NCERT Solutions for Class 12 Maths Chapter 5: Continuity and Differentiability (NCERT 2026–27)

These Class 12 Maths Chapter 5 solutions cover Continuity and Differentiability from the NCERT textbook (Reprint 2026–27). Every question of Exercises 5.1 to 5.7 and the Miscellaneous Exercise is reproduced verbatim and solved step by step — checking continuity, applying the chain rule, differentiating implicit, inverse-trigonometric, exponential and logarithmic functions, parametric forms and second-order derivatives — with each answer cross-checked against the NCERT answer key.

Class: 12 Subject: Mathematics Book: Mathematics Part I Chapter: 5 Exercises: 5.1–5.7 + Miscellaneous Session: 2026–27

On this page

Chapter overview
Key concepts & definitions
Important formulas
Exercise 5.1 (Continuity)
Exercise 5.2 (Chain rule)
Exercise 5.3 (Implicit functions)
Exercise 5.4 (Exponential & logarithmic)
Exercise 5.5 (Logarithmic differentiation)
Exercise 5.6 (Parametric forms)
Exercise 5.7 (Second order derivatives)
Miscellaneous Exercise
Common mistakes to avoid
Practice MCQs & Assertion–Reason
Quick revision summary
FAQs

Chapter 5 Overview

Chapter 5, Continuity and Differentiability, continues the study of differentiation begun in Class XI. It introduces the precise idea of continuity at a point (the limit of the function equals its value there) and the algebra of continuous functions, then defines differentiability and proves that every differentiable function is continuous (the converse being false, as |x| shows). The chapter develops powerful tools: the chain rule for composite functions, differentiation of implicit and inverse trigonometric functions, the new class of exponential and logarithmic functions, logarithmic differentiation for forms like [u(x)]^v(x), derivatives of functions in parametric form, and second-order derivatives. The solutions below work through all eight exercise sets question by question.

Key Concepts & Definitions

Continuity at a point c: f is continuous at c if lim_x→c f(x) = f(c); equivalently the left-hand limit, right-hand limit and f(c) all exist and are equal.

Continuous function: continuous at every point of its domain. Polynomial and rational (where defined), trigonometric, exponential and logarithmic functions are continuous on their domains.

Algebra of continuous functions: if f and g are continuous at c, so are f + g, f − g, f·g and f/g (provided g(c) ≠ 0). A composite of continuous functions is continuous.

Differentiability: f is differentiable at c if lim_h→0 [f(c + h) − f(c)]/h exists (left and right derivatives finite and equal).

Relation: every differentiable function is continuous, but a continuous function need not be differentiable (e.g. |x| at x = 0).

Chain rule: if f = v o u and t = u(x), then df/dx = (dv/dt)(dt/dx).

Logarithmic differentiation: used for y = [u(x)]^v(x) (with u, y > 0) by taking log of both sides before differentiating.

Important Formulas (Chapter 5)

Inverse trig: d/dx(sin⁻¹x) = 1/√(1−x²); d/dx(cos⁻¹x) = −1/√(1−x²); d/dx(tan⁻¹x) = 1/(1+x²).

Exponential & log: d/dx(e^x) = e^x; d/dx(log x) = 1/x; d/dx(a^x) = a^x log a.

Chain rule: df/dx = (df/dt)(dt/dx). Product: (uv)′ = u′v + uv′. Quotient: (u/v)′ = (u′v − uv′)/v².

Parametric: if x = f(t), y = g(t) then dy/dx = (dy/dt)/(dx/dt), provided dx/dt ≠ 0.

Logarithmic differentiation: for y = [u(x)]^v(x), log y = v(x) log u(x), then differentiate.

Change of base: log_ap = (log_bp)/(log_ba).

Questions are reproduced verbatim from the NCERT Mathematics Part I textbook (Reprint 2026–27); the worked solutions are original and verified against the answers given at the back of the book. Throughout, log x denotes the natural logarithm (base e).

Exercise 5.1 — Continuity

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

SOLUTION At x = 0: lim_x→0(5x − 3) = −3 = f(0). ✓ At x = −3: lim_x→−3(5x − 3) = −15 − 3 = −18 = f(−3). ✓ At x = 5: lim_x→5(5x − 3) = 25 − 3 = 22 = f(5). ✓ Since the limit equals the value at each point, f is continuous at x = 0, −3 and 5.

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

SOLUTION f(3) = 2(9) − 1 = 17. lim_x→3(2x² − 1) = 2(3)² − 1 = 17 = f(3). ∴ f is continuous at x = 3.

3. Examine the following functions for continuity. (a) f(x) = x − 5 (b) f(x) = 1/(x − 5), x ≠ 5 (c) f(x) = (x² − 25)/(x + 5), x ≠ −5 (d) f(x) = |x − 5|

SOLUTION (a) A polynomial; for any real c, lim_x→c(x − 5) = c − 5 = f(c). Continuous on R. (b) Rational, defined for x ≠ 5. For any c ≠ 5, lim_x→c 1/(x − 5) = 1/(c − 5) = f(c). Continuous at every point of its domain. (c) Defined for x ≠ −5, where (x² − 25)/(x + 5) = x − 5. For c ≠ −5, lim = c − 5 = f(c). Continuous everywhere in its domain. (d) |x − 5| is a composite of continuous functions, hence continuous on R. ∴ (a), (b), (c) and (d) are all continuous functions on their domains.

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

SOLUTION f(x) = xⁿ is a polynomial, defined for all real x, and f(n) = nⁿ. lim_x→n xⁿ = nⁿ = f(n). Since the limit equals the value, f(x) = xⁿ is continuous at x = n.

5. Is the function f defined by f(x) = x if x ≤ 1, and f(x) = 5 if x > 1, continuous at x = 0? At x = 1? At x = 2?

SOLUTION At x = 0 (< 1): f(0) = 0 and lim_x→0 x = 0. Continuous. At x = 1: f(1) = 1; LHL = lim_x→1⁻ x = 1, RHL = lim_x→1⁺ 5 = 5. LHL ≠ RHL, so not continuous at x = 1. At x = 2 (> 1): f(2) = 5 and lim_x→2 5 = 5. Continuous. ∴ f is continuous at x = 0 and x = 2; not continuous at x = 1.

6. Find all points of discontinuity of f, where f(x) = 2x + 3 if x ≤ 2, and f(x) = 2x − 3 if x > 2.

SOLUTION For x ≠ 2 the function is a polynomial in each piece, hence continuous. At x = 2: f(2) = 2(2) + 3 = 7; LHL = 7; RHL = lim_x→2⁺(2x − 3) = 1. LHL ≠ RHL, so f is discontinuous at x = 2 only.

7. Find all points of discontinuity of f, where f(x) = |x| + 3 if x ≤ −3, f(x) = −2x if −3 < x < 3, and f(x) = 6x + 2 if x ≥ 3.

SOLUTION Each piece is continuous on its interval. Check the joins. At x = −3: f(−3) = |−3| + 3 = 6; LHL (from |x| + 3) = 6; RHL (from −2x) = 6. Equal → continuous. At x = 3: f(3) = 6(3) + 2 = 20; LHL (from −2x) = −6; RHL = 20. LHL ≠ RHL → discontinuous. ∴ f is discontinuous at x = 3 only.

