What is the difference between a function being differentiable and continuous?

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CBSE Class 12 Mathematics★ 8-10 marks · CBSE Class 12 Mathematics Chapter 5

Continuity and Differentiability

A function is CONTINUOUS if you can draw it without lifting your pen. It is DIFFERENTIABLE if it has a derivative at a point. This chapter deepens the calculus from Class 11 — with chain rule, implicit functions, logarithmic differentiation, and second-order derivatives. CBSE Class 12 Mathematics Chapter 5.

⏱ 24 min readHard difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Define continuity at a point and on an interval; check continuity using the three-condition definition
2Apply the chain rule for derivatives of composite functions
3Differentiate implicit functions and parametric functions
4Apply logarithmic differentiation for products/powers
5State and apply Rolle's Theorem and the Mean Value Theorem (MVT)

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Why this chapter matters

Continuity and Differentiability is the theoretical backbone of all calculus. Derivatives of composite functions (chain rule), implicit differentiation, logarithmic differentiation, and the mean value theorem are all tested. Rolle's Theorem and MVT are reliable short-answer questions every year.

Continuity and Differentiability

"You can't differentiate a function that isn't continuous. But a continuous function isn't always differentiable."

1. Chapter Overview

This chapter formalises the concepts of CONTINUITY (a function is continuous at a point if limit = function value: lim(x→a) f(x) = f(a)) and DIFFERENTIABILITY (the derivative EXISTS at a point). It introduces POWERFUL differentiation techniques: chain rule, implicit differentiation, logarithmic differentiation, parametric forms, and second-order derivatives. Rolle's Theorem and the Mean Value Theorem provide the theoretical foundation.

2. Continuity

A function f is CONTINUOUS at x = a if: lim(x→a⁻) f(x) = lim(x→a⁺) f(x) = f(a)
'You can draw the graph without lifting your pen.'
Algebra of continuous functions: sum, difference, product, quotient (where denominator ≠ 0) of continuous functions are continuous.
Every polynomial, trigonometric, exponential, and logarithmic function is continuous on its domain.

3. Differentiability

f is DIFFERENTIABLE at x = a if the limit lim(h→0) [f(a+h) — f(a)]/h EXISTS (both left and right must be equal)
Relationship: Differentiability ⇒ Continuity. BUT Continuity ⇏ Differentiability. 'Every differentiable function is continuous. But a continuous function may not be differentiable — e.g., |x| at x = 0 (the "corner").'

4. Differentiation Techniques

Chain Rule

If y = f(u) and u = g(x): dy/dx = (dy/du) × (du/dx). 'Differentiate the outer function, multiply by the derivative of the inner function.'

Implicit Differentiation

When the relationship between x and y is given IMPLICITLY (f(x,y) = 0): differentiate both sides w.r.t. x, treating y as a function of x. Solve for dy/dx.

Logarithmic Differentiation

Take LOG of both sides. Useful when: function is of the form [f(x)]ᵍ⁽ˣ⁾, product/quotient of many terms. Simplify first, then differentiate.

Parametric Forms

x = f(t), y = g(t). dy/dx = (dy/dt) / (dx/dt).

Second-Order Derivatives

d²y/dx² = d/dx (dy/dx). 'The derivative of the derivative.'

5. Rolle's Theorem and Mean Value Theorem (MVT)

Rolle's Theorem

If f is: (a) continuous on [a,b], (b) differentiable on (a,b), (c) f(a) = f(b) → there exists at least one c ∈ (a,b) such that f'(c) = 0. 'The derivative is zero somewhere between two points with equal function values.'

Mean Value Theorem (MVT)

If f is: (a) continuous on [a,b], (b) differentiable on (a,b) → there exists at least one c ∈ (a,b) such that f'(c) = [f(b) — f(a)]/(b — a). 'The instantaneous rate of change equals the average rate of change — somewhere.'

