1. Chapter at a Glance
A function is continuous if its graph can be drawn without lifting the pencil — no jumps, no holes. It is differentiable if, in addition, it has no sharp corners, so a unique tangent exists at every point. This chapter makes these ideas precise and then builds a complete toolkit for differentiation: the chain rule, derivatives of implicit and inverse trigonometric functions, exponential and logarithmic functions, logarithmic differentiation, parametric forms and second-order derivatives. It is one of the most heavily weighted chapters of Class 12 (Calculus carries about 35 marks in CBSE).
- Continuity at a point and on an interval; algebra of continuous functions.
- Differentiability; every differentiable function is continuous (not conversely).
- Chain rule; implicit differentiation.
- Derivatives of inverse trigonometric functions.
- Exponential and logarithmic functions and their derivatives.
- Logarithmic differentiation; parametric differentiation; second-order derivatives.
2. Continuity
f is continuous at x = c if lim (x→c) f(x) = f(c), i.e. LHL = RHL = f(c). f is continuous on its domain if it is continuous at every point.
- Polynomials, sin x, cos x, eˣ, |x| are continuous everywhere; rational functions, tan x, log x are continuous on their domains.
- Sums, differences, products, quotients (denominator ≠ 0) and compositions of continuous functions are continuous.
Find k so that f(x) = kx + 1 for x ≤ 5 and 3x − 5 for x > 5 is continuous at x = 5.
Solution: 5k + 1 = 15 − 5 = 10 → k = 9/5.
Is f(x) = (sin x)/x for x ≠ 0 and f(0) = 1 continuous at 0? What if f(0) = 2?
Solution: lim (sin x)/x = 1. With f(0) = 1 → continuous; with f(0) = 2 → not continuous.
3. Differentiability
f is differentiable at c if the limit lim (h→0) [f(c + h) − f(c)]/h exists (left and right derivatives are equal).
If f is differentiable at c, then f is continuous at c. The converse is false: |x| is continuous at 0, but the left derivative (−1) ≠ right derivative (+1), so it is not differentiable there.
4. Chain Rule and Implicit Differentiation
If y = f(u) and u = g(x), then dy/dx = (dy/du)(du/dx). "Differentiate the outside, keep the inside, multiply by the derivative of the inside."
Differentiate (a) sin(x²) (b) (2x + 3)⁵ (c) √(cos x).
Solution: (a) 2x cos(x²). (b) 10(2x + 3)⁴. (c) −sin x / (2√(cos x)).
Implicit differentiation: when y is not isolated, differentiate both sides with respect to x, treating y as a function of x (so d/dx of y² is 2y dy/dx), then solve for dy/dx.
Find dy/dx if x² + xy + y² = 100.
Solution: 2x + (y + x dy/dx) + 2y dy/dx = 0 → dy/dx = −(2x + y)/(x + 2y).
5. Inverse Trigonometric Functions
| f(x) | f′(x) | Valid for |
|---|---|---|
| sin⁻¹x | 1/√(1 − x²) | −1 < x < 1 |
| cos⁻¹x | −1/√(1 − x²) | −1 < x < 1 |
| tan⁻¹x | 1/(1 + x²) | all x |
| cot⁻¹x | −1/(1 + x²) | all x |
| sec⁻¹x | 1/(|x|√(x² − 1)) | |x| > 1 |
| cosec⁻¹x | −1/(|x|√(x² − 1)) | |x| > 1 |
Derivation of the first: y = sin⁻¹x ⇒ sin y = x ⇒ cos y · dy/dx = 1 ⇒ dy/dx = 1/cos y = 1/√(1 − x²) (cos y ≥ 0 on the branch).
Differentiate y = sin⁻¹(2x√(1 − x²)), −1/√2 < x < 1/√2.
Solution: Put x = sin θ, θ ∈ (−π/4, π/4): 2x√(1 − x²) = sin 2θ and 2θ ∈ (−π/2, π/2), so y = 2θ = 2 sin⁻¹x → dy/dx = 2/√(1 − x²).
