Continuity and Differentiability Class 12 Notes: The Complete Revision Guide

These continuity and differentiability class 12 notes compress every idea in Chapter 5 of the NCERT textbook into the revision version you actually need the night before a test. For the 2026-27 session, this page maps the chapter in the order ideas build: limits (from Class 11) → continuity → differentiability → then every derivative technique real exam questions use.

The chapter has five clear blocks: the three-part continuity test (5.2), the definition of differentiability and the theorem linking it to continuity (5.3), inverse trigonometric derivatives (5.3.3), the new exponential and logarithmic functions (5.4), and then logarithmic (5.5), parametric (5.6) and second-order differentiation (5.7). Once you see that the chain rule does the work inside almost every technique, the whole chapter clicks into one idea.

Work through chapters in order and you’ll want the Determinants chapter for the background, and the very next topic, Application of Derivatives, builds directly on the techniques here. For more subject help, see the Class 12 Mathematics notes hub and the full Class 12 notes collection, or browse the CBSE notes index.

Continuity: The Three-Part Test

Definition 1 (NCERT, p. 106): a function is continuous at c when \( \lim_{x \to c} f(x) = f(c) \). Informally, you can draw its graph around \( c \) without lifting the pen from the paper.

Every continuity question reduces to this three-part test:

1. Does \( f(c) \) exist? The function must be defined at \( c \).
2. Does \( \lim_{x \to c} f(x) \) exist? The left-hand limit (LHL) and right-hand limit (RHL) must be equal.
3. Do they match? The common limit must equal the value \( f(c) \).

All three conditions must hold at once. If any one fails, \( f \) is discontinuous at \( c \), and \( c \) is a point of discontinuity (NCERT, p. 106).

The two ways continuity fails

Jump (LHL ≠ RHL). NCERT opens with \( f(x) = 1 \) for \( x \le 0 \), \( f(x) = 2 \) for \( x \gt 0 \) (p. 105). Left and right limits at 0 differ (1 vs 2), so the limit does not exist — the graph forces you to lift the pen at 0.

Hole (limit exists but value differs). In the next example the function equals 1 everywhere except \( f(0) = 2 \). Both one-sided limits are 1, so the limit exists, but \( \lim_{x \to 0} f(x) = 1 \ne 2 = f(0) \). Same verdict: not continuous (p. 105-106).

Continuity on an interval and the surprising cases

Definition 2 (NCERT, p. 107-108): a function is continuous if it is continuous at every point of its domain.

On a closed interval \( [a, b] \), continuity at the left endpoint \( a \) means \( \lim_{x \to a^+} f(x) = f(a) \), and at \( b \) it means \( \lim_{x \to b^-} f(x) = f(b) \) — one-sided limits only, because the limit from outside the interval does not exist.

Two facts follow that surprise students:

• If the domain is a single point, the function is continuous there automatically (p. 108).
• The greatest integer function \( f(x) = [x] \) is continuous at every non-integer, but discontinuous at every integer — at \( c \) an integer, \( [c-r] = c-1 \) while \( [c+r] = c \), so the one-sided limits never match (Example 15, p. 112-113).

Standard families you can quote without re-checking limits: polynomials, rational functions where defined, \( \sin x \), \( \cos x \), \( |x| \), exponential and logarithmic functions (Examples 5, 6, 14, 16, 17, 18).

Table 1: Continuity at a Glance

Algebra of Continuous Functions

Theorem 1 (NCERT, p. 114): if \( f \) and \( g \) are continuous at \( c \), then so are \( f + g \), \( f – g \), and \( f \cdot g \). The quotient \( \frac{f}{g} \) is continuous at \( c \) provided \( g(c) \ne 0 \).

Two special cases from the Remarks (p. 114):

• \( \lambda \cdot g \) is continuous for any constant \( \lambda \) (put \( f(x) = \lambda \)).
• \( \frac{1}{g} \) is continuous wherever \( g(x) \ne 0 \) (put \( f(x) = 1 \)).