6. Exam Focus

Continuity — definition. Algebra of continuous functions.
Differentiability — definition. Continuity vs. differentiability.
Chain rule. Implicit. Logarithmic. Parametric differentiation.
Second-order derivatives.
Rolle's Theorem and MVT — conditions and statement.

7. Conclusion

Continuity and differentiability are the FOUNDATIONS on which all of calculus rests:

CONTINUITY: No sudden jumps. The graph holds together.
DIFFERENTIABILITY: No sharp corners. The slope exists and is well-behaved.
TECHNIQUES: Chain rule, implicit, logarithmic, parametric — powerful tools for any function.
THEOREMS: Rolle's and MVT — 'somewhere between the start and the end, the derivative hits the average.'

'Calculus is the study of change. Continuity and differentiability are the conditions under which change can be analysed.'

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Continuity at a Point — Three Conditions

f is continuous at x = a if: (1) f(a) is DEFINED. (2) lim(x→a) f(x) EXISTS (left-hand limit = right-hand limit). (3) lim(x→a) f(x) = f(a). If any condition fails → f is discontinuous at a. LHL = lim(x→a⁻) f(x). RHL = lim(x→a⁺) f(x). Continuous at a ⟺ LHL = RHL = f(a).

For piecewise functions: ALWAYS check continuity at the BREAKPOINT. Compute LHL (using the expression for x < a), RHL (using x > a), and f(a) (using the actual defined value). If LHL = RHL = f(a), continuous. If LHL ≠ RHL, jump discontinuity.

Standard Derivatives

d/dx(xⁿ) = nxⁿ⁻¹. d/dx(eˣ) = eˣ. d/dx(aˣ) = aˣ ln a. d/dx(ln x) = 1/x. d/dx(sin x) = cos x. d/dx(cos x) = −sin x. d/dx(tan x) = sec²x. d/dx(sin⁻¹x) = 1/√(1−x²). d/dx(cos⁻¹x) = −1/√(1−x²). d/dx(tan⁻¹x) = 1/(1+x²).

Inverse trig derivatives are heavily tested in Ch. 5 problems. Key: d/dx(sin⁻¹x) = +1/√(1−x²); d/dx(cos⁻¹x) = −1/√(1−x²) (just the negative). d/dx(tan⁻¹x) = 1/(1+x²).

Chain Rule, Implicit, and Logarithmic Differentiation

CHAIN RULE: d/dx[f(g(x))] = f'(g(x)) × g'(x). If y = f(u) and u = g(x): dy/dx = (dy/du)(du/dx). IMPLICIT: For F(x,y) = 0: differentiate BOTH sides w.r.t. x (using chain rule for y terms: d/dx[y²] = 2y·dy/dx). Solve for dy/dx. LOGARITHMIC DIFFERENTIATION: For y = [f(x)]^g(x): take ln both sides: ln y = g(x) ln f(x). Differentiate: (1/y)(dy/dx) = g'(x) ln f(x) + g(x) f'(x)/f(x). Multiply by y.

Logarithmic differentiation is ESSENTIAL for functions like y = xˢⁱⁿˣ or y = (sin x)ˡⁿˣ where the BASE is variable AND the exponent is variable. Take ln, differentiate, multiply by y.

Rolle's Theorem and Mean Value Theorem

ROLLE'S THEOREM: If f is (i) continuous on [a,b], (ii) differentiable on (a,b), and (iii) f(a) = f(b), then there exists c ∈ (a,b) such that f'(c) = 0. MEAN VALUE THEOREM (MVT): If f is (i) continuous on [a,b] and (ii) differentiable on (a,b), then there exists c ∈ (a,b) such that f'(c) = [f(b)−f(a)]/(b−a). Rolle's is a SPECIAL CASE of MVT (when f(a) = f(b), slope of chord = 0 = f'(c)).

For exam problems asking to verify MVT: (1) Check continuity on [a,b]. (2) Check differentiability on (a,b). (3) Compute [f(b)−f(a)]/(b−a). (4) Set f'(c) = this value. (5) Solve for c. (6) Verify c ∈ (a,b).