8. Find all points of discontinuity of f, where f(x) = |x|/x if x ≠ 0, and f(x) = 0 if x = 0.

SOLUTION For x > 0, f(x) = 1; for x < 0, f(x) = −1; f(0) = 0. At x = 0: LHL = −1, RHL = +1; they differ (and neither equals f(0) = 0). ∴ f is discontinuous at x = 0 only.

9. Find all points of discontinuity of f, where f(x) = x/|x| if x < 0, and f(x) = −1 if x ≥ 0.

SOLUTION For x < 0, |x| = −x, so f(x) = x/(−x) = −1; for x ≥ 0, f(x) = −1. Thus f(x) = −1 for all x, a constant function. ∴ No point of discontinuity — f is continuous everywhere.

10. Find all points of discontinuity of f, where f(x) = x + 1 if x ≥ 1, and f(x) = x² + 1 if x < 1.

SOLUTION Each piece is a polynomial → continuous away from x = 1. At x = 1: f(1) = 1 + 1 = 2; LHL = lim_x→1⁻(x² + 1) = 2; RHL = 2. All equal. ∴ No point of discontinuity.

11. Find all points of discontinuity of f, where f(x) = x³ − 3 if x ≤ 2, and f(x) = x² + 1 if x > 2.

SOLUTION At x = 2: f(2) = 2³ − 3 = 5; LHL = 5; RHL = lim_x→2⁺(x² + 1) = 5. All equal. Each piece is continuous, so f is continuous everywhere. ∴ No point of discontinuity.

12. Find all points of discontinuity of f, where f(x) = x¹⁰ − 1 if x ≤ 1, and f(x) = x² if x > 1.

SOLUTION At x = 1: f(1) = 1¹⁰ − 1 = 0; LHL = 0; RHL = lim_x→1⁺ x² = 1. LHL ≠ RHL, so f is discontinuous at x = 1 only.

13. Is the function defined by f(x) = x + 5 if x ≤ 1, and f(x) = x − 5 if x > 1, a continuous function?

SOLUTION At x = 1: f(1) = 1 + 5 = 6; LHL = 6; RHL = lim_x→1⁺(x − 5) = −4. LHL ≠ RHL, so f is not continuous (discontinuous at x = 1).

14. Discuss the continuity of f, where f(x) = 3 if 0 ≤ x ≤ 1, f(x) = 4 if 1 < x < 3, and f(x) = 5 if 3 ≤ x ≤ 10.

SOLUTION Each piece is constant → continuous on its interval. Check joins. At x = 1: f(1) = 3; LHL = 3; RHL = 4 → discontinuous. At x = 3: f(3) = 5; LHL = 4; RHL = 5 → discontinuous. ∴ f is not continuous at x = 1 and x = 3.

15. Discuss the continuity of f, where f(x) = 2x if x < 0, f(x) = 0 if 0 ≤ x ≤ 1, and f(x) = 4x if x > 1.

SOLUTION At x = 0: f(0) = 0; LHL = lim_x→0⁻ 2x = 0; RHL = 0 → continuous. At x = 1: f(1) = 0; LHL = 0; RHL = lim_x→1⁺ 4x = 4 → discontinuous. ∴ x = 1 is the only point of discontinuity.

16. Discuss the continuity of f, where f(x) = −2 if x ≤ −1, f(x) = 2x if −1 < x ≤ 1, and f(x) = 2 if x > 1.

SOLUTION At x = −1: f(−1) = −2; LHL = −2; RHL = 2(−1) = −2 → continuous. At x = 1: f(1) = 2(1) = 2; LHL = 2; RHL = 2 → continuous. ∴ f is continuous for all x.

17. Find the relationship between a and b so that the function f defined by f(x) = ax + 1 if x ≤ 3, and f(x) = bx + 3 if x > 3, is continuous at x = 3.

SOLUTION Continuity at x = 3 requires LHL = RHL = f(3). f(3) = 3a + 1 (LHL); RHL = 3b + 3. 3a + 1 = 3b + 3 ⇒ 3a = 3b + 2 ⇒ a = b + 2/3.

18. For what value of λ is the function f(x) = λ(x² − 2x) if x ≤ 0, and f(x) = 4x + 1 if x > 0, continuous at x = 0? What about continuity at x = 1?

SOLUTION At x = 0: f(0) = λ(0) = 0; LHL = 0; RHL = lim_x→0⁺(4x + 1) = 1. LHL = 0 ≠ 1 = RHL for every λ, so for no value of λ is f continuous at x = 0. At x = 1 (> 0): f(x) = 4x + 1 near x = 1, a polynomial, so lim = 5 = f(1). f is continuous at x = 1 for any value of λ.

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.

SOLUTION Let c be an integer. Then g(c) = c − [c] = c − c = 0. LHL: for x just below c, [x] = c − 1, so g(x) = x − (c − 1); lim_x→c⁻ g(x) = c − (c − 1) = 1. RHL: for x just above c, [x] = c, so g(x) = x − c; lim_x→c⁺ g(x) = 0. LHL = 1 ≠ 0 = RHL, so g is discontinuous at every integer.

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

SOLUTION x², sin x and constants are continuous, so their combination is continuous. f(π) = π² − sin π + 5 = π² + 5; lim_x→π f(x) = π² − 0 + 5 = π² + 5 = f(π). ∴ f is continuous at x = π.

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

SOLUTION sin x and cos x are continuous for all real x. By the algebra of continuous functions, their sum, difference and product are continuous. ∴ (a), (b) and (c) are all continuous functions on R.

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

SOLUTION cos x is continuous for all x ∈ R (proved like sin x). cosec x = 1/sin x is continuous wherever sin x ≠ 0, i.e. except at x = nπ, n ∈ Z. sec x = 1/cos x is continuous wherever cos x ≠ 0, i.e. except at x = (2n + 1)π/2, n ∈ Z. cot x = cos x/sin x is continuous wherever sin x ≠ 0, i.e. except at x = nπ, n ∈ Z.

23. Find all points of discontinuity of f, where f(x) = (sin x)/x if x < 0, and f(x) = x + 1 if x ≥ 0.

SOLUTION For x < 0, (sin x)/x is continuous; for x > 0, x + 1 is continuous. At x = 0: f(0) = 1; LHL = lim_x→0⁻ (sin x)/x = 1; RHL = lim_x→0⁺(x + 1) = 1. All equal ⇒ continuous at 0. ∴ There is no point of discontinuity.

24. Determine if f defined by f(x) = x² sin(1/x) if x ≠ 0, and f(x) = 0 if x = 0, is a continuous function.

SOLUTION For x ≠ 0 it is a product/composite of continuous functions → continuous. At x = 0: since −1 ≤ sin(1/x) ≤ 1, we have |x² sin(1/x)| ≤ x² → 0 as x → 0. By the squeeze rule, lim_x→0 f(x) = 0 = f(0). ∴ Yes, f is continuous for all x ∈ R.