This is how we know every rational function is continuous: it is a quotient of two continuous polynomials, so Theorem 1(4) applies (Example 16).

Theorem 2 (Composition, p. 116): if \( g \) is continuous at \( c \) and \( f \) is continuous at \( g(c) \), then the composite \( (f \circ g)(x) = f(g(x)) \) is continuous at \( c \). This is the fast proof for functions like \( \sin(x^2) \), a composite of \( \sin x \) with \( x^2 \) (Example 19).

For the harder Example 20, \( f(x) = |1 – x + |x|| \), think of it as \( | \cdot | \circ g \) where \( g(x) = 1 – x + |x| \): the inner function is a sum of a polynomial and \( |x| \), both continuous, and the outer \( | \cdot | \) is continuous — so the whole composite is continuous.

Differentiability and the Derivative

The derivative of \( f \) at \( c \) is defined as (NCERT, p. 119):

\[ f'(c) = \lim_{h \to 0} \frac{f(c+h) – f(c)}{h} \]

provided this limit exists. If both the left and right hand limits of this difference quotient are finite and equal, the function is differentiable at \( c \); if the limit does not exist, it is not differentiable there.

Four notations for the derivative (p. 119): \( f'(x) \), \( \frac{d}{dx} f(x) \), \( \frac{dy}{dx} \), and \( y’ \). The process is called differentiation.

The algebra of derivatives from Class 11, restated on p. 119:

• Sum rule: \( (u \pm v)’ = u’ \pm v’ \)
• Product rule (Leibnitz): \( (uv)’ = u’v + uv’ \)
• Quotient rule: \( \left( \frac{u}{v} \right)’ = \frac{u’v – uv’}{v^2} \), wherever \( v \ne 0 \)

The differentiability ⇒ continuity theorem

Theorem 3 (NCERT, p. 120): if \( f \) is differentiable at \( c \), then \( f \) is continuous at \( c \). The proof is one line of algebra: rewrite \[ f(x) – f(c) = \frac{f(x) – f(c)}{x – c} \cdot (x – c) \]

As \( x \to c \), the first factor tends to \( f'(c) \) and the second to 0, so the product tends to 0 — hence \( \lim_{x \to c} f(x) = f(c) \).

The converse is false. The classic counterexample is \( f(x) = |x| \) at 0 (p. 120). It is continuous at 0, but the left hand derivative is \( \frac{-h}{h} = -1 \) while the right hand derivative is \( \frac{h}{h} = 1 \). The two one-sided limits differ, so \( f \) is not differentiable at 0. This is the single most quoted example in the chapter.

Standard Derivatives Table (memorise before the exam)

A single table carries every derivative the exam expects, combining Table 5.3 (p. 119), the inverse trig table (p. 124), Theorem 5 (p. 129) and Example 28 (p. 131). Learn each one together with its domain restriction:

The Chain Rule and Its Three-Level Extension

Theorem 4 (Chain Rule) (NCERT, p. 122): if \( f = v \circ u \) with \( t = u(x) \), and both \( \frac{dv}{dt} \) and \( \frac{dt}{dx} \) exist, then \[ \frac{df}{dx} = \frac{dv}{dt} \cdot \frac{dt}{dx} \]

Why it exists. The textbook motivates it with \( (2x+1)^3 \): expanding works, but no one wants to expand \( (2x+1)^{100} \). The chain rule differentiates the outer function and multiplies by the derivative of the inner function, which is exactly the step students drop.