6. Exponential and Logarithmic Functions
The number e ≈ 2.71828 is special: the function eˣ is its own derivative. The natural logarithm log x (= logₑ x) is the inverse of eˣ, defined for x > 0.
| f(x) | f′(x) |
|---|---|
| eˣ | eˣ |
| log x | 1/x (x > 0) |
| aˣ | aˣ log a |
| logₐ x | 1/(x log a) |
log(mn) = log m + log n; log(m/n) = log m − log n; log(mⁿ) = n log m; logₐ x = log x / log a.
7. Logarithmic, Parametric and Second-order Derivatives
Logarithmic differentiation is used for products/quotients of many factors and for powers like u(x)^v(x): take log of both sides, then differentiate.
Differentiate y = xˣ (x > 0).
Solution: log y = x log x → (1/y) dy/dx = log x + 1 → dy/dx = xˣ(1 + log x).
If x = f(t), y = g(t): dy/dx = (dy/dt)/(dx/dt), dx/dt ≠ 0.
Find dy/dx if x = a cos θ, y = a sin θ.
Solution: dy/dx = (a cos θ)/(−a sin θ) = −cot θ.
d²y/dx² = d/dx (dy/dx), also written y″ or y₂.
If y = A sin x + B cos x, show that d²y/dx² + y = 0.
Solution: y′ = A cos x − B sin x; y″ = −A sin x − B cos x = −y. ∎
8. Solved Problems (Board Pattern)
f(x) = 5 for x ≤ 2, ax + b for 2 < x < 10, 21 for x ≥ 10. Find a, b for continuity.
Solution: 2a + b = 5 and 10a + b = 21 → a = 2, b = 1.
Differentiate y = (log x)^(cos x).
Solution: log y = cos x · log(log x) → y′/y = −sin x log(log x) + cos x/(x log x). y′ = (log x)^(cos x) [cos x/(x log x) − sin x log(log x)].
If y = 3e²ˣ + 2e³ˣ, prove y″ − 5y′ + 6y = 0.
Solution: y′ = 6e²ˣ + 6e³ˣ; y″ = 12e²ˣ + 18e³ˣ. y″ − 5y′ + 6y = (12 − 30 + 18)e²ˣ + (18 − 30 + 12)e³ˣ = 0. ∎
9. Master Formula Table
| Rule | Formula |
|---|---|
| Continuity at c | lim (x→c) f(x) = f(c) |
| Chain rule | dy/dx = (dy/du)(du/dx) |
| sin⁻¹x | 1/√(1 − x²) |
| tan⁻¹x | 1/(1 + x²) |
| eˣ, aˣ | eˣ, aˣ log a |
| log x | 1/x |
| u^v | u^v [v′ log u + v u′/u] |
| Parametric | dy/dx = (dy/dt)/(dx/dt) |
| Second derivative | d²y/dx² = d/dx(dy/dx) |
10. Master Summary Table
| Situation | Technique |
|---|---|
| Function of a function | Chain rule |
| y mixed with x | Implicit differentiation |
| Inverse trig of algebraic expression | Trig substitution then simplify |
| Variable power or many factors | Logarithmic differentiation |
| x and y in terms of t | Parametric |
| Prove a relation with y″ | Differentiate twice, substitute |
11. Mind Map
- LHL = RHL = f(c)
- Find k
- ⇒ continuity
- |x| counterexample
- Chain, implicit
- Log, parametric
- Inverse trig
- eˣ, log x, y″
12. Tips and Tricks
- For inverse trig of expressions like (1 − x²)/(1 + x²), try x = tan θ; for 2x√(1 − x²), try x = sin θ.
- Whenever the exponent contains x, take logs.
- In parametric questions, simplify dy/dt and dx/dt before dividing.
- Forgetting the inner derivative in the chain rule.
- d/dx(aˣ) = x aˣ⁻¹ — wrong; that rule is for xⁿ.
- Treating y as a constant in implicit differentiation.
13. Important One-liners
- Every differentiable function is continuous; |x| shows the converse fails.
- d/dx(eˣ) = eˣ.
- d/dx(log x) = 1/x.
