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Mathematics Grade XII

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Chapter 5: Continuity and Differentiability

Class 12 • Mathematics • Chapter 5

Continuity and Differentiability

When a graph flows without a break, and when it is smooth enough to have a slope at every point.

CBSE / NCERT Aligned

School Revise

Chapter Roadmap

Continuity at a Point • Algebra of Continuous Functions • Differentiability • Chain Rule for Composites • Implicit and Logarithmic Differentiation • Derivatives of Standard Functions • Parametric Form • Second Order Derivatives

From Unbroken Graphs to Smooth Slopes

Some graphs you can draw without lifting your pen; others jump suddenly. The first kind are continuous. Continuity is the precise way of saying a function has no gaps, jumps or holes, and it is the foundation on which all of calculus is built. Once a function is continuous, the next question is whether it is smooth enough to have a definite slope at each point. That property is differentiability, and the slope itself is the derivative.

This chapter sharpens both ideas. You will test continuity using limits, learn which functions are automatically continuous, and find derivatives of composite, implicit, exponential, logarithmic and parametric functions, right up to second order derivatives. These tools power everything from physics and economics to machine learning, where rates of change are everywhere.

Foundation

A function f is continuous at x = a when three things agree: f(a) exists, the limit of f(x) as x → a exists, and that limit equals f(a). A function is differentiable at a when the limit defining f′(a) exists; differentiability is the stronger condition.

Key Terms You Must Know

Term

Meaning

Example

Continuous at a

f(a) exists, the limit exists, and they are equal.

x² is continuous everywhere

Discontinuity

A point where the graph breaks, jumps or has a hole.

A step function jumps

Derivative f′(x)

The limit giving the slope of the tangent at each point.

f(x)=x² → f′(x)=2x

Chain rule

Rule for differentiating a function inside another.

d/dx sin(x²) = 2x cos(x²)

Implicit function

y tied to x by an equation, not solved for y.

x² + y² = 25

Second derivative

The derivative of the derivative.

y=x³ → y″ = 6x

Core Concepts, Step by Step

1. Continuity at a Point

To test continuity at x = a, check the three-part rule: the value f(a) must exist, the limit as x approaches a must exist (the left side and right side must agree), and the limit must equal the value. If any part fails, the function is discontinuous there. Polynomials are continuous everywhere; rational functions are continuous wherever the denominator is non-zero.

A continuous curve has no gaps, so it passes the three-part test at every point

Smooth continuous parabola y equals x squared with no breaks, point (1,1) marked

2. Algebra of Continuous Functions

If two functions are continuous at a point, then so are their sum, difference, product and (where the denominator is non-zero) their quotient. The composition of continuous functions is also continuous. This means most functions you build from polynomials, roots, trigonometric, exponential and logarithmic pieces are continuous on their natural domains, so you rarely test every point by hand.

3. Differentiability and the Tangent Slope

The derivative f′(a) is the limit of the average rate of change as the interval shrinks to zero. Geometrically it is the slope of the tangent line to the curve at that point. A key relationship holds: every differentiable function is continuous, but not every continuous function is differentiable. A graph with a sharp corner, such as |x| at 0, is continuous yet has no single tangent slope there.

The derivative at a point is the slope of the tangent line there

Parabola y equals x squared with tangent line at point (1,1) having slope 2

4. The Chain Rule for Composite Functions

When one function sits inside another, such as sin(x²), we differentiate the outer function and multiply by the derivative of the inner function. Symbolically, if y = f(g(x)) then dy/dx = f′(g(x)) · g′(x). The chain rule is the workhorse of differentiation and is used constantly alongside the product and quotient rules.

5. Implicit and Logarithmic Differentiation

When y is not written explicitly, as in x² + y² = 25, we differentiate both sides treating y as a function of x and then solve for dy/dx. Logarithmic differentiation tames awkward expressions such as x raised to the power x: take natural logs of both sides first, which turns powers into products, then differentiate. Both techniques widen the range of functions you can handle.

6. Parametric Form and Second Order Derivatives

Sometimes x and y are each given in terms of a third variable t. Then dy/dx = (dy/dt) ÷ (dx/dt). The second derivative is found by differentiating the first derivative again; it measures how the slope itself is changing and tells us about concavity. Standard derivatives of trigonometric, inverse trigonometric, exponential and logarithmic functions should be at your fingertips for all of this.

