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Class 12 Mathematics Chapter 5 Continuity and Differentiability — Formulas & Key Points

Chapter 5 of NCERT Class 12 Mathematics builds the rigorous foundation for calculus by defining continuity at a point and on intervals, then extending to differentiability and advanced differentiation techniques. Mastery of these formulas is non-negotiable for the CBSE board exam, where 10–13 marks typically come from this chapter. This sheet organizes every key formula, definition, theorem statement, and application rule so you can revise efficiently in the final weeks before your exam.

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Key takeaways

✓A function is continuous at x=a if limit equals function value: lim(x→a) f(x) = f(a), with no jumps or breaks.
✓Every differentiable function is continuous, but a continuous function need not be differentiable (sharp corners exist).
✓Chain rule is central: d/dx[f(g(x))] = f'(g(x)) · g'(x); apply it layer by layer for composite functions.
✓Derivatives of inverse trigonometric functions carry specific signs and denominators involving square roots.
✓Logarithmic differentiation simplifies products, quotients, and power-power forms like [f(x)]^[g(x)].
✓Rolle's Theorem and Mean Value Theorem connect continuity, differentiability, and the existence of a tangent parallel to a chord.
✓Common errors include forgetting the chain rule multiplier, sign mistakes in inverse trig derivatives, and assuming continuity implies differentiability.

Continuity: Definitions and Conditions

A function f is continuous at x = a if three conditions hold simultaneously: f(a) is defined, the limit lim(x→a) f(x) exists, and that limit equals f(a). If any one condition fails, the function has a discontinuity at a. For continuity on an interval [a, b], f must be continuous at every point in (a, b), right-continuous at a, and left-continuous at b. Piecewise functions are a favourite board-exam question type: you equate left-hand and right-hand limits at the junction point to find unknown constants.

f is continuous at a ⇔ lim(x→a⁺) f(x) = lim(x→a⁻) f(x) = f(a)
Right-continuous at a: lim(x→a⁺) f(x) = f(a); left-continuous: lim(x→a⁻) f(x) = f(a)
Discontinuity types: removable (limit exists but ≠ f(a)), jump (LHL ≠ RHL), infinite (limit → ±∞)
Sum, difference, product, quotient (denominator ≠ 0), and composition of continuous functions are continuous

Differentiability: Definition and Relationship to Continuity

A function f is differentiable at x = a if the derivative f'(a) = lim(h→0) [f(a+h)−f(a)]/h exists and is finite. Geometrically, the graph has a unique non-vertical tangent at that point. A crucial theorem states every differentiable function is continuous, but the converse is false: f(x)=|x| at x=0 is continuous yet not differentiable due to a sharp corner. For piecewise functions, you must verify that left-hand derivative equals right-hand derivative at the junction. This concept appears in 4–6 mark board questions where you find the value of a parameter ensuring differentiability.

f'(a) = lim(h→0) [f(a+h)−f(a)]/h must exist and be finite
Differentiable ⇒ continuous; continuous ⇏ differentiable
Non-differentiable scenarios: sharp corner (|x|), vertical tangent (x^(1/3) at 0), discontinuity
For piecewise f, check: lim(x→a⁻) f'(x) = lim(x→a⁺) f'(x)

Standard Derivatives and Chain Rule

The chain rule is the workhorse of differentiation: if y=f(u) and u=g(x), then dy/dx = (dy/du)·(du/dx). In compact notation, d/dx[f(g(x))] = f'(g(x))·g'(x). Apply it layer by layer for nested functions like sin(x³) or e^(tan x). The table below lists standard derivatives you must memorize verbatim, because board examiners expect exact forms. Errors in signs, especially for cotangent and cosecant derivatives, cost marks. Product rule and quotient rule extend these basics: (uv)' = u'v + uv' and (u/v)' = (u'v−uv')/v². Practise combining these rules in one problem to build speed.