- d/dx(tan⁻¹x) = 1/(1 + x²).
- Polynomials are continuous and differentiable everywhere.
14. Exam-pattern Questions with Detailed Answers
The derivative of e^(sin x) is:
- e^(sin x)
- cos x · e^(sin x)
- sin x · e^(sin x)
- e^(cos x)
Show answer
Chain rule: e^(sin x) × cos x.
Answer: (B)
Differentiate log(sin x).
Show answer
cos x / sin x = cot x.
Answer: cot x
Is f(x) = |x − 3| continuous at x = 3?
Show answer
LHL = RHL = f(3) = 0. Yes (though not differentiable there).
Answer: Yes
Find k if f(x) = (1 − cos 4x)/(8x²) for x ≠ 0 and f(0) = k is continuous at 0.
Show answer
(1 − cos 4x) = 2sin²2x → 2sin²2x/(8x²) = ¼ × 4 × (sin 2x/2x)² → 1. k = 1.
Answer: k = 1
Find dy/dx if sin²y + cos xy = π.
Show answer
2 sin y cos y y′ − sin xy (y + xy′) = 0 → y′ (sin 2y − x sin xy) = y sin xy → y′ = y sin xy/(sin 2y − x sin xy).
Answer: y sin(xy)/(sin 2y − x sin(xy))
Differentiate tan⁻¹[(3x − x³)/(1 − 3x²)], −1/√3 < x < 1/√3.
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x = tan θ: expression = tan 3θ with 3θ ∈ (−π/2, π/2) → y = 3 tan⁻¹x → 3/(1 + x²).
Answer: 3/(1 + x²)
Find dy/dx if x = a(θ − sin θ), y = a(1 + cos θ).
Show answer
dx/dθ = a(1 − cos θ), dy/dθ = −a sin θ. dy/dx = −sin θ/(1 − cos θ) = −cot(θ/2).
Answer: −cot(θ/2)
Differentiate (sin x)ˣ.
Show answer
log y = x log sin x → y′/y = log sin x + x cot x → y′ = (sin x)ˣ (log sin x + x cot x).
Answer: (sin x)ˣ(log sin x + x cot x)
If y = e^(a cos⁻¹x), −1 ≤ x ≤ 1, show that (1 − x²)y″ − xy′ − a²y = 0.
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y′ = −a y/√(1 − x²) → √(1 − x²) y′ = −a y. Square: (1 − x²)y′² = a²y². Differentiate: (1 − x²)2y′y″ − 2xy′² = 2a²yy′. Divide by 2y′: (1 − x²)y″ − xy′ − a²y = 0. ∎
Answer: Proved
If xʸ = eˣ⁻ʸ, prove dy/dx = log x/(1 + log x)².
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y log x = x − y → y = x/(1 + log x). dy/dx = [(1 + log x) − x(1/x)]/(1 + log x)² = log x/(1 + log x)². ∎
Answer: Proved
Show that f(x) = |x − 1| + |x − 2| is continuous everywhere but not differentiable at x = 1 and x = 2.
Show answer
Sum of continuous functions → continuous. At x = 1: left derivative −2, right derivative 0 → not differentiable. At x = 2: left 0, right 2 → not differentiable.
Answer: Proved
If y = (tan⁻¹x)², show that (x² + 1)²y″ + 2x(x² + 1)y′ = 2.
Show answer
y′ = 2 tan⁻¹x/(1 + x²) → (1 + x²)y′ = 2 tan⁻¹x. Differentiate: (1 + x²)y″ + 2xy′ = 2/(1 + x²). Multiply by (1 + x²). ∎
Answer: Proved
15. Quick Revision Summary
- Continuous at c ⇔ LHL = RHL = f(c); differentiable ⇒ continuous.
- Chain rule for compositions; implicit differentiation for mixed x, y.
- Inverse trig derivatives; use trig substitutions to simplify first.
- eˣ → eˣ, log x → 1/x, aˣ → aˣ log a; take logs for variable powers.
- Parametric: dy/dx = (dy/dt)/(dx/dt); second derivatives for proofs.