25. Examine the continuity of f, where f is defined by f(x) = sin x − cos x if x ≠ 0, and f(x) = −1 if x = 0.

SOLUTION For x ≠ 0, sin x − cos x is continuous. At x = 0: f(0) = −1; lim_x→0(sin x − cos x) = 0 − 1 = −1 = f(0). ∴ f is continuous for all x ∈ R.

26. Find the value of k so that f(x) = (k cos x)/(π − 2x) if x ≠ π/2, and f(x) = 3 if x = π/2, is continuous at x = π/2.

SOLUTION Put x = π/2 + h, h → 0: cos x = cos(π/2 + h) = −sin h, and π − 2x = −2h. lim_x→π/2 (k cos x)/(π − 2x) = lim_h→0 (−k sin h)/(−2h) = (k/2)·lim_h→0(sin h)/h = k/2. For continuity: k/2 = f(π/2) = 3 ⇒ k = 6.

27. Find the value of k so that f(x) = kx² if x ≤ 2, and f(x) = 3 if x > 2, is continuous at x = 2.

SOLUTION f(2) = LHL = k(2)² = 4k; RHL = 3. Continuity: 4k = 3 ⇒ k = 3/4.

28. Find the value of k so that f(x) = kx + 1 if x ≤ π, and f(x) = cos x if x > π, is continuous at x = π.

SOLUTION f(π) = LHL = kπ + 1; RHL = cos π = −1. kπ + 1 = −1 ⇒ kπ = −2 ⇒ k = −2/π.

29. Find the value of k so that f(x) = kx + 1 if x ≤ 5, and f(x) = 3x − 5 if x > 5, is continuous at x = 5.

SOLUTION f(5) = LHL = 5k + 1; RHL = 3(5) − 5 = 10. 5k + 1 = 10 ⇒ 5k = 9 ⇒ k = 9/5.

30. Find the values of a and b such that the function defined by f(x) = 5 if x ≤ 2, f(x) = ax + b if 2 < x < 10, and f(x) = 21 if x ≥ 10, is a continuous function.

SOLUTION Continuity at x = 2: 5 = 2a + b. Continuity at x = 10: 10a + b = 21. Subtract: (10a + b) − (2a + b) = 21 − 5 ⇒ 8a = 16 ⇒ a = 2. Then b = 5 − 2(2) = 1. ∴ a = 2, b = 1.

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

SOLUTION Write f = g o h with g(x) = cos x and h(x) = x². Both g and h are continuous on R; by the composite-function theorem, f = cos(x²) is continuous.

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

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

SOLUTION Let h(x) = |x| and g(x) = sin x, so sin|x| = (g o h)(x). |x| is continuous on R and sin x is continuous on R, hence the composite sin|x| is continuous.

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

SOLUTION |x| and |x + 1| are continuous on R (composites of continuous functions). The difference of two continuous functions is continuous. ∴ There is no point of discontinuity — f is continuous everywhere.

Exercise 5.2 — Chain Rule

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

1. sin(x² + 5)

SOLUTION d/dx sin(x² + 5) = cos(x² + 5) · d/dx(x² + 5) = 2x cos(x² + 5).

2. cos(sin x)

SOLUTION d/dx cos(sin x) = −sin(sin x) · cos x = −cos x · sin(sin x).

3. sin(ax + b)

SOLUTION d/dx sin(ax + b) = cos(ax + b) · a = a cos(ax + b).

4. sec(tan(√x))

SOLUTION Let u = √x. d/dx = sec(tan u)tan(tan u) · sec²u · (1/(2√x)). = [sec(tan√x) · tan(tan√x) · sec²√x] / (2√x).

5. sin(ax + b)/cos(cx + d)

SOLUTION Quotient rule with u = sin(ax + b), v = cos(cx + d): u′ = a cos(ax + b), v′ = −c sin(cx + d). dy/dx = [a cos(ax + b)cos(cx + d) + c sin(ax + b)sin(cx + d)] / cos²(cx + d). = a cos(ax + b)sec(cx + d) + c sin(ax + b)tan(cx + d)sec(cx + d).

6. cos x³ · sin²(x⁵)

SOLUTION Product rule. d/dx(cos x³) = −3x² sin x³; d/dx(sin²x⁵) = 2 sin x⁵ cos x⁵ · 5x⁴ = 10x⁴ sin x⁵ cos x⁵. dy/dx = 10x⁴ sin x⁵ cos x⁵ cos x³ − 3x² sin x³ sin²(x⁵).

7. 2√(cot(x²))

SOLUTION Let y = 2(cot x²)^1/2. dy/dx = 2 · (1/2)(cot x²)^−1/2 · (−cosec²x²) · 2x = −2x cosec²x² / √(cot x²). Writing cosec²x² = 1/sin²x² and √(cot x²) = √(cos x²)/√(sin x²) and using sin 2x² = 2 sin x² cos x², this is the NCERT form dy/dx = −2√2 · x / (sin x² · √(sin 2x²)). (Equivalently −2x cosec²(x²)/√(cot x²).)

8. cos(√x)

SOLUTION d/dx cos(√x) = −sin(√x) · (1/(2√x)) = −sin(√x) / (2√x).

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

SOLUTION LHD at 1: lim_h→0⁻ [f(1 + h) − f(1)]/h = lim [|h| − 0]/h = lim (−h)/h = −1. RHD at 1: lim_h→0⁺ [|h|]/h = lim h/h = +1. LHD = −1 ≠ +1 = RHD, so the derivative does not exist. ∴ 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.

SOLUTION A function must be continuous to be differentiable. [x] jumps at every integer. At x = 1: LHL = [x] = 0, RHL = 1, so [x] is discontinuous at 1; hence not differentiable at x = 1. At x = 2: LHL = 1, RHL = 2, so [x] is discontinuous at 2; hence not differentiable at x = 2. ∴ f(x) = [x] is not differentiable at x = 1 and x = 2.

Exercise 5.3 — Implicit Functions

Find dy/dx in the following.

1. 2x + 3y = sin x

SOLUTION Differentiate: 2 + 3 dy/dx = cos x ⇒ dy/dx = (cos x − 2)/3.

2. 2x + 3y = sin y

SOLUTION 2 + 3 dy/dx = cos y dy/dx ⇒ 2 = (cos y − 3) dy/dx ⇒ dy/dx = 2/(cos y − 3).

3. ax + by² = cos y

SOLUTION a + 2by dy/dx = −sin y dy/dx ⇒ a = −(2by + sin y) dy/dx ⇒ dy/dx = −a/(2by + sin y).

4. xy + y² = tan x + y

SOLUTION y + x dy/dx + 2y dy/dx = sec²x + dy/dx. (x + 2y − 1) dy/dx = sec²x − y ⇒ dy/dx = (sec²x − y)/(x + 2y − 1).

5. x² + xy + y² = 100

SOLUTION 2x + (y + x dy/dx) + 2y dy/dx = 0 ⇒ (x + 2y) dy/dx = −(2x + y). dy/dx = −(2x + y)/(x + 2y).

6. x³ + x²y + xy² + y³ = 81

SOLUTION 3x² + (2xy + x²y′) + (y² + 2xy y′) + 3y²y′ = 0. (x² + 2xy + 3y²)y′ = −(3x² + 2xy + y²) ⇒ dy/dx = −(3x² + 2xy + y²)/(x² + 2xy + 3y²).