Table of Standard Derivatives

Memorize this table for instant recall during the exam. Each entry is tested repeatedly in CBSE papers, often in combination with chain rule or product rule. Pay special attention to the signs in trigonometric derivatives (tangent and cotangent have positive coefficients in sec² and −csc² respectively) and the domains (inverse trig functions have restricted domains, e.g., sin⁻¹ x requires |x|≤1). Constants differentiate to zero, and the power rule d/dx(xⁿ)=n·xⁿ⁻¹ applies for any real n. Exponential and logarithmic derivatives are also shown; note that (aˣ)'=aˣ·ln a, which reduces to eˣ when a=e because ln e=1.

d/dx(c) = 0; d/dx(xⁿ) = n·xⁿ⁻¹
d/dx(eˣ) = eˣ; d/dx(aˣ) = aˣ·ln a
d/dx(ln x) = 1/x; d/dx(logₐ x) = 1/(x·ln a)
d/dx(sin x) = cos x; d/dx(cos x) = −sin x
d/dx(tan x) = sec²x; d/dx(cot x) = −csc²x
d/dx(sec x) = sec x·tan x; d/dx(csc x) = −csc x·cot x

Derivatives of Inverse Trigonometric Functions

Inverse trigonometric functions have derivatives involving square roots and specific signs. The most common exam errors are sign flips and forgetting the square root in the denominator. For instance, d/dx(sin⁻¹ x)=1/√(1−x²) is positive, whereas d/dx(cos⁻¹ x)=−1/√(1−x²) is negative. Similarly, d/dx(tan⁻¹ x)=1/(1+x²) is always positive, but d/dx(cot⁻¹ x)=−1/(1+x²) is negative. These six derivatives must be memorized with their exact forms because board questions often ask you to differentiate composite functions like sin⁻¹(2x) or tan⁻¹(√x), where the chain rule multiplies the standard derivative by the inner derivative. Always write the domain restrictions when solving: sin⁻¹ and cos⁻¹ require |x|<1 inside the square root, while tan⁻¹ and cot⁻¹ accept all real x.

d/dx(sin⁻¹ x) = 1/√(1−x²), |x|<1
d/dx(cos⁻¹ x) = −1/√(1−x²), |x|<1
d/dx(tan⁻¹ x) = 1/(1+x²), x ∈ ℝ
d/dx(cot⁻¹ x) = −1/(1+x²), x ∈ ℝ
d/dx(sec⁻¹ x) = 1/(|x|√(x²−1)), |x|>1
d/dx(csc⁻¹ x) = −1/(|x|√(x²−1)), |x|>1

Logarithmic Differentiation Technique

Logarithmic differentiation is a powerful method for functions involving products, quotients, or variable exponents like [f(x)]^[g(x)]. Take the natural logarithm of both sides, apply log properties to simplify products into sums and powers into multiples, then differentiate implicitly. Finally, multiply both sides by y to isolate dy/dx. This technique is especially clean for expressions like y=(x²·sin x)/eˣ or y=xˣ. Board examiners love these because they test multiple skills: log rules, implicit differentiation, and algebraic manipulation. Always write 'taking log on both sides' as the first step to earn method marks even if you make a later computational slip. Remember that ln(ab)=ln a + ln b, ln(a/b)=ln a − ln b, and ln(aⁿ)=n·ln a are the three log properties you will use repeatedly in this method.

Step 1: y = [expression]; take ln both sides: ln y = ln[expression]
Step 2: Use log properties to expand: ln(ab)=ln a+ln b, ln(a/b)=ln a−ln b, ln(aⁿ)=n·ln a
Step 3: Differentiate implicitly: (1/y)·(dy/dx) = [derivative of RHS]
Step 4: Multiply by y: dy/dx = y·[derivative of RHS]
Useful for y=uᵛ where u,v are functions of x: ln y=v·ln u, then differentiate

