $\frac{sin(log x)}{x}$

     (D) $-\frac{cos(log x)}{x}$

V. Inverse Trigonometric Functions

110. $\frac{d}{dx} [\sin^{-1} (3x – 4x^3)] =$ (BSEB 2021 A)

     (A) $\frac{1}{\sqrt{1-x^2}}$

     (B) $\frac{1}{\sqrt{1-(3x-4x^3)^2}}$

     (C) $\frac{3}{\sqrt{1-x^2}}$

     (D) $\frac{2}{\sqrt{1-x^2}}$

111. If $y = \tan^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}$, then $\frac{dy}{dx} =$ (BSEB 2018 C)

     (A) $\frac{1}{2}$

     (B) $-\frac{1}{2}$

     (C) $\frac{1}{1 + \tan^2 \frac{x}{2}}$

     (D) $\frac{1}{1+x^2}$

112. $\frac{d}{dx} (\sec^{-1} x + \csc^{-1} x) =$ (BSEB 2024 A)

     (A) $\frac{1}{\sqrt{1-x^2}}$

     (B) $\frac{-1}{\sqrt{1-x^2}}$

     (C) $\frac{1}{\sqrt{1-x^2}}$

     (D) $0$

113. $\frac{d}{dx} (\cot^{-1} x) =$ (BSEB 2020 A)

     (A) $\frac{1}{1+x^2}$

     (B) $\frac{1}{1-x^2}$

     (C) $-\frac{1}{1+x}$

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

114. If $y = \tan^{-1} (\frac{\sin x + \cos x}{\cos x – \sin x})$, then $\frac{dy}{dx} =$ (BSEB 2022)

     (A) $0$

     (B) $1$

     (C) $1/2$

     (D) $x$

115. $\frac{d}{dx} (2\tan^{-1} x) =$ (BSEB 2021 A)

     (A) $\frac{1}{1+x^2}$

     (B) $\frac{2}{1+x^2}$

     (C) $\frac{1}{2(1+x^2)}$

     (D) $\frac{1}{2(1-x^2)}$

116. If $y = \tan^{-1} \sqrt{\frac{1+\cos x}{1-\cos x}}$, then $\frac{dy}{dx} =$ (BSEB Previous Year)

     (A) $-\frac{1}{2}$

     (B) $0$

     (C) $\frac{1}{2}$

     (D) $1$

117. $\frac{d}{dx} (\sin^{-1} x) =$ (BSEB Previous Year)

     (A) $\frac{1}{1+x^2}$

     (B) $\frac{1}{1-x^2}$

     (C) $\frac{1}{\sqrt{1-x^2}}$

     (D) $\sqrt{\frac{1}{1+x^2}}$

118. $\frac{d}{dx} (\sec^{-1} x) =$ (BSEB Previous Year)

     (A) $\frac{1}{\sqrt{1-x^2}}$

     (B) $\frac{1}{x\sqrt{x^2-1}}$

     (C) $\frac{1}{1+x^2}$

     (D) $\frac{1}{x\sqrt{x^2-1}}$

119. If $y = \tan^{-1} \sqrt{x}$, then $\frac{dy}{dx} =$ (BSEB 2017 C)

     (A) $\frac{1}{2\sqrt{x}(1-x)}$

     (B) $\frac{1}{x(1+x)}$

     (C) $\frac{1}{x^2(1+x)}$

     (D) $\frac{1}{2\sqrt{x}(1+x)}$

120. $\frac{d}{dx} (\sin^{-1} \sqrt{x} + \cos^{-1} \sqrt{x}) =$ (BSEB Previous Year)

     (A) $\pi/2$

     (B) $0$

     (C) $1$

     (D) $\sqrt{x} \pi/2$

121. $\frac{d}{dx} (\tan^{-1} x + \cot^{-1} x) =$ (BSEB 2016 A)

     (A) $\frac{2}{1+x^2}$

     (B) $0$

     (C) $1$

     (D) $2$

122. $\frac{d}{dx} (\cos^{-1} x) =$ (BSEB 2016 A)