Key Results with Proofs

Result 1: Differentiable Implies Continuous

Statement. If f is differentiable at x = a, then f is continuous at x = a.

Proof

We turn the difference f(x) − f(a) into a product whose limit is clearly zero.

Differentiability means the limit of [f(x) − f(a)]/(x − a) exists as x → a; call it f′(a). – start from the definition

Write f(x) − f(a) = [f(x) − f(a)]/(x − a) × (x − a). – multiply and divide by (x − a)

As x → a the first factor → f′(a) and the second → 0, so the product → 0. – take limits

Hence f(x) → f(a), which is exactly continuity at a. – conclusion

The converse fails: |x| is continuous at 0 but has a corner there, so it is not differentiable.

Result 2: The Power Rule

Statement. For any positive integer n, the derivative of xⁿ is n xⁿ⁻¹.

Proof

The binomial theorem makes the first-principles limit collapse neatly.

By definition f′(x) = limit of [(x + h)ⁿ − xⁿ]/h as h → 0. – first principles

Expand (x + h)ⁿ by the binomial theorem: xⁿ + n xⁿ⁻¹ h + (terms with h² and higher). – binomial expansion

Subtract xⁿ and divide by h: n xⁿ⁻¹ + (terms still containing h). – simplify

Let h → 0; every leftover term vanishes, leaving n xⁿ⁻¹. – take the limit

So d/dx of x³ is 3x², and the rule extends to all real powers.

Result 3: Rolle’s Theorem

Statement. If f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then there is a point c in (a, b) with f′(c) = 0.

Proof

Equal end values force the curve to turn somewhere inside, where the slope is zero.

Since f is continuous on a closed interval it attains a maximum and a minimum value there. – extreme value idea

If both extremes occur only at the ends, then f is constant and f′ = 0 everywhere inside. – constant case

Otherwise a maximum or minimum occurs at an interior point c. – interior extremum

At a smooth interior extremum the tangent is horizontal, so f′(c) = 0. – slope is zero there

The Mean Value Theorem generalises this to f′(c) = [f(b) − f(a)]/(b − a).

Worked Examples

Example 1

Question: Is f(x) = x² + 1 continuous at x = 2?

▶ Show full working

Check value and limit, then compare.

f(2) = 4 + 1 = 5, so the value exists. – value

As x → 2, x² + 1 → 5, so the limit exists and equals 5. – limit

Value and limit agree, so f is continuous at 2. – compare

Answer: Yes, f is continuous at x = 2.

Example 2

Question: Find k so that f(x) = kx + 1 for x ≤ 2 and 3x − 1 for x > 2 is continuous at x = 2.

▶ Show full working

Match the two pieces at x = 2.

Left piece at 2: k(2) + 1 = 2k + 1. – value from the left

Right limit at 2: 3(2) − 1 = 5. – value from the right

For continuity 2k + 1 = 5, so k = 2. – equate and solve

Answer: k = 2.

Example 3

Question: Differentiate f(x) = x² from first principles.

▶ Show full working

Use the limit definition.

f′(x) = limit of [(x + h)² − x²]/h. – set up the limit

Numerator = 2xh + h², so the quotient is 2x + h. – expand and simplify

As h → 0 this tends to 2x. – take the limit

Answer: f′(x) = 2x.

Example 4

Question: Differentiate y = sin(x²).

▶ Show full working

Use the chain rule.

Outer function sin has derivative cos; inner function x² has derivative 2x. – identify inner and outer

Multiply: dy/dx = cos(x²) · 2x. – apply the chain rule

Answer: dy/dx = 2x cos(x²).

Example 5

Question: Differentiate y = x² sin x.

▶ Show full working

Use the product rule.

Let u = x² (u′ = 2x) and v = sin x (v′ = cos x). – split into two factors

Product rule: u′v + uv′ = 2x sin x + x² cos x. – apply the rule

Answer: dy/dx = 2x sin x + x² cos x.

Example 6

Question: Differentiate y = (sin x)/x.

▶ Show full working

Use the quotient rule.

Quotient rule: (v u′ − u v′)/v² with u = sin x, v = x. – set up

= (x cos x − sin x)/x². – substitute and tidy

Answer: dy/dx = (x cos x − sin x)/x².