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Differentiating d/dx(cos x) as +sin x

✓ d/dx(cos x) = −sin x (NEGATIVE). Whereas d/dx(sin x) = +cos x. The cosine derivative has a negative sign. A memory aid: differentiation rotates sin → cos → −sin → −cos → sin (each step involves one function and the derivative is the next with appropriate sign changes).

WATCH OUT

✗ Forgetting the chain rule factor when differentiating sin(x²)

✓ d/dx[sin(x²)] = cos(x²) × 2x (not just cos(x²)). The chain rule requires multiplying by the derivative of the inner function (x² → 2x). Every composite function requires the chain rule. Whenever you differentiate f(g(x)), you MUST multiply by g'(x).

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex s 5.1–5.8

Exercises 5.1–5.8

Continuity check for piecewise functions; chain rule applications; implicit differentiation; parametric differentiation; logarithmic differentiation; Rolle's/MVT problems; second-order derivatives

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· continuity-check

Find the value of k that makes f(x) = {kx + 1 if x ≤ π; cos x if x > π} continuous at x = π.

▶ Show solution

For f to be continuous at x = π: LHL = RHL = f(π). f(π) = kπ + 1 (from the first piece, since π ≤ π). LHL: lim(x→π⁻) f(x) = lim(x→π⁻) (kx + 1) = kπ + 1. RHL: lim(x→π⁺) f(x) = lim(x→π⁺) cos x = cos π = −1. For continuity: LHL = RHL → kπ + 1 = −1 → kπ = −2 → k = −2/π. VERIFY: f(π) = (−2/π)π + 1 = −2 + 1 = −1 = RHL ✓.

Q2MEDIUM· logarithmic-diff

Differentiate y = xˢⁱⁿˣ + (sin x)ˣ with respect to x.

▶ Show solution

Let y = u + v where u = xˢⁱⁿˣ and v = (sin x)ˣ. DIFFERENTIATING u = xˢⁱⁿˣ: Take ln: ln u = sin x · ln x. Differentiate: (1/u)(du/dx) = cos x · ln x + sin x · (1/x). du/dx = u[cos x · ln x + sin x/x] = xˢⁱⁿˣ [cos x · ln x + sin x/x]. DIFFERENTIATING v = (sin x)ˣ: Take ln: ln v = x · ln(sin x). Differentiate: (1/v)(dv/dx) = ln(sin x) + x · (cos x/sin x) = ln(sin x) + x cot x. dv/dx = v[ln(sin x) + x cot x] = (sin x)ˣ [ln(sin x) + x cot x]. RESULT: dy/dx = du/dx + dv/dx = xˢⁱⁿˣ [cos x · ln x + sin x/x] + (sin x)ˣ [ln(sin x) + x cot x].

Q3HARD· mvt

Verify the Mean Value Theorem for f(x) = x(x−1)(x−2) on [0, 1/2].