For a composite of three functions — \( f = w \circ u \circ v \) with \( t = v(x) \) and \( s = u(t) \) — the rule extends (p. 122):

\[ \frac{df}{dx} = \frac{dw}{ds} \cdot \frac{ds}{dt} \cdot \frac{dt}{dx} \]

Worked example: \( f(x) = \sin(4x^3) \)

Worked example: \( f(x) = (3x+7)^{100} \)

Implicit Differentiation: When You Cannot Solve for y

The textbook contrasts two relations (p. 122): \( x – y – \pi = 0 \) can be solved for \( y = x – \pi \) (an explicit function), but \( x + \sin xy – y = 0 \) cannot be easily solved, even though \( y \) clearly depends on \( x \). That second type defines \( y \) implicitly.

The rule: differentiate the whole equation term by term with respect to \( x \). Every time you differentiate a term containing \( y \), multiply by \( \frac{dy}{dx} \) — that is just the chain rule applied to \( y \), which is itself a function of \( x \).

Worked example: \( x^2 + y^2 = 25 \)

Worked example: \( y + \cos y = x^2 \)

Notice both answers still contain \( y \). That is normal and correct — the derivative of an implicit function is typically written in terms of both \( x \) and \( y \).

Derivatives of Inverse Trigonometric Functions

The inverse trig derivatives drop straight out of implicit differentiation (Example 24, p. 124). Set \( y = \sin^{-1} x \), which converts to \( x = \sin y \).

Questions 9–15 of Exercise 5.3 are not about applying this formula directly: they hide a standard identity inside the inverse trig function. Recognise the pattern and substitute before differentiating:

• An argument like \( \frac{2x}{1+x^2} \) inside \( \sin^{-1} \) or \( \cos^{-1} \) suggests \( x = \tan \theta \), since \( \sin 2\theta = \frac{2\tan\theta}{1+\tan^2\theta} \) and \( \cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta} \).
• An argument of the form \( \frac{1-x^2}{1+x^2} \) also fits the \( \cos 2\theta \) identity.
• An argument like \( 2x\sqrt{1-x^2} \) suggests \( x = \sin \theta \), since \( \sin 2\theta = 2 \sin\theta \cos\theta \).
• An argument \( \frac{3x – x^3}{1 – 3x^2} \) suggests \( x = \tan \theta \) again, using \( \tan 3\theta = \frac{3\tan\theta – \tan^3\theta}{1 – 3\tan^2\theta} \).

Substituting \( x = \tan \theta \) is also how you simplify any expression involving \( \sqrt{1 + x^2} \).

Exponential and Logarithmic Functions

Definition 3 (NCERT, p. 126): the exponential function with positive base \( b \gt 1 \) is \( y = f(x) = b^x \). Its five salient features (p. 126-127):

• Domain: all real numbers \( \mathbb{R} \).
• Range: all positive real numbers.
• The point \( (0, 1) \) is always on the graph, since \( b^0 = 1 \).
• It is ever increasing — the graph rises as \( x \) increases.
• For large negative \( x \), the graph approaches the \( x \)-axis but never meets it.

The number \( e \) (between 2 and 3) comes from the series \( 1 + \frac{1}{1!} + \frac{1}{2!} + \dots \) (p. 127). Using \( e \) as the base gives the natural exponential function \( e^x \).

Definition 4 (p. 127): the logarithm of \( a \) to base \( b \) is \( x \) if \( b^x = a \), written \( \log_b a = x \). For \( b = 10 \) we get common logarithms; for \( b = e \) we get natural logarithms. Convention for this whole chapter (p. 127): \( \log x \) means the natural logarithm to base \( e \), i.e. \( \ln x \) is written simply as \( \log x \).

Key properties of the log function for any base \( b \gt 1 \) (p. 128): domain is the positive reals, range is all reals, the point \( (1, 0) \) is always on the graph, and the function is ever increasing.

The log laws (p. 128-129) you will use constantly in differentiation:

• \( \log_b (pq) = \log_b p + \log_b q \)
• \( \log_b \left( \frac{p}{q} \right) = \log_b p – \log_b q \)
• \( \log_b (p^n) = n \log_b p \)
• Change of base: \( \log_a p = \frac{\log_b p}{\log_b a} \) (p. 128)

Also note \( x = e^{\log x} \) is true only for positive \( x \), because the domain of the log is the positive reals (Example 25, p. 129).