Example 7

Question: Find dy/dx for x² + y² = 25.

▶ Show full working

Differentiate implicitly.

Differentiate both sides: 2x + 2y (dy/dx) = 0. – treat y as a function of x

Solve: dy/dx = −x/y. – make dy/dx the subject

Answer: dy/dx = −x/y.

Example 8

Question: Differentiate y = xₓ (x to the power x).

▶ Show full working

Use logarithmic differentiation.

Take natural logs: ln y = x ln x. – log both sides

Differentiate: (1/y) dy/dx = ln x + 1. – product rule on the right

So dy/dx = y(1 + ln x) = xₓ(1 + ln x). – multiply back by y

Answer: dy/dx = xₓ(1 + ln x).

Example 9

Question: If x = t² and y = t³, find dy/dx.

▶ Show full working

Use the parametric rule.

dx/dt = 2t and dy/dt = 3t². – differentiate each

dy/dx = (3t²)/(2t) = (3/2)t. – divide

Answer: dy/dx = (3/2)t.

Example 10

Question: Find the second derivative of y = x³.

▶ Show full working

Differentiate twice.

First derivative: y′ = 3x². – power rule

Differentiate again: y″ = 6x. – second derivative

Answer: y″ = 6x.

Example 11

Question: Differentiate y = tan⁻¹(x).

▶ Show full working

Use the standard inverse-trig derivative.

The standard result is d/dx of tan⁻¹(x) = 1/(1 + x²). – recall the formula

Answer: dy/dx = 1/(1 + x²).

Example 12

Question: Differentiate y = ln(x² + 1).

▶ Show full working

Use the chain rule with the log derivative.

d/dx of ln u is (1/u) u′, with u = x² + 1 and u′ = 2x. – chain rule on a log

= 2x/(x² + 1). – substitute

Answer: dy/dx = 2x/(x² + 1).

Where You Meet This in Real Life

Physics

Velocity is the derivative of position and acceleration is the derivative of velocity, so motion is described entirely by derivatives.

Economics

Marginal cost and marginal revenue are derivatives that tell a business how profit changes with one more unit produced.

Medicine

Rates at which a drug concentration rises and falls in the blood are modelled with derivatives to set safe doses.

Engineering

Smoothness conditions, which are continuity and differentiability, decide whether a designed curve such as a road or a cam runs without jolts.

Machine learning

Training a neural network relies on the chain rule, applied millions of times, to adjust the model and reduce error.

Practice Sets A–D

Practice Set A – Basics

A1. Differentiate y = x⁵.

▶ Reveal full working

Use the power rule.

Bring the power down and reduce it by one: 5x⁴.

Answer: 5x⁴.

A2. Differentiate y = cos x.

▶ Reveal full working

Standard derivative.

d/dx of cos x is −sin x.

Answer: −sin x.

A3. Differentiate y = eₓ.

▶ Reveal full working

Standard derivative.

The exponential is its own derivative: eₓ.

Answer: eₓ.

A4. Is f(x) = 3x − 4 continuous at x = 1?

▶ Reveal full working

Polynomials are continuous everywhere.

f(1) = −1 and the limit as x → 1 is also −1, so it is continuous.

Answer: Yes.

Practice Set B – Conceptual

B1. Give a function that is continuous but not differentiable at a point.

▶ Reveal full working

Think of a sharp corner.

f(x) = |x| has no break, so it is continuous at 0. – continuous

But the slope from the left is −1 and from the right is +1, so no single tangent exists. – corner

Answer: f(x) = |x| at x = 0.

B2. Why does differentiable always imply continuous?

▶ Reveal full working

Link the difference to a product whose limit is 0.

f(x) − f(a) = [the difference quotient] × (x − a). – rewrite

As x → a the second factor → 0 while the first stays finite, so f(x) → f(a). – limit

Answer: Because the change in f shrinks to 0 as x approaches a.

B3. State the chain rule in words.

▶ Reveal full working

Outer then inner.

Differentiate the outer function keeping the inner one inside, then multiply by the derivative of the inner function.

Answer: Outer derivative times inner derivative.

B4. What does the second derivative tell you about a curve?

▶ Reveal full working

Think about how the slope changes.