     (A) $\frac{1}{2\sqrt{1-x^2}}$

     (B) $\frac{1}{\sqrt{1-x^2}}$

     (C) $\frac{-1}{\sqrt{1-x^2}}$

     (D) $\frac{-1}{2\sqrt{1-x^2}}$

123. $\frac{d}{dx} [\tan^{-1} (\frac{1-x}{1+x})] =$ (BSEB 2017)

     (A) $\frac{1}{1+x^2}$

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

     (C) $\frac{1}{1-x^2}$

     (D) $\frac{-1}{1-x^2}$

124. $\frac{d}{dx} [\cos^{-1}(\sin x)] =$ (BSEB Previous Year)

     (A) $-1$

     (B) $\frac{x}{\sqrt{1-x^2}}$

     (C) $\frac{\sin x}{\sqrt{1-x^2}}$

     (D) $\frac{\pi-x}{2}$

125. If $y = \tan^{-1} (\frac{1+x}{1-x})$, then $\frac{dy}{dx} =$ (BSEB Previous Year)

     (A) $\frac{2}{1+x^2}$

     (B) $\frac{1}{1+x^2}$

     (C) $\frac{1}{1-x^2}$

     (D) None

126. If $y = \sec(\tan^{-1} x)$, then $\frac{dy}{dx} =$ (BSEB Previous Year)

     (A) $\frac{xy}{\sqrt{1+x^2}}$

     (B) $\frac{-x}{\sqrt{1+x^2}}$

     (C) $\frac{x}{\sqrt{1-x^2}}$

     (D) None

127. $\frac{d}{dx} [\tan^{-1} (\frac{\cos x + \sin x}{\cos x – \sin x})] =$ (BSEB Previous Year)

     (A) 0

     (B) 1

     (C) $1/2$

     (D) $-1$

128. $\frac{d}{dx} [\sin^{-1} x] =$ (BSEB 2015, 2019)

     (A) $\frac{1}{\sqrt{1-x^2}}$

     (B) $\frac{-1}{\sqrt{1-x^2}}$

     (C) $\frac{1}{1-x^2}$

     (D) $\frac{1}{1+x^2}$

129. $\frac{d}{dx} [\tan^{-1} \sqrt{1+x^2} – \cot^{-1} (-\sqrt{1+x^2})] =$ (BSEB 2018 A)

     (A) $\pi$

     (B) $1$

     (C) $0$

     (D) $\frac{2x}{\sqrt{1+x^2}}$

130. $\frac{d}{dx} (\sin^{-1} x + \cos^{-1} x) =$ (BSEB 2021)

     (A) $0$

     (B) $1$

     (C) $\pi/2$

     (D) $\frac{1}{\sqrt{1-x^2}}$

131. $\frac{d}{dx} \sin^{-1}(3x – 4x^3) =$ (2026)

     (A) $\frac{3}{\sqrt{1-x^2}}$

     (B) $\frac{-3}{\sqrt{1-x^2}}$

     (C) $\frac{1}{\sqrt{1-x^2}}$

     (D) $\frac{-1}{\sqrt{1-x^2}}$

132. If $y = \tan^{-1} x$, then $\frac{dy}{dx} =$ (2026)

     (A) $\frac{1}{1+x^2}$

     (B) $\frac{1}{1-x^2}$

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

     (D) $\frac{1}{\sqrt{1-x^2}}$

133. The derivative of $\cos^{-1} x$ with respect to $x$ is: (2026)

     (A) $\frac{1}{\sqrt{1-x^2}}$

     (B) $\frac{-1}{\sqrt{1-x^2}}$

     (C) $\frac{1}{1+x^2}$

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

134. $\frac{d}{dx}(\sin^{-1} x + \cos^{-1} x) =$ (2026)

     (A) $\frac{2}{\sqrt{1-x^2}}$

     (B) 0

     (C) $\pi/2$

     (D) 1

135. $\frac{d}{dx}(\tan^{-1} \sqrt{x}) =$ (2026)

