Class 12 Mathematics (041) · Chapter 5 NotesContinuity and Differentiability

Comprehensive Class 12 Mathematics notes for Chapter 5: Continuity and Differentiability. Learn chain rule, implicit functions, and logarithmic differentiation.

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Chapter contents

Chapter summary

Chapter 5 of Class 12 Mathematics, Continuity and Differentiability, serves as a vital bridge between basic calculus and advanced mathematical analysis. Building upon the foundations laid in Class 11, this chapter explores the precise conditions under which a function is considered 'smooth' or 'unbroken'. Students will move beyond simple polynomial differentiation to master complex techniques involving composite, implicit, and parametric functions. A significant portion of the chapter introduces exponential and logarithmic functions, which are essential for modeling growth and decay in various scientific fields. By understanding the relationship between continuity and differentiability, students learn that while every differentiable function is necessarily continuous, the reverse is not always true. These concepts are fundamental for solving real-world problems involving rates of change and are prerequisite for the subsequent study of integration and differential equations.

What you'll learn

1Define continuity of a real function at a point and over an interval
2Establish the relationship between differentiability and continuity
3Apply the Chain Rule to find derivatives of composite functions
4Differentiate implicit functions and inverse trigonometric functions
5Understand the properties and derivatives of exponential and logarithmic functions
6Utilize logarithmic differentiation for functions of the form u(x) raised to v(x)
7Calculate derivatives of functions expressed in parametric forms
8Determine the second-order derivative of various functions

Chapter at a glance

01Continuity of Functions at a Point
02Continuity on an Interval
03Derivatives of Composite Functions
04Derivatives of Implicit and Parametric Functions
05Second Order Derivatives

Detailed chapter notes

01

Continuity at a Point and on an Interval

A function is continuous at a specific point 'c' if the function is defined at that point and the limit of the function as x approaches c equals the value of the function at c. Informally, this means the graph of the function can be drawn around that point without lifting the pen. For a function to be continuous on a closed interval [a, b], it must be continuous at every point within the interval, including the endpoints where one-sided limits are considered. If the left-hand limit, right-hand limit, and the function value do not all coincide, the function is said to be discontinuous at that point.

  • Continuity conditionlimit of f(x) as x approaches c = f(c)
  • A function is continuous if it is continuous at every point in its domain
  • Polynomial and constant functions are continuous everywhere
  • Algebra of continuitySums, products, and quotients of continuous functions are also continuous
02

Differentiability and its Relation to Continuity

Differentiability refers to the existence of a unique derivative at a point. A function f is differentiable at c if the limit of [f(c+h) - f(c)] / h exists as h approaches zero. This chapter establishes a crucial theorem: if a function is differentiable at a point, it must also be continuous at that point. However, the converse is not true; for example, the modulus function f(x) = |x| is continuous at x = 0 but not differentiable there because the left-hand and right-hand derivatives are not equal. Differentiability requires a 'smooth' curve without sharp corners or vertical tangents.

  • Derivative at cf'(c) = limit as h approaches 0 of [f(c+h) - f(c)] / h
  • Differentiability implies continuity
  • Continuity does not necessarily imply differentiability
  • A function is differentiable in an interval if it is differentiable at every point therein
03

The Chain Rule and Composite Functions

The Chain Rule is the primary tool for differentiating composite functions, often described as 'functions of functions'. If a function f is composed of two functions v and u such that f = v(u(x)), the derivative is found by differentiating the outer function with respect to the inner function and then multiplying it by the derivative of the inner function with respect to x. This rule can be extended to compositions of three or more functions. It allows for the differentiation of complex expressions like sin(x^2) or log(cos x) by breaking them down into simpler, nested components.

  • Chain Ruledy/dx = (dy/dt) * (dt/dx)
  • Used for functions like f(x) = v(u(x))
  • Essential for differentiating trigonometric functions with non-linear arguments
  • Fundamental for all advanced differentiation techniques
04

Implicit and Inverse Trigonometric Functions

Implicit functions occur when x and y are mixed in an equation such that y is not isolated on one side, such as x + sin(xy) - y = 0. To differentiate these, we apply the derivative operator to every term in the equation with respect to x, treating y as a function of x and applying the chain rule where necessary. Additionally, the chapter introduces derivatives for inverse trigonometric functions. These derivatives are derived using the chain rule and trigonometric identities, resulting in algebraic expressions. For instance, the derivative of sin^-1(x) is 1 / sqrt(1 - x^2) for x between -1 and 1.

  • Implicit differentiationDifferentiate both sides of the equation w.r.t. x
  • d/dx (sin^-1 x) = 1 / sqrt(1 - x^2)
  • d/dx (cos^-1 x) = -1 / sqrt(1 - x^2)
  • d/dx (tan^-1 x) = 1 / (1 + x^2)
05

Exponential and Logarithmic Functions

Exponential functions with a positive base b > 1, particularly the natural exponential function e^x, grow faster than any polynomial function. The logarithmic function is the inverse of the exponential function. The natural logarithm (base e) is denoted as log x or ln x. Key properties include log(pq) = log p + log q and log(p^n) = n log p. A unique property of e^x is that its derivative is itself. The derivative of log x is 1/x. These functions are critical for logarithmic differentiation, a technique used to simplify the differentiation of products, quotients, or functions where the variable is in the exponent.