7. sin²y + cos xy = κ

SOLUTION 2 sin y cos y · y′ + (−sin xy)(y + x y′) = 0, i.e. sin 2y · y′ − sin xy(y + x y′) = 0. (sin 2y − x sin xy) y′ = y sin xy ⇒ dy/dx = (y sin xy)/(sin 2y − x sin xy).

8. sin²x + cos²y = 1

SOLUTION 2 sin x cos x + 2 cos y(−sin y)y′ = 0 ⇒ sin 2x − sin 2y · y′ = 0. dy/dx = (sin 2x)/(sin 2y).

9. y = sin⁻¹(2x/(1 + x²))

SOLUTION Put x = tan θ, so 2x/(1 + x²) = sin 2θ and y = sin⁻¹(sin 2θ) = 2θ = 2 tan⁻¹x. dy/dx = 2 · 1/(1 + x²) = 2/(1 + x²).

10. y = tan⁻¹((3x − x³)/(1 − 3x²)), −1/√3 < x < 1/√3

SOLUTION Put x = tan θ: (3 tanθ − tan³θ)/(1 − 3 tan²θ) = tan 3θ, so y = tan⁻¹(tan 3θ) = 3θ = 3 tan⁻¹x. dy/dx = 3 · 1/(1 + x²) = 3/(1 + x²).

11. y = cos⁻¹((1 − x²)/(1 + x²)), 0 < x < 1

SOLUTION Put x = tan θ (0 < θ < π/4): (1 − x²)/(1 + x²) = cos 2θ, so y = cos⁻¹(cos 2θ) = 2θ = 2 tan⁻¹x. dy/dx = 2/(1 + x²).

12. y = sin⁻¹((1 − x²)/(1 + x²)), 0 < x < 1

SOLUTION Put x = tan θ: (1 − x²)/(1 + x²) = cos 2θ = sin(π/2 − 2θ), so y = π/2 − 2θ = π/2 − 2 tan⁻¹x. dy/dx = −2 · 1/(1 + x²) = −2/(1 + x²).

13. y = cos⁻¹(2x/(1 + x²)), −1 < x < 1

SOLUTION Put x = tan θ: 2x/(1 + x²) = sin 2θ = cos(π/2 − 2θ), so y = π/2 − 2θ = π/2 − 2 tan⁻¹x. dy/dx = −2/(1 + x²).

14. y = sin⁻¹(2x√(1 − x²)), −1/√2 < x < 1/√2

SOLUTION Put x = sin θ: 2x√(1 − x²) = 2 sinθ cosθ = sin 2θ, so y = sin⁻¹(sin 2θ) = 2θ = 2 sin⁻¹x. dy/dx = 2 · 1/√(1 − x²) = 2/√(1 − x²).

15. y = sec⁻¹(1/(2x² − 1)), 0 < x < 1/√2

SOLUTION sec⁻¹(1/(2x² − 1)) = cos⁻¹(2x² − 1). Put x = cos θ: 2x² − 1 = cos 2θ, so y = cos⁻¹(cos 2θ) = 2θ = 2 cos⁻¹x. dy/dx = 2 · (−1/√(1 − x²)) = −2/√(1 − x²).

Exercise 5.4 — Exponential and Logarithmic Functions

Differentiate the following w.r.t. x.

1. e^x/sin x

SOLUTION Quotient rule: dy/dx = (e^x sin x − e^x cos x)/sin²x = e^x(sin x − cos x)/sin²x, x ≠ nπ, n ∈ Z.

2. e^sin⁻¹x

SOLUTION dy/dx = e^sin⁻¹x · d/dx(sin⁻¹x) = e^sin⁻¹x/√(1 − x²), x ∈ (−1, 1).

3. e^x³

SOLUTION dy/dx = e^x³ · 3x² = 3x² e^x³.

4. sin(tan⁻¹e^−x)

SOLUTION dy/dx = cos(tan⁻¹e^−x) · 1/(1 + e^−2x) · (−e^−x) = −e^−xcos(tan⁻¹e^−x)/(1 + e^−2x).

5. log(cos e^x)

SOLUTION dy/dx = (1/cos e^x) · (−sin e^x) · e^x = −e^x tan(e^x), e^x ≠ (2n + 1)π/2.

6. e^x + e^x² + … + e^x⁵

SOLUTION Differentiate term by term: dy/dx = e^x + 2x e^x² + 3x² e^x³ + 4x³ e^x⁴ + 5x⁴ e^x⁵.

7. √(e^√x), x > 0

SOLUTION y = √(e^√x) = e^√x/2. dy/dx = e^√x/2 · (1/2) · (1/(2√x)) = e^√x/2/(4√x). Since e^√x/2 = √(e^√x), this equals the NCERT form dy/dx = e^√x / (4√(x e^√x)), x > 0.

8. log(log x), x > 1

SOLUTION dy/dx = (1/log x) · (1/x) = 1/(x log x), x > 1.

9. cos x/log x, x > 0

SOLUTION Quotient rule: dy/dx = [(−sin x)(log x) − cos x(1/x)] / (log x)². = −(x sin x · log x + cos x)/(x(log x)²), x > 0.

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

SOLUTION dy/dx = −sin(log x + e^x) · (1/x + e^x) = −(1/x + e^x) sin(log x + e^x), x > 0.

Exercise 5.5 — Logarithmic Differentiation

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

1. cos x · cos 2x · cos 3x

SOLUTION Take log: log y = log cos x + log cos 2x + log cos 3x. (1/y)y′ = −tan x − 2 tan 2x − 3 tan 3x. dy/dx = −cos x cos 2x cos 3x [tan x + 2 tan 2x + 3 tan 3x].

2. √((x − 1)(x − 2)/((x − 3)(x − 4)(x − 5)))

SOLUTION log y = ½[log(x−1) + log(x−2) − log(x−3) − log(x−4) − log(x−5)]. (1/y)y′ = ½[1/(x−1) + 1/(x−2) − 1/(x−3) − 1/(x−4) − 1/(x−5)]. dy/dx = ½√((x−1)(x−2)/((x−3)(x−4)(x−5))) [1/(x−1) + 1/(x−2) − 1/(x−3) − 1/(x−4) − 1/(x−5)].

3. (log x)^{cos x}

SOLUTION log y = cos x · log(log x). (1/y)y′ = −sin x · log(log x) + cos x · (1/log x)(1/x). dy/dx = (log x)^{cos x} [cos x/(x log x) − sin x · log(log x)].

4. x^x − 2^{sin x}

SOLUTION For x^x: log u = x log x ⇒ u′ = x^x(1 + log x). For 2^{sin x}: derivative = 2^{sin x} log 2 · cos x. dy/dx = x^x(1 + log x) − 2^{sin x} cos x · log 2.

5. (x + 3)² · (x + 4)³ · (x + 5)⁴

SOLUTION log y = 2 log(x+3) + 3 log(x+4) + 4 log(x+5). (1/y)y′ = 2/(x+3) + 3/(x+4) + 4/(x+5). Simplifying with common factor (x+3)(x+4)²(x+5)³: dy/dx = (x+3)(x+4)²(x+5)³(9x² + 70x + 133).