Geometrically, the graphs of \( y = e^x \) and \( y = \ln x \) are mirror images of each other reflected in the line \( y = x \) (Fig 5.11, p. 128). That reflection is exactly why the two functions are inverses.

Theorem 5 (p. 129) gives the two golden derivatives:

\[ \frac{d}{dx}(e^x) = e^x \qquad \text{and} \qquad \frac{d}{dx}(\log x) = \frac{1}{x} \]

The striking fact: \( e^x \) is the only function that is its own derivative — differentiate it and it does not change. This is the chapter’s best memory anchor: every other exponential \( a^x \) carries a \( \log a \) multiplier, but \( e^x \) carries nothing.

Worked example: differentiate \( e^{3x} \cdot \log x \)

Logarithmic Differentiation: Taming \( [u(x)]^{v(x)} \)

Functions like \( x^{\sin x} \) or \( x^x \) are neither \( x^n \) (the exponent is not constant) nor \( a^x \) (the base is not constant). The only reliable route is logarithmic differentiation (p. 130):

1. Take the natural log of both sides: \( \log y = v(x) \log[u(x)] \).
2. Differentiate implicitly with respect to \( x \):

\[ \frac{1}{y}\cdot\frac{dy}{dx} = v(x)\cdot\frac{u'(x)}{u(x)} + v'(x)\log[u(x)] \]

3. Multiply through by \( y \) and resubstitute \( y = [u(x)]^{v(x)} \).

Condition: \( f(x) \) and \( u(x) \) must be positive, because logs of non-positive numbers are not defined (p. 130). Always state this before taking logs.

Worked example: \( y = x^{3x} \), \( x \gt 0 \)

Worked example: \( y = (\sin x)^x \), \( x \in (0, \pi) \)

This same machinery produces \( \frac{d}{dx}(a^x) = a^x \log a \) (Example 28, p. 131). Take \( y = a^x \), so \( \log y = x\log a \); differentiating gives \( \frac{1}{y}\frac{dy}{dx} = \log a \). The alternative route writes \( a^x = e^{x\log a} \) and uses the chain rule — both give the same answer.

Parametric Differentiation: \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \)

When both \( x \) and \( y \) are given in terms of a third variable — \( x = f(t) \), \( y = g(t) \) — the connection is said to be in parametric form with \( t \) as the parameter (p. 135). Applying the chain rule:

\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{g'(t)}{f'(t)} \qquad \text{provided } \frac{dx}{dt} \ne 0 \]

Worked example: \( x = 3t^2 \), \( y = 6t \)

Worked example: \( x = 5\sin\theta \), \( y = 5\cos\theta \)

The note to Example 33 (p. 136): it is perfectly acceptable to leave the answer in terms of the parameter \( t \) or \( \theta \), without eliminating it to get \( y \) as a function of \( x \).

Parametric forms are sometimes chosen to fit a given relation. Example 34 (p. 136-137) shows that \( x = a\cos^3\theta \), \( y = a\sin^3\theta \) satisfies \( x^{2/3} + y^{2/3} = a^{2/3} \), and then \[ \frac{dx}{d\theta} = -3a\cos^2\theta\sin\theta, \qquad \frac{dy}{d\theta} = 3a\sin^2\theta\cos\theta \quad \Rightarrow \quad \frac{dy}{dx} = -\tan\theta = -\sqrt[3]{\frac{y}{x}} \]

Second Order Derivatives

If \( y = f(x) \), the derivative of \( \frac{dy}{dx} \) with respect to \( x \) is the second order derivative, written \( \frac{d^2 y}{dx^2} \), \( f”(x) \), \( D^2y \), \( y” \) or \( y_2 \) (p. 137).