  • d/dx (e^x) = e^x
  • d/dx (log x) = 1/x
  • Logarithmic differentiationTake log on both sides before differentiating
  • The domain of log x is all positive real numbers
06

Parametric Forms and Second Order Derivatives

Sometimes x and y are both expressed in terms of a third variable, called a parameter (usually t or theta). To find dy/dx in parametric form, we calculate dy/dt and dx/dt separately and then divide them. This is represented as dy/dx = (dy/dt) / (dx/dt), provided dx/dt is not zero. Finally, the chapter introduces second-order derivatives, which represent the rate of change of the first derivative. If y = f(x), the second-order derivative is denoted as d^2y/dx^2 or f''(x). This is found by differentiating the first derivative dy/dx once more with respect to x.

  • Parametric derivativedy/dx = (dy/dt) / (dx/dt)
  • Second order derivatived/dx (dy/dx) = d^2y/dx^2
  • Parametric equations are common in physics and geometry
  • Higher order derivatives follow the same logic of repeated differentiation
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Quick revision: key points

  • A function is continuous at c if LHL = RHL = f(c).
  • Every differentiable function is continuous, but not every continuous function is differentiable.
  • The Chain Rule is used for differentiating composite functions: dy/dx = dy/dt * dt/dx.
  • The derivative of e^x is e^x and the derivative of log x is 1/x.
  • Logarithmic differentiation is necessary for functions of the form y = [u(x)]^v(x).
  • For parametric equations x = f(t) and y = g(t), dy/dx = g'(t) / f'(t).
  • The second-order derivative is the derivative of the first derivative.
  • Inverse trigonometric functions have specific algebraic derivatives defined on restricted domains.

Test yourself

Try each question first, then reveal the answer.

Question 01

A function f(x) is continuous at x = a if and only if which of the following conditions holds?

  • Af(a) exists and lim(x→a) f(x) = f(a)
  • Blim(x→a⁻) f(x) = lim(x→a⁺) f(x)
  • Cf(x) is defined in a neighbourhood of a
  • Dlim(x→a) f(x) exists
Show answer
Answer: (A) f(a) exists and lim(x→a) f(x) = f(a)

Continuity at a point requires both that f(a) is defined and that the limit equals f(a). This combines existence, definition, and the matching of the limit to the function value.

Question 02

A function f(x) is continuous on a closed interval [a, b] if it is continuous at every point in that interval. Which of the following is a necessary condition for continuity at a point c in [a, b]?

  • Af(c) exists and lim(x→c) f(x) = f(c)
  • Bf(c) > 0
  • Cf'(c) exists
  • Df(x) is defined only at c
Show answer
Answer: (A) f(c) exists and lim(x→c) f(x) = f(c)

By definition, a function is continuous at point c if f(c) is defined and the limit of f(x) as x approaches c equals f(c). This is the fundamental criterion for continuity at a point.

Question 03

If f(x) = 5x + 3, then f'(x) is equal to:

  • A5
  • B3
  • C5x
  • D8
Show answer
Answer: (A) 5

The derivative of a linear function f(x) = mx + c is simply the slope m. Here, f'(x) = 5.

Question 04

If f(x) = (3x + 2)^5, then f'(x) using the chain rule is:

  • A5(3x + 2)^4 × 3
  • B5(3x + 2)^4
  • C(3x + 2)^4 × 3
  • D15(3x + 2)^5
Show answer
Answer: (A) 5(3x + 2)^4 × 3

By the chain rule, d/dx[(3x + 2)^5] = 5(3x + 2)^4 × d/dx(3x + 2) = 5(3x + 2)^4 × 3.

Question 05

If x² + y² = 25, find dy/dx using implicit differentiation.

  • A-x/y
  • Bx/y
  • Cy/x
  • D-y/x
Show answer
Answer: (A) -x/y

Differentiating x² + y² = 25 implicitly: 2x + 2y(dy/dx) = 0, so dy/dx = -x/y.

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Frequently asked questions

What is the main difference between continuity and differentiability?

Continuity means the function's graph has no breaks or jumps at a point. Differentiability is a stricter condition; it means the graph is smooth and has a defined tangent line at that point. While every differentiable function is continuous, a continuous function may not be differentiable if it has a sharp corner or cusp.

Why is the modulus function not differentiable at x = 0?

At x = 0, the graph of f(x) = |x| has a sharp corner. The left-hand derivative is -1 and the right-hand derivative is 1. Since these two limits are not equal, the derivative does not exist at that specific point, even though the function is continuous there.

When should I use logarithmic differentiation?

You should use logarithmic differentiation when the function is a complex product or quotient of multiple terms, or when the function has a variable in both the base and the exponent, such as y = x^sin x. Taking the natural log simplifies these into sums and products that are easier to differentiate.

How do you find the derivative of a parametric function?

To find dy/dx for parametric equations x = f(t) and y = g(t), you first differentiate both x and y with respect to the parameter t to get dx/dt and dy/dt. Then, you divide dy/dt by dx/dt to obtain the final derivative in terms of t.

What does the second-order derivative represent?

The second-order derivative, d^2y/dx^2, is the derivative of the first derivative. It measures the rate at which the slope of the function is changing. In physical terms, if the first derivative represents velocity, the second-order derivative represents acceleration.

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