6. (x + 1/x)^x + x^{(1 + 1/x)}

SOLUTION Let u = (x + 1/x)^x: log u = x log(x + 1/x). u′ = u[log(x + 1/x) + x · (1 − 1/x²)/(x + 1/x)] = (x + 1/x)^x[(x²−1)/(x²+1) + log(x + 1/x)]. Let v = x^{(1 + 1/x)}: log v = (1 + 1/x)log x. v′ = v[(1 + 1/x)/x − (log x)/x²] = x^(1+1/x)[(x + 1 − log x)/x²]. dy/dx = (x + 1/x)^x[(x² − 1)/(x² + 1) + log(x + 1/x)] + x^{(1 + 1/x)}[(x + 1 − log x)/x²].

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

SOLUTION u = (log x)^x: log u = x log(log x); u′ = (log x)^x[log(log x) + 1/log x]. v = x^{log x}: log v = (log x)²; v′ = x^{log x} · 2(log x)/x = 2 x^{log x − 1} log x. dy/dx = (log x)^x−1[1 + log x · log(log x)] + 2 x^{log x−1} log x.

8. (sin x)^x + sin⁻¹√x

SOLUTION u = (sin x)^x: log u = x log sin x; u′ = (sin x)^x[log sin x + x cot x]. d/dx sin⁻¹√x = 1/√(1 − x) · 1/(2√x) = 1/(2√(x − x²)). dy/dx = (sin x)^x(x cot x + log sin x) + 1/(2√(x − x²)).

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

SOLUTION u = x^{sin x}: log u = sin x log x; u′ = x^{sin x}[(sin x)/x + cos x log x]. v = (sin x)^{cos x}: log v = cos x log sin x; v′ = (sin x)^{cos x}[cos x cot x − sin x log sin x]. dy/dx = x^{sin x}[(sin x)/x + cos x log x] + (sin x)^{cos x}[cos x cot x − sin x log sin x].

10. x^{x cos x} + (x² + 1)/(x² − 1)

SOLUTION u = x^{x cos x}: log u = x cos x log x; u′ = x^{x cos x}[cos x(1 + log x) − x sin x log x]. d/dx[(x² + 1)/(x² − 1)] = [2x(x²−1) − (x²+1)2x]/(x²−1)² = −4x/(x² − 1)². dy/dx = x^{x cos x}[cos x(1 + log x) − x sin x log x] − 4x/(x² − 1)².

11. (x cos x)^x + (x sin x)^1/x

SOLUTION u = (x cos x)^x: log u = x log(x cos x); u′ = (x cos x)^x[log(x cos x) + x(1/x − tan x)] = (x cos x)^x[1 − x tan x + log(x cos x)]. v = (x sin x)^1/x: log v = (1/x)log(x sin x); v′ = (x sin x)^1/x[(1/x)(1/x + cot x) − (log(x sin x))/x²]. dy/dx = (x cos x)^x[1 − x tan x + log(x cos x)] + (x sin x)^1/x[(x cot x + 1 − log(x sin x))/x²].

Find dy/dx of the functions given in Exercises 12 to 15.

12. x^y + y^x = 1

SOLUTION d/dx(x^y) = x^y[(y/x) + log x · y′]; d/dx(y^x) = y^x[log y + (x/y)y′]. Summing to 0 and collecting y′: (x^y log x + x y^x−1) y′ = −(y x^y−1 + y^x log y). dy/dx = −(y x^y−1 + y^x log y)/(x^y log x + x y^x−1).

13. y^x = x^y

SOLUTION Take log: x log y = y log x. log y + (x/y)y′ = (y/x) + log x · y′ ⇒ (x/y − log x)y′ = y/x − log y. dy/dx = (y/x)(y − x log y)/(x − y log x).

14. (cos x)^y = (cos y)^x

SOLUTION Take log: y log cos x = x log cos y. y′ log cos x + y(−tan x) = log cos y + x(−tan y)y′. (log cos x + x tan y)y′ = log cos y + y tan x ⇒ dy/dx = (y tan x + log cos y)/(x tan y + log cos x).

15. x^y = e^{(x − y)}

SOLUTION Take log: y log x = x − y. y′ log x + y/x = 1 − y′ ⇒ (log x + 1)y′ = 1 − y/x = (x − y)/x. Since x − y = y log x, (x − y)/x = (y log x)/x, so dy/dx = (y log x)/(x(1 + log x)) = y(x − 1)/(x(y + 1)) using y log x = x − y ⇒ the book form dy/dx = y(x − 1)/[x(y + 1)].

16. Find the derivative of f(x) = (1 + x)(1 + x²)(1 + x⁴)(1 + x⁸) and hence find f′(1).

SOLUTION log f = log(1+x) + log(1+x²) + log(1+x⁴) + log(1+x⁸). f′(x) = (1+x)(1+x²)(1+x⁴)(1+x⁸)[1/(1+x) + 2x/(1+x²) + 4x³/(1+x⁴) + 8x⁷/(1+x⁸)]. At x = 1: product = 2·2·2·2 = 16; bracket = 1/2 + 2/2 + 4/2 + 8/2 = 1/2 + 1 + 2 + 4 = 15/2. f′(1) = 16 × 15/2 = 120.

17. Differentiate (x² − 5x + 8)(x³ + 7x + 9) in three ways: (i) product rule, (ii) expanding the product, (iii) logarithmic differentiation. Do they give the same answer?

SOLUTION (i) Product rule: (2x − 5)(x³ + 7x + 9) + (x² − 5x + 8)(3x² + 7). Expanding gives 5x⁴ − 20x³ + 45x² − 52x + 11. (ii) Expanding first: (x²−5x+8)(x³+7x+9) = x⁵ − 5x⁴ + 15x³ − 26x² + 11x + 72; its derivative = 5x⁴ − 20x³ + 45x² − 52x + 11. (iii) Logarithmic: (1/y)y′ = (2x−5)/(x²−5x+8) + (3x²+7)/(x³+7x+9); multiplying by y gives the same expression. ∴ all three methods give 5x⁴ − 20x³ + 45x² − 52x + 11 — yes, the answers agree.

18. If u, v and w are functions of x, then show that d/dx(u·v·w) = (du/dx)v·w + u(dv/dx)w + u·v(dw/dx), in two ways: first by repeated product rule, second by logarithmic differentiation.

SOLUTION Product rule: let p = uv. Then d/dx(pw) = p′w + pw′ = (u′v + uv′)w + uvw′ = u′vw + uv′w + uvw′. Logarithmic: log(uvw) = log u + log v + log w. Differentiating, (1/(uvw))(uvw)′ = u′/u + v′/v + w′/w. Multiply by uvw: (uvw)′ = u′vw + uv′w + uvw′. Both methods agree.

Exercise 5.6 — Parametric Forms

Without eliminating the parameter, find dy/dx (Exercises 1 to 10).

1. x = 2at², y = at⁴

SOLUTION dx/dt = 4at, dy/dt = 4at³. dy/dx = 4at³/(4at) = t².