     (A) $\frac{1}{2\sqrt{x}(1+x)}$

     (B) $\frac{1}{1+x}$

     (C) $\frac{\sqrt{x}}{1+x}$

     (D) $\frac{2}{\sqrt{x}(1+x)}$

VI. Parametric, Implicit & Infinite Series

136. If $x^m y^n = (x+y)^{m+n}$, then $\frac{dy}{dx} =$ (BSEB 2018 C)

     (A) $\frac{x}{y}$

     (B) $\frac{y}{x}$

     (C) $\frac{-y}{x}$

     (D) $\frac{-x}{y}$

137. If $x = a \sec \theta, y = b \tan \theta$, then $\frac{dy}{dx} =$ (BSEB Previous Year)

     (A) $\frac{b}{a} \sec \theta$

     (B) $\frac{b}{a} \csc \theta$

     (C) $\frac{b}{a} \cot \theta$

     (D) $\frac{b}{a}$

138. If $y = e^{x + e^{x + e^{x + \dots \infty}}}$, then $\frac{dy}{dx} =$ (BSEB Previous Year)

     (A) $\frac{y}{y+1}$

     (B) $\frac{y}{y-1}$

     (C) $\frac{y}{1-y}$

     (D) None

139. If $x = a \cos \theta, y = a \sin \theta$, then $\frac{dy}{dx} =$ (BSEB Previous Year)

     (A) $\tan \theta$

     (B) $-\cot \theta$

     (C) $-\tan \theta$

     (D) $\sec^2 \theta$

140. If $x = \sin \theta, y = \cos \theta$, then $\frac{dy}{dx} =$ (BSEB 2019 C)

     (A) $\tan \theta$

     (B) $-\tan \theta$

     (C) $\cot \theta$

     (D) None of these

141. If $y + x = \sin(y+x)$, then $\frac{dy}{dx} =$ (BSEB 2018 C)

     (A) $\frac{1-\cos(y+x)}{1+\cos(y+x)}$

     (B) $1$

     (C) $-1$

     (D) $0$

142. If $y = \sqrt{\sin x + \sqrt{\sin x + \dots \infty}}$, then $\frac{dy}{dx} =$ (BSEB 2024 A)

     (A) $\frac{\cos x}{2y + 1}$

     (B) $\frac{\cos x}{2y – 1}$

     (C) $\frac{\sin x}{2y + 1}$

     (D) $\frac{\sin x}{2y – 1}$

143. If $x^2y^3 = (x+y)^5$, then $\frac{dy}{dx} =$ (BSEB 2018 A)

     (A) $\frac{x}{y}$

     (B) $\frac{y}{x}$

     (C) $-\frac{y}{x}$

     (D) $-\frac{x}{y}$

144. If $x = a \cos^2 \theta, y = b \sin^2 \theta$, then $\frac{dy}{dx} =$ (BSEB Previous Year)

     (A) $\frac{b}{a}$

     (B) $-\frac{b}{a}$

     (C) $\frac{b}{a} \sin 2\theta$

     (D) $-\frac{b}{a} \tan^2 \theta$

145. If $x = a \cos \theta, y = b \cos \theta$, then $\frac{dy}{dx} =$ (BSEB 2017)

     (A) $b/a$

     (B) $-b/a$

     (C) $a/b$

     (D) $-a/b$

146. $\frac{d}{dx} (x^x) =$ (BSEB Previous Year)