Worked example: find \( \frac{d^2y}{dx^2} \) for \( y = 4x^3 + \cos 2x \)

The standard “show that” pattern

Boards love asking you to prove a differential equation. With \( y = 2\sin x + 5\cos x \):

A tougher recurring type is Example 38 (p. 139-140): if \( y = \sin^{-1}x \), show \( (1-x^2)y_2 – xy_1 = 0 \). The clean route: \( y_1 = \frac{1}{\sqrt{1-x^2}} \), so \( (1-x^2)y_1^2 = 1 \). Differentiate this identity implicitly:

\[ (1-x^2) \cdot 2y_1 y_2 + y_1^2 (-2x) = 0 \]

Dividing by \( 2y_1 \) gives \( (1-x^2)y_2 – xy_1 = 0 \). This is the pattern where you square the first derivative to avoid differentiating the square root directly.

Worked Examples: Five Exam-Style Problems

These five problems together exercise every technique in the chapter. Work them on paper before reading the steps.

Problem 1: Continuity test (original numbers)

Check continuity of \( f(x) = 3x^2 – 2x + 1 \) at \( x = 2 \).

Problem 2: Chain rule

Differentiate \( y = \cos(5x^2 + 3) \).

Problem 3: Implicit differentiation

Find \( \frac{dy}{dx} \) if \( x^2 + 4y^2 = 16 \).

Problem 4: Logarithmic differentiation

Differentiate \( y = (2x+1)^x \), \( x \gt -\frac{1}{2} \).

Problem 5: Second derivative proof

If \( y = e^{-x}\cos x \), show that \( \frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 2y = 0 \).

Common Mistakes Students Make in This Chapter

Exam Notes: What the Examiner Looks For

These are observed board patterns, written with an examiner’s mindset — no year claims, no guarantees, just what earns the mark.

• Continuity proofs need the three-mark skeleton: state \( f(c) = \dots \), then \( \lim_{x \to c} f(x) = \dots \), then the explicit conclusion “hence continuous”. Skipping the equality statement loses the final mark.
• The inner-derivative multiplier is the earned mark in chain rule work. Most dropped steps are the \( 2x \), \( 10x \) or \( 3 \) factors — write them explicitly and you hold the mark.
• Piecewise functions at the junction require both LHL and RHL shown separately. Writing only \( \lim_{x\to c} \) is treated as incomplete; show \( \lim_{x\to c^-} \) and \( \lim_{x\to c^+} \) before concluding.
• In log differentiation, state the positivity condition (\( x \gt 0 \), or \( u(x) \gt 0 \)) before taking logs — it is part of the method, not decoration.
• “Show that” second-derivative questions are won by substitution. Compute \( y_1 \) and \( y_2 \) cleanly, then substitute into the expression; 2–3 marks are reserved for the substitution step itself.
• Learn the inverse trig domains exactly. One-mark questions test \( x \in (-1, 1) \) for \( \sin^{-1}x \) and \( \cos^{-1}x \), all reals for \( \tan^{-1}x \).

For reference, the official NCERT Class 12 Mathematics Part I textbook — Chapter 5 Continuity and Differentiability (pages 104–139) lets you verify any formula directly against the source.