2. x = a cos θ, y = b cos θ

SOLUTION dx/dθ = −a sin θ, dy/dθ = −b sin θ. dy/dx = (−b sinθ)/(−a sinθ) = b/a.

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

SOLUTION dx/dt = cos t, dy/dt = −2 sin 2t = −4 sin t cos t. dy/dx = −4 sin t cos t/cos t = −4 sin t.

4. x = 4t, y = 4/t

SOLUTION dx/dt = 4, dy/dt = −4/t². dy/dx = (−4/t²)/4 = −1/t².

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

SOLUTION dx/dθ = −sin θ + 2 sin 2θ, dy/dθ = cos θ − 2 cos 2θ. dy/dx = (cos θ − 2 cos 2θ)/(2 sin 2θ − sin θ).

6. x = a(θ − sin θ), y = a(1 + cos θ)

SOLUTION dx/dθ = a(1 − cos θ), dy/dθ = −a sin θ. dy/dx = −sin θ/(1 − cos θ) = −(2 sin(θ/2)cos(θ/2))/(2 sin²(θ/2)) = −cot(θ/2).

7. x = sin³t/√(cos 2t), y = cos³t/√(cos 2t)

SOLUTION Differentiating each (cos 2t = cos²t − sin²t) and simplifying gives dx/dt and dy/dt; the ratio reduces to dy/dx = −cot 3t.

8. x = a(cos t + log tan(t/2)), y = a sin t

SOLUTION dx/dt = a(−sin t + 1/sin t) = a cos²t/sin t; dy/dt = a cos t. dy/dx = a cos t · sin t/(a cos²t) = tan t.

9. x = a sec θ, y = b tan θ

SOLUTION dx/dθ = a sec θ tan θ, dy/dθ = b sec²θ. dy/dx = b sec²θ/(a sec θ tan θ) = (b/a)(sec θ/tan θ) = (b/a)cosec θ.

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

SOLUTION dx/dθ = a(−sinθ + sinθ + θ cosθ) = aθ cos θ; dy/dθ = a(cosθ − cosθ + θ sinθ) = aθ sin θ. dy/dx = aθ sinθ/(aθ cosθ) = tan θ.

11. If x = a^sin⁻¹t, y = a^cos⁻¹t, show that dy/dx = −y/x.

SOLUTION dx/dt = a^sin⁻¹t log a · 1/√(1−t²); dy/dt = a^cos⁻¹t log a · (−1/√(1−t²)). dy/dx = (dy/dt)/(dx/dt) = −a^cos⁻¹t/a^sin⁻¹t = −y/x. (Proved.)

Exercise 5.7 — Second Order Derivatives

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

1. x² + 3x + 2

SOLUTION y′ = 2x + 3; y″ = 2.

2. x²⁰

SOLUTION y′ = 20x¹⁹; y″ = 20·19 x¹⁸ = 380 x¹⁸.

3. x · cos x

SOLUTION y′ = cos x − x sin x; y″ = −sin x − (sin x + x cos x) = −x cos x − 2 sin x.

4. log x

SOLUTION y′ = 1/x; y″ = −1/x².

5. x³ log x

SOLUTION y′ = 3x² log x + x³(1/x) = 3x² log x + x². y″ = 6x log x + 3x²(1/x) + 2x = 6x log x + 3x + 2x = x(5 + 6 log x).

6. e^x sin 5x

SOLUTION y′ = e^x(sin 5x + 5 cos 5x). y″ = e^x(sin 5x + 5 cos 5x) + e^x(5 cos 5x − 25 sin 5x) = 2e^x(5 cos 5x − 12 sin 5x).

7. e^6x cos 3x

SOLUTION y′ = e^6x(6 cos 3x − 3 sin 3x). y″ = e^6x[6(6 cos 3x − 3 sin 3x) + (−18 sin 3x − 9 cos 3x)] = 9e^6x(3 cos 3x − 4 sin 3x).

8. tan⁻¹x

SOLUTION y′ = 1/(1 + x²); y″ = −(2x)/(1 + x²)² = −2x/(1 + x²)².

9. log(log x)

SOLUTION y′ = 1/(x log x). y″ = −[d/dx(x log x)]/(x log x)² = −(log x + 1)/(x log x)². y″ = −(1 + log x)/(x log x)².

10. sin(log x)

SOLUTION y′ = cos(log x)/x. y″ = [−sin(log x)(1/x)·x − cos(log x)]/x². y″ = −[sin(log x) + cos(log x)]/x².

11. If y = 5 cos x − 3 sin x, prove that d²y/dx² + y = 0.

SOLUTION y′ = −5 sin x − 3 cos x; y″ = −5 cos x + 3 sin x = −(5 cos x − 3 sin x) = −y. ∴ y″ + y = 0. (Proved.)

12. If y = cos⁻¹x, find d²y/dx² in terms of y alone.

SOLUTION y′ = −1/√(1 − x²) = −(1 − x²)^−1/2. y″ = −(−1/2)(1 − x²)^−3/2(−2x) = −x/(1 − x²)^3/2. Since x = cos y and 1 − x² = sin²y: y″ = −cos y/(sin²y)^3/2 = −cot y cosec²y.

13. If y = 3 cos(log x) + 4 sin(log x), show that x²y₂ + xy₁ + y = 0.

SOLUTION y₁ = [−3 sin(log x) + 4 cos(log x)]/x ⇒ xy₁ = −3 sin(log x) + 4 cos(log x). Differentiate again: y₁ + xy₂ = [−3 cos(log x) − 4 sin(log x)]/x = −y/x. Multiply by x: xy₁ + x²y₂ = −y ⇒ x²y₂ + xy₁ + y = 0. (Proved.)

14. If y = Ae^mx + Be^nx, show that d²y/dx² − (m + n)dy/dx + mny = 0.

SOLUTION y′ = Am e^mx + Bn e^nx; y″ = Am²e^mx + Bn²e^nx. y″ − (m+n)y′ + mny = Ae^mx(m² − (m+n)m + mn) + Be^nx(n² − (m+n)n + mn) = 0 + 0 = 0. (Proved.)

15. If y = 500e^7x + 600e^−7x, show that d²y/dx² = 49y.

SOLUTION y′ = 3500e^7x − 4200e^−7x; y″ = 24500e^7x + 29400e^−7x. = 49(500e^7x + 600e^−7x) = 49y. (Proved.)

16. If e^y(x + 1) = 1, show that d²y/dx² = (dy/dx)².

SOLUTION e^y = 1/(x + 1) ⇒ y = −log(x + 1). So y′ = −1/(x + 1). y″ = 1/(x + 1)² = [−1/(x + 1)]² = (y′)². (Proved.)

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

SOLUTION y₁ = 2 tan⁻¹x · 1/(1 + x²) ⇒ (1 + x²)y₁ = 2 tan⁻¹x. Differentiate: 2x y₁ + (1 + x²)y₂ = 2/(1 + x²). Multiply by (1 + x²): (x² + 1)²y₂ + 2x(x² + 1)y₁ = 2. (Proved.)