▶ Show solution

VERIFY CONDITIONS: f(x) = x(x−1)(x−2) = x³ − 3x² + 2x is a polynomial → continuous on [0, 1/2] and differentiable on (0, 1/2). Conditions satisfied. FIND THE MVT SLOPE: f(0) = 0(−1)(−2) = 0. f(1/2) = (1/2)(−1/2)(−3/2) = 3/8. [f(1/2) − f(0)] / (1/2 − 0) = (3/8 − 0)/(1/2) = (3/8)(2) = 3/4. FIND c: f'(x) = 3x² − 6x + 2. Set f'(c) = 3/4: 3c² − 6c + 2 = 3/4. 3c² − 6c + 2 − 3/4 = 0. 3c² − 6c + 5/4 = 0. 12c² − 24c + 5 = 0. c = [24 ± √(576 − 240)] / 24 = [24 ± √336] / 24 = [24 ± 4√21] / 24 = (6 ± √21)/6. √21 ≈ 4.58. c₁ = (6 − 4.58)/6 ≈ 0.237. c₂ = (6 + 4.58)/6 ≈ 1.763. Only c₁ ≈ 0.237 ∈ (0, 0.5). Hence c = (6 − √21)/6 ∈ (0, 1/2). MVT VERIFIED.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Continuity at a: LHL = RHL = f(a). For piecewise functions, check breakpoint.
•d/dx(sin x) = cos x. d/dx(cos x) = −sin x. d/dx(tan x) = sec²x.
•d/dx(sin⁻¹x) = 1/√(1−x²). d/dx(cos⁻¹x) = −1/√(1−x²). d/dx(tan⁻¹x) = 1/(1+x²).
•Chain rule: d/dx[f(g(x))] = f'(g(x))·g'(x). Must be applied to every composite.
•Logarithmic differentiation: take ln, differentiate implicitly (1/y)(dy/dx) = ..., then dy/dx = y×(...).
•Rolle's Theorem: continuous on [a,b], differentiable on (a,b), f(a)=f(b) ⟹ ∃c: f'(c)=0.
•MVT: continuous on [a,b], differentiable on (a,b) ⟹ ∃c: f'(c) = [f(b)−f(a)]/(b−a).
•Rolle's is a special case of MVT where f(a) = f(b) → chord slope = 0.

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 8-10 marks

Question type	Marks each	Typical count	What it tests
Continuity / Finding k	2-3	1	Find k for continuity of piecewise function; verify continuity
Differentiation	4-6	2	Implicit/logarithmic differentiation; chain rule applications; second-order derivatives
Rolle's / MVT	3-4	1	Verify Rolle's or MVT; find c

Prep strategy

For continuity of piecewise functions: ALWAYS compute LHL, RHL, and f(a) separately, then set them equal. Show three steps explicitly.
For logarithmic differentiation of xˢⁱⁿˣ or u˵: take ln → differentiate → multiply by y. Show each step for full marks.
MVT problems: check conditions first (continuous + differentiable), then compute slope of chord, set f'(c) = slope, solve for c, verify c is in the open interval.

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Rate of Change in Engineering

The chain rule is the foundation of every engineering simulation. In thermodynamics: if temperature depends on pressure, and pressure depends on time, the chain rule gives dT/dt = (dT/dP)(dP/dt). In fluid mechanics, chemical engineering, and electrical engineering, every dynamic system modelling 'how one quantity changes as another changes as a third changes' uses the chain rule. The Mean Value Theorem underpins error analysis in numerical computation.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

For any differentiation problem: identify the TYPE first (direct, chain rule, product, quotient, implicit, logarithmic), then apply the correct method. Wrong method = zero marks even with correct working.
For Rolle's/MVT: write THREE steps in order — (1) Verify continuity/differentiability conditions. (2) Calculate target value (f'(c) = 0 or = slope). (3) Solve for c and verify c is in the open interval. Missing step 1 loses marks.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Explore the MEAN VALUE THEOREM in higher dimensions (Green's Theorem, Stokes' Theorem) — the idea that 'the instantaneous rate somewhere equals the average rate over an interval' generalises to surfaces and volumes in multivariable calculus
Study L'Hôpital's Rule — it uses derivatives to evaluate indeterminate limits (0/0, ∞/∞ forms). It is a direct application of the MVT and is standard in JEE Advanced problems though not in the CBSE syllabus

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 12 Board (Mathematics)High

JEE Main (Calculus — Limits, Continuity, Differentiability)High

CUET (Mathematics)High

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

DIFFERENTIABLE ⟹ CONTINUOUS (always). But CONTINUOUS does NOT imply differentiable. Example: f(x) = |x| is continuous at x = 0 but NOT differentiable at x = 0 (the left and right derivatives differ: LHD = −1, RHD = +1). The function has a 'sharp corner' — it is continuous but the derivative doesn't exist at the corner. Every differentiable function is continuous, but not every continuous function is differentiable.

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