Continuity and Differentiability Class 12 Notes: One-Page Revision Summary

• Continuity at \( c \): \( \lim_{x\to c} f(x) = f(c) \). Two failure modes — jump (LHL ≠ RHL) and hole (limit exists but value differs).
• Differentiable ⇒ continuous, but not conversely. \( |x| \) is continuous at 0 yet not differentiable there (LHD = \(-1\), RHD = \( +1 \)).
• Chain rule: \( \frac{df}{dx} = \frac{dv}{dt}\frac{dt}{dx} \); three-level form \( \frac{df}{dx} = \frac{dw}{ds}\frac{ds}{dt}\frac{dt}{dx} \).
• Inverse trig: \( \sin^{-1}x \to \frac{1}{\sqrt{1-x^2}} \), \( \cos^{-1}x \to -\frac{1}{\sqrt{1-x^2}} \) on \( (-1,1) \); \( \tan^{-1}x \to \frac{1}{1+x^2} \) on \( \mathbb{R} \).
• Golden derivatives: \( \frac{d}{dx}e^x = e^x \), \( \frac{d}{dx}\log x = \frac{1}{x} \), \( \frac{d}{dx}a^x = a^x\log a \). Only \( e^x \) is its own derivative.
• Log differentiation for \( [u(x)]^{v(x)} \): take logs, differentiate implicitly, multiply by \( y \). Requires positivity.
• Parametric: \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \), \( \frac{dx}{dt} \ne 0 \); answer may stay in terms of the parameter.
• Second order: \( \frac{d^2y}{dx^2} = \frac{d}{dx}\left( \frac{dy}{dx} \right) \) with notations \( f”(x) \), \( y” \), \( y_2 \), \( D^2y \). Board favourite: prove \( \frac{d^2y}{dx^2} + y = 0 \) for \( y = A\sin x + B\cos x \).
• Greatest integer function is continuous at non-integers, discontinuous at every integer.

Reference: NCERT Class 12 Mathematics Part I textbook, chapter 5: Continuity and Differentiability.

Frequently Asked Questions

What is the difference between continuity and differentiability?

Continuity means the graph can be drawn without lifting the pen — \( \lim_{x\to c} f(x) = f(c) \). Differentiability is stronger: the slope of the tangent must exist, meaning the left and right derivatives are equal. Every differentiable function is continuous, but a continuous function need not be differentiable — \( |x| \) at 0 is the standard example.

Why is \( |x| \) continuous at 0 but not differentiable at 0?

As \( x \) approaches 0 from the left, \( |x| = -x \), so the left hand derivative is \( -1 \); from the right, \( |x| = x \), giving right hand derivative \( +1 \). The two one-sided derivatives differ, so \( \lim_{h\to 0}\frac{f(0+h)-f(0)}{h} \) does not exist — the graph has a sharp corner at 0, and a corner has no single tangent line.

When should I use logarithmic differentiation?

Whenever the variable appears in both the base and the exponent, as in \( x^x \), \( (\sin x)^x \) or \( x^{\sin x} \). Since these are neither \( x^n \) nor \( a^x \), no direct rule applies — take logs, differentiate the product, then multiply by \( y \).

It is also useful before differentiating messy products and quotients such as \( \sqrt{\frac{(x-3)(x^2+4)}{3x^2+4x+5}} \), where logs convert multiplication into addition.

Why is the derivative of \( \sin^{-1}x \) only valid for \( x \in (-1, 1) \)?

The derivative is \( \frac{1}{\sqrt{1-x^2}} \), and the square root must be real. At \( x = \pm 1 \), \( 1 – x^2 = 0 \) gives division by zero, so the derivative is undefined there. Inside the open interval \( (-1, 1) \), it exists because \( \cos y \) is positive and the original implicit equation \( 1 = \cos y \frac{dy}{dx} \) is well defined.

How do I recognise which technique to use for a differentiation problem?

Ask what form the function takes. Pure composition — chain rule. Both \( x \) and \( y \) mixed in one equation — implicit differentiation. An inverse trig argument that collides with a standard identity — substitute \( x = \sin\theta \) or \( x = \tan\theta \) first. Base and exponent both variables — log differentiation. Both variables given in terms of a parameter — divide the two \( dt \)-derivatives.

Growing and shrinking functions of the form \( e^{kx} \sin x \) — product rule plus careful second derivatives.

Is every continuous function differentiable? What about the converse?

The converse holds: every differentiable function is continuous (Theorem 3). But not every continuous function is differentiable — \( f(x) = |x| \) is continuous everywhere yet fails differentiability at 0, and the greatest integer function fails both at every integer. Differentiability is a strictly stronger condition than continuity.