Miscellaneous Exercise on Chapter 5

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

1. (3x² − 9x + 5)⁹

SOLUTION dy/dx = 9(3x² − 9x + 5)⁸(6x − 9) = 27(3x² − 9x + 5)⁸(2x − 3).

2. sin³x + cos⁶x

SOLUTION dy/dx = 3 sin²x cos x + 6 cos⁵x(−sin x) = 3 sin x cos x(sin x − 2 cos⁴x). = 3 sin x cos x(sin x − 2 cos⁴x).

3. (5x)^{3 cos 2x}

SOLUTION log y = 3 cos 2x · log 5x. (1/y)y′ = −6 sin 2x log 5x + 3 cos 2x · (1/x). dy/dx = (5x)^{3 cos 2x}[3 cos 2x/x − 6 sin 2x · log 5x].

4. sin⁻¹(x√x), 0 ≤ x ≤ 1

SOLUTION Let u = x√x = x^3/2; du/dx = (3/2)x^1/2. dy/dx = (1/√(1 − x³)) · (3/2)√x = 3√x/(2√(1 − x³)).

5. cos⁻¹(x/2)/√(2x + 7), −2 < x < 2

SOLUTION Quotient rule with u = cos⁻¹(x/2), u′ = −1/(2√(1 − x²/4)) = −1/√(4 − x²); v = (2x + 7)^1/2, v′ = 1/√(2x + 7). dy/dx = −1/(√(4 − x²)√(2x + 7)) − cos⁻¹(x/2)/(2x + 7)^3/2.

6. cot⁻¹[(√(1 + sin x) + √(1 − sin x))/(√(1 + sin x) − √(1 − sin x))], 0 < x < π/2

SOLUTION With 1 ± sin x = (cos(x/2) ± sin(x/2))² and 0 < x < π/2, the bracket simplifies to cot(x/2), so y = cot⁻¹(cot(x/2)) = x/2. dy/dx = 1/2.

7. (log x)^{log x}, x > 1

SOLUTION log y = log x · log(log x). (1/y)y′ = (1/x)log(log x) + log x · (1/(x log x)) = (1/x)[log(log x) + 1]. dy/dx = (log x)^{log x}[1/x + log(log x)/x], x > 1.

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

SOLUTION dy/dx = −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).

9. (sin x − cos x)^{(sin x − cos x)}, π/4 < x < 3π/4

SOLUTION log y = (sin x − cos x)log(sin x − cos x). (1/y)y′ = (cos x + sin x)[1 + log(sin x − cos x)]. dy/dx = (sin x − cos x)^{(sin x − cos x)}(cos x + sin x)[1 + log(sin x − cos x)], sin x > cos x.

10. x^x + x^a + a^x + a^a, for fixed a > 0 and x > 0.

SOLUTION d/dx(x^x) = x^x(1 + log x); d/dx(x^a) = a x^a−1; d/dx(a^x) = a^x log a; d/dx(a^a) = 0. dy/dx = x^x(1 + log x) + a x^a−1 + a^x log a.

11. x^{(x² − 3)} + (x − 3)^x², for x > 3

SOLUTION u = x^(x²−3): log u = (x²−3)log x; u′ = x^(x²−3)[(x²−3)/x + 2x log x]. v = (x−3)^x²: log v = x² log(x−3); v′ = (x−3)^x²[x²/(x−3) + 2x log(x−3)]. dy/dx = x^(x²−3)[(x² − 3)/x + 2x log x] + (x − 3)^x²[x²/(x − 3) + 2x log(x − 3)].

12. Find dy/dx, if y = 12(1 − cos t), x = 10(t − sin t), −π/2 < t < π/2.

SOLUTION dy/dt = 12 sin t; dx/dt = 10(1 − cos t). dy/dx = 12 sin t/(10(1 − cos t)) = (6/5) · (2 sin(t/2)cos(t/2))/(2 sin²(t/2)) = (6/5)cot(t/2).

13. Find dy/dx, if y = sin⁻¹x + sin⁻¹√(1 − x²), 0 < x < 1.

SOLUTION For 0 < x < 1, put x = sinθ (0 < θ < π/2): √(1 − x²) = cosθ, and sin⁻¹(cosθ) = π/2 − θ. So y = θ + (π/2 − θ) = π/2, a constant. dy/dx = 0.

14. If x√(1 + y) + y√(1 + x) = 0, for −1 < x < 1, prove that dy/dx = −1/(1 + x)².

SOLUTION From x√(1 + y) = −y√(1 + x), squaring: x²(1 + y) = y²(1 + x) ⇒ x² − y² + x²y − xy² = 0. (x − y)(x + y) + xy(x − y) = 0 ⇒ (x − y)(x + y + xy) = 0. Since x ≠ y, x + y + xy = 0 ⇒ y = −x/(1 + x). dy/dx = −[(1 + x) − x]/(1 + x)² = −1/(1 + x)². (Proved.)

15. If (x − a)² + (y − b)² = c², for some c > 0, prove that [1 + (dy/dx)²]^3/2/(d²y/dx²) is a constant independent of a and b.

SOLUTION Differentiate: 2(x − a) + 2(y − b)y′ = 0 ⇒ y′ = −(x − a)/(y − b). Differentiating again: y″ = −[(y − b) − (x − a)y′]/(y − b)². Substituting y′ gives y″ = −c²/(y − b)³. Also 1 + (y′)² = c²/(y − b)², so [1 + (y′)²]^3/2 = c³/|y − b|³. Ratio = (c³/|y − b|³)/(−c²/(y − b)³) = −c, a constant independent of a and b. (Proved.)

16. If cos y = x cos(a + y), with cos a ≠ ±1, prove that dy/dx = cos²(a + y)/sin a.

SOLUTION From cos y = x cos(a + y), x = cos y/cos(a + y). dx/dy = [−sin y cos(a + y) + cos y sin(a + y)]/cos²(a + y) = sin(a + y − y)/cos²(a + y) = sin a/cos²(a + y). ∴ dy/dx = 1/(dx/dy) = cos²(a + y)/sin a. (Proved.)

17. If x = a(cos t + t sin t) and y = a(sin t − t cos t), find d²y/dx².

SOLUTION dx/dt = a(−sin t + sin t + t cos t) = at cos t; dy/dt = a(cos t − cos t + t sin t) = at sin t. dy/dx = at sin t/(at cos t) = tan t. Then d²y/dx² = d(tan t)/dx = sec²t · dt/dx = sec²t/(at cos t). d²y/dx² = sec³t/(at), 0 < t < π/2.

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

SOLUTION For x ≥ 0: f(x) = x³, f′(x) = 3x², f″(x) = 6x. For x < 0: f(x) = −x³, f′(x) = −3x², f″(x) = −6x. At x = 0 both one-sided values of f′ and f″ agree (= 0), so f″ exists everywhere. f″(x) = 6x for x ≥ 0 and −6x for x < 0, i.e. f″(x) = 6|x|.

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

SOLUTION Treat B as constant and differentiate both sides of sin(A + B) = sin A cos B + cos A sin B w.r.t. A. LHS: cos(A + B); RHS: cos A cos B − sin A sin B. ∴ cos(A + B) = cos A cos B − sin A sin B. (Derived.)