     (A) $x \cdot x^{x-1}$

     (B) $x^x (1 + \log x)$

     (C) $x^x \log x$

     (D) None

147. If $x = \sin t, y = \cos t$, then $\frac{dy}{dx} =$ (BSEB Previous Year)

     (A) $-\tan t$

     (B) $-\cot t$

     (C) $\tan t$

     (D) $\cot t$

148. If $x = t + \frac{1}{t}, y = t – \frac{1}{t}$, then $\frac{dy}{dx} =$ (BSEB Previous Year)

     (A) $\frac{2t}{t^2+1}$

     (B) $\frac{t^2+1}{t^2-1}$

     (C) $\frac{t^2-1}{t^2+1}$

     (D) $\frac{2t}{1-t^2}$

149. If $x\sqrt{1+y} + y\sqrt{1+x} = 0$, then $\frac{dy}{dx} =$ (BSEB Previous Year)

     (A) $\frac{x+1}{x}$

     (B) $\frac{1}{1+x}$

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

     (D) None

150. If $y = \sqrt{\sin x + y}$, then $\frac{dy}{dx} =$ (BSEB Previous Year)

     (A) $\frac{\cos x}{2y – 1}$

     (B) $\frac{\cos x}{1 – 2y}$

     (C) $\frac{\sin x}{1 – 2y}$

     (D) $\frac{\sin x}{2y – 1}$

151. If $y = x \tan y$, then $\frac{dy}{dx} =$ (BSEB Previous Year)

     (A) $\frac{\tan y}{x – x^2 – y^2}$

     (B) $\frac{y^2}{x – x^2 – y^2}$

     (C) $\frac{\tan y}{y – x}$

     (D) $\frac{\tan x}{x – y}$

152. If $y = \sqrt{\sin x + \sqrt{\sin x + \dots \infty}}$, then $\frac{dy}{dx} =$ (2026)

     (A) $\frac{1}{2y-1}$

     (B) $\frac{\cos x}{2y-1}$

     (C) $\frac{\sin x}{2y-1}$

     (D) $\frac{2y-1}{\cos x}$

153. If $x^n + y^n = a^n$ then $\frac{dy}{dx} =$ (2026)

     (A) $-\frac{x^{n-1}}{y^{n-1}}$

     (B) $\frac{x^{n-1}}{y^{n-1}}$

     (C) $-\frac{y^{n-1}}{x^{n-1}}$

     (D) $nx^{n-1}$

154. If $x = a(1 – \cos \theta)$, $y = a(\theta + \sin \theta)$, then $\frac{dy}{dx} =$ (2026)

     (A) $\tan \frac{\theta}{2}$

     (B) $-\tan \frac{\theta}{2}$

     (C) $\cot \frac{\theta}{2}$

     (D) $-\cot \frac{\theta}{2}$

155. If $y = x^x$ then $\frac{dy}{dx} =$ (2026)

     (A) $x^x(\log x + 1)$

     (B) $\log x$

     (C) $(\log x + 1)$

     (D) $nx^{n-1}$

156. If $x = a cos \theta, y = a sin \theta$, then $\frac{dy}{dx} =$ (2026)

     (A) $tan \theta$

     (B) $cot \theta$

     (C) $-cot \theta$

     (D) $-tan \theta$

157. If $x = at^2, y = 2at$, then $\frac{dy}{dx} =$ (2026)

     (A) $t$

     (B) $\frac{1}{t}$

     (C) $-t$

     (D) $-\frac{1}{t}$

VII. Limits, Continuity & Constant Functions

158. $\frac{d}{dx} [\lim_{x \to 0} \cos 3x]$ (BSEB Previous Year)

     (A) $-\sin 3x$

     (B) $1$

     (C) $-\sin 3x$

     (D) $0$

159. $\frac{d}{dx} [\lim_{x \to 1} \frac{x^n – 1}{x – 1}] =$ (BSEB 2018 C)

     (A) $n$

     (B) $nx^{n-1}$

     (C) $1$

     (D) $0$

160. $\frac{d^2}{dx^2} [x^2 + 3x + 2] =$ (BSEB Previous Year)

     (A) $4$

     (B) $4x$

     (C) $2x+3$

     (D) $2$

161. $\frac{d}{dx} [\lim_{x \to a} \frac{x^n – a^n}{x – a}] =$ (BSEB 2021 A)

     (A) $na^{n-1}$

     (B) $1$

     (C) $0$

     (D) $n$

162. If $y = \log(\sin x)$, then $\frac{dy}{dx} =$ (2026)

     (A) $\cot x$

     (B) $\tan x$

     (C) $\frac{1}{\sin x}$

     (D) $\cos x$

163. If $y = \sin(\log x)$, then $\frac{dy}{dx} =$ (2026)

     (A) $\frac{\cos(\log x)}{x}$

     (B) $\cos(\log x)$

     (C) $\frac{\sin(\log x)}{x}$

     (D) $-\frac{\cos(\log x)}{x}$

164. $\frac{d}{dx}(\sin^{-1} x + \cos^{-1} x) =$ (2026)

     (A) $\frac{2}{\sqrt{1-x^2}}$

     (B) 0

     (C) $\pi/2$

     (D) 1