It measures how the slope itself is changing, which describes the bending or concavity of the curve.

Answer: It describes the concavity of the curve.

Practice Set C – Application / Numerical

C1. Differentiate y = cos(3x).

▶ Reveal full working

Chain rule.

Outer cos gives −sin; inner 3x gives 3. – identify pieces

dy/dx = −3 sin(3x). – multiply

Answer: −3 sin(3x).

C2. Differentiate y = x² eₓ.

▶ Reveal full working

Product rule.

u = x² (u′ = 2x), v = eₓ (v′ = eₓ). – split

u′v + uv′ = 2x eₓ + x² eₓ = eₓ(x² + 2x). – combine

Answer: eₓ(x² + 2x).

C3. Find dy/dx if x = 2t and y = t².

▶ Reveal full working

Parametric rule.

dx/dt = 2 and dy/dt = 2t. – differentiate

dy/dx = 2t/2 = t. – divide

Answer: dy/dx = t.

C4. Find the second derivative of y = sin x.

▶ Reveal full working

Differentiate twice.

y′ = cos x. – first

y″ = −sin x. – second

Answer: y″ = −sin x.

Practice Set D – HOTS / Multi-step

D1. Find dy/dx for y = (1 + x²)⁵.

▶ Reveal full working

Chain rule on a power.

Outer power 5 gives 5(1 + x²)⁴; inner 1 + x² gives 2x. – two layers

dy/dx = 5(1 + x²)⁴ · 2x = 10x(1 + x²)⁴. – multiply

Answer: 10x(1 + x²)⁴.

D2. Differentiate y = xₓ for x > 0 and evaluate dy/dx at x = 1.

▶ Reveal full working

Use logarithmic differentiation, then substitute.

From ln y = x ln x we get dy/dx = xₓ(1 + ln x). – logarithmic differentiation

At x = 1: 1¹(1 + ln 1) = 1(1 + 0) = 1. – substitute x = 1

Answer: dy/dx = xₓ(1 + ln x); at x = 1 it equals 1.

D3. For x² + y² = 25 find the slope of the tangent at the point (3, 4).

▶ Reveal full working

Use the implicit derivative.

Implicit differentiation gives dy/dx = −x/y. – derivative

At (3, 4): −3/4. – substitute

Answer: Slope = −3/4.

D4. Verify Rolle’s Theorem for f(x) = x² − 4x on [0, 4] and find c.

▶ Reveal full working

Check the conditions, then solve f′(c) = 0.

f is a polynomial, so continuous and differentiable; f(0) = 0 and f(4) = 0, so the end values match. – conditions hold

f′(x) = 2x − 4; set 2c − 4 = 0. – set derivative to zero

c = 2, which lies in (0, 4). – solve

Answer: c = 2.

Chapter Summary

Everything in One Glance

Continuity

f(a) exists, the limit exists, and the two are equal; polynomials are continuous everywhere.

Differentiability

The derivative is the tangent slope; differentiable functions are always continuous, but not the reverse.

Chain Rule

For y = f(g(x)), dy/dx = f′(g(x)) g′(x).

Implicit & Log

Differentiate both sides treating y as a function of x; take logs first for powers like xₓ.

Standard Derivatives

Know the derivatives of trig, inverse trig, exponential and logarithmic functions.

Higher Order

dy/dx in parametric form is (dy/dt)/(dx/dt); the second derivative shows concavity.

Are You Exam-Ready?

8-Point Exam Quick-Check

Is x³ − 2x continuous at x = 1?

Differentiate y = x⁷.

Differentiate y = sin(5x).

Differentiate y = x cos x.

Find dy/dx for x² + y² = 1.

Differentiate y = ln(3x).

If x = t and y = t², find dy/dx.

Find the second derivative of y = x⁴.

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Class 12 Mathematics Chapter 5: Continuity and Differentiability, Complete Notes and Practice

These free Class 12 Maths Continuity and Differentiability notes follow the NCERT 2026 to 27 syllabus and cover continuity at a point, the algebra of continuous functions, differentiability, the chain rule, implicit and logarithmic differentiation, parametric and second order derivatives, with twelve worked examples and sixteen graded practice questions. Includes worked examples, step-by-step proofs and graded practice, free on SchoolRevise.com.