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

SOLUTION Yes. Consider f(x) = |x| + |x − 1|. It is a sum of continuous functions, hence continuous everywhere. It has sharp corners at x = 0 and x = 1, where the left and right derivatives differ, so it is not differentiable at exactly those two points. ∴ such a function exists.

21. If y = |f(x) g(x) h(x) ; l m n ; a b c| (a 3×3 determinant with rows [f g h], [l m n], [a b c]), prove that dy/dx = |f′ g′ h′ ; l m n ; a b c|.

SOLUTION Expand y along the first row: y = f(mc − nb) − g(lc − na) + h(lb − ma), where l, m, n, a, b, c are constants. Differentiate w.r.t. x: dy/dx = f′(mc − nb) − g′(lc − na) + h′(lb − ma). This is exactly the determinant with the first row replaced by [f′ g′ h′]. (Proved.)

22. If y = e^{a cos⁻¹x}, −1 ≤ x ≤ 1, show that (1 − x²)d²y/dx² − x dy/dx − a²y = 0.

SOLUTION y′ = e^{a cos⁻¹x} · (−a/√(1 − x²)) = −ay/√(1 − x²) ⇒ √(1 − x²)y′ = −ay. Square: (1 − x²)(y′)² = a²y². Differentiate: (1 − x²)2y′y″ − 2x(y′)² = 2a²y y′. Divide by 2y′: (1 − x²)y″ − x y′ − a²y = 0. (Proved.)

Common Mistakes to Avoid

Watch out for these

For piecewise functions, forgetting to check the value f(c) as well as both one-sided limits at the join — all three must be equal for continuity.
Confusing continuity with differentiability — |x| is continuous at 0 but not differentiable there; continuity does not imply differentiability.
Dropping the inner derivative in the chain rule (e.g. writing d/dx sin(x²) = cos(x²) instead of 2x cos(x²)).
In logarithmic differentiation, forgetting to multiply the bracket by y at the end, or applying it where the base can be non-positive.
In implicit differentiation, omitting dy/dx when differentiating terms containing y.
For inverse-trig simplifications, ignoring the domain restriction that justifies sin⁻¹(sinθ) = θ or cos⁻¹(cosθ) = θ.
In parametric problems, computing dy/dx as (dx/dt)/(dy/dt) instead of (dy/dt)/(dx/dt).

Practice MCQs & Assertion–Reason

1. The function f(x) = [x] (greatest integer function) is discontinuous at:

(a) all real numbers (b) all integers (c) x = 0 only (d) no point

2. d/dx(sin(x² + 5)) equals:

(a) cos(x² + 5) (b) 2x cos(x² + 5) (c) 2x sin(x² + 5) (d) −2x cos(x² + 5)

3. If f is differentiable at a point c, then f is:

(a) discontinuous at c (b) continuous at c (c) constant near c (d) undefined at c

4. d/dx(tan⁻¹x) is:

(a) 1/√(1 − x²) (b) 1/(1 + x²) (c) −1/(1 + x²) (d) 1/(1 − x²)

5. d/dx(a^x), a > 0, equals:

(a) x a^x−1 (b) a^x (c) a^x log a (d) a^x/log a

6. For x = at², y = 2at, dy/dx equals:

(a) t (b) 1/t (c) 2t (d) 1/(2t)

7. If y = sin⁻¹(2x/(1 + x²)), then dy/dx is:

(a) 1/(1 + x²) (b) 2/(1 + x²) (c) 2/√(1 − x²) (d) 1/√(1 − x²)

8. The function f(x) = |x| is not differentiable at:

(a) x = 1 (b) x = −1 (c) x = 0 (d) every point

9. The second order derivative of y = log x is:

(a) 1/x (b) −1/x² (c) 1/x² (d) −1/x

10. The value of k making f(x) = kx² (x ≤ 2), 3 (x > 2) continuous at x = 2 is:

(a) 3 (b) 3/4 (c) 4/3 (d) 6

Answer key: 1-(b), 2-(b), 3-(b), 4-(b), 5-(c), 6-(b), 7-(b), 8-(c), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The function f(x) = |x| is continuous at x = 0.

Reason: Every continuous function is differentiable.

A-R 2. Assertion: If a function is differentiable at a point, it is continuous there.

Reason: Differentiability at c implies the limit defining f′(c) exists, which forces lim_x→c f(x) = f(c).

A-R 3. Assertion: d/dx(sin x · cos x) exists for all x ∈ R.

Reason: The product of two functions continuous and differentiable on R is differentiable on R.

A-R 4. Assertion: The greatest integer function [x] is differentiable at x = 2.

Reason: [x] is discontinuous at every integer.

A-R 5. Assertion: If y = e^x, then dy/dx = e^x.

Reason: The natural exponential function is its own derivative.

Answer key: 1-(C), 2-(A), 3-(A), 4-(D), 5-(A).

Quick Revision Summary

f is continuous at c if LHL = RHL = f(c); polynomial, rational (where defined), trig, exp and log functions are continuous on their domains.
Every differentiable function is continuous; the converse is false (|x| at 0).
Chain rule: df/dx = (df/dt)(dt/dx); product and quotient rules extend differentiation to combinations.
Standard derivatives: sin⁻¹x → 1/√(1−x²), cos⁻¹x → −1/√(1−x²), tan⁻¹x → 1/(1+x²), e^x → e^x, log x → 1/x.
Logarithmic differentiation handles y = [u(x)]^v(x) (both u, y > 0).
For x = f(t), y = g(t): dy/dx = (dy/dt)/(dx/dt), provided dx/dt ≠ 0.
Second-order derivative y″ = d/dx(dy/dx); used to verify differential equations.

How to score full marks in this chapter

Always test continuity with the three-part check (LHL, RHL, f(c)) and state the rule you use. In differentiation, name the technique — chain, product, quotient, implicit, logarithmic or parametric — and never drop the inner derivative. For inverse-trig problems, substitute x = tanθ or x = sinθ and quote the domain that makes the simplification valid. In second-order-derivative proofs, work the relation step by step and substitute back to reach the required differential equation. Keep each line tidy so every step earns its mark.

Frequently Asked Questions

What is Class 12 Maths Chapter 5 Continuity and Differentiability about?

Chapter 5 covers continuity at a point and the algebra of continuous functions, differentiability and its link with continuity, the chain rule, differentiation of implicit, inverse-trigonometric, exponential and logarithmic functions, logarithmic differentiation, parametric forms and second-order derivatives.

How many exercises are there in Class 12 Maths Chapter 5?

There are seven numbered exercises (5.1 to 5.7) plus a Miscellaneous Exercise on Chapter 5. Every question of each set is solved step by step on this page.

Is every continuous function differentiable?

No. Every differentiable function is continuous, but the converse is false. For example, f(x) = |x| is continuous at x = 0 yet not differentiable there, because the left and right derivatives are −1 and +1.

Are these Class 12 Maths Chapter 5 solutions free?

Yes. All solutions are free and follow the official NCERT Mathematics Part I textbook for the 2026–27 session, with answers verified against the book’s answer key.

← Chapter 4: Determinants Class 12 Maths Chapter 6: Application of Derivatives →