Parametric Differentiation and Implicit Differentiation

When a curve is given parametrically as x=f(t), y=g(t), the derivative dy/dx = (dy/dt)/(dx/dt) provided dx/dt ≠ 0. This formula is direct and appears in 2–3 mark short-answer questions. For implicit relations like x²+y²=25 where y is not isolated, differentiate both sides with respect to x term-by-term, treating y as y(x) and applying chain rule to y terms. For example, d/dx(y²)=2y·(dy/dx). Collect all dy/dx terms on one side and solve. Implicit differentiation is tested in 4-mark questions, often combined with finding the slope of a tangent at a given point. Remember to substitute the point coordinates only after you have the general expression for dy/dx, not before differentiating.

Parametric: dy/dx = (dy/dt) ÷ (dx/dt)
Implicit: differentiate both sides w.r.t. x; treat y as y(x); use chain rule on y terms
Collect dy/dx terms, factor out, solve
Second derivative for parametric: d²y/dx² = d/dx(dy/dx) = [d/dt(dy/dx)] ÷ (dx/dt)

Rolle's Theorem and Mean Value Theorem

Rolle's Theorem states that if f is continuous on [a,b], differentiable on (a,b), and f(a)=f(b), then there exists at least one c in (a,b) such that f'(c)=0. Geometrically, the tangent is horizontal somewhere between a and b. The Mean Value Theorem generalizes this: under the same continuity and differentiability conditions but without requiring f(a)=f(b), there exists c in (a,b) where f'(c)=[f(b)−f(a)]/(b−a). This c is where the tangent is parallel to the secant joining (a,f(a)) and (b,f(b)). Board questions ask you to verify the hypotheses and find the value of c. Always check all three conditions explicitly in your solution to earn full marks, even if the question says 'verify the theorem' without spelling them out.

Rolle's: f continuous on [a,b], differentiable on (a,b), f(a)=f(b) ⇒ ∃c: f'(c)=0
MVT: same conditions but f(a) may ≠ f(b) ⇒ ∃c: f'(c)=[f(b)−f(a)]/(b−a)
Geometric meaning MVT: tangent at c is parallel to chord AB
LMVT (Lagrange): another name for MVT; Cauchy MVT involves two functions

Common Mistakes and Memory Tricks

Students lose marks on sign errors in derivatives of cot, csc, cos⁻¹, cot⁻¹, and csc⁻¹—all carry a negative sign. A mnemonic: 'Co-functions are negative'. For chain rule, remember to multiply by the derivative of the inner function every time; forgetting that multiplier is the number-one error. In logarithmic differentiation, write ln on both sides explicitly; do not skip steps. When applying Rolle's or MVT, always state all three hypotheses before concluding. For parametric forms, never write dy/dx = (dx/dt)/(dy/dt); the correct order is (dy/dt)/(dx/dt). In implicit differentiation, every term involving y picks up a factor dy/dx; treat y as an implicit function y(x). Finally, check domains: sin⁻¹ and cos⁻¹ require |x|≤1, and logarithms require positive arguments. Write these checks in your board answer to show rigour and earn method marks even if you make a numerical slip later.

Mnemonic for negative derivatives: 'Co-functions carry minus' (cos, cot, csc, cos⁻¹, cot⁻¹, csc⁻¹)
Chain rule multiplier: ALWAYS multiply by derivative of inner function
Logarithmic differentiation: write 'ln both sides' explicitly; expand before differentiating
Rolle's/MVT: state all three conditions (continuous, differentiable, endpoint condition)
Parametric: dy/dx = (dy/dt)/(dx/dt), NOT the reciprocal
Implicit: every y term → multiply by dy/dx; do not isolate y first
Domain checks: |x|<1 for sin⁻¹, cos⁻¹; x>0 for ln x

Three Solved Mini-Examples Applying Formulas

Example 1: Differentiate y=e^(sin x). Let u=sin x. Then y=eᵘ. By chain rule, dy/dx=(eᵘ)·(cos x)=e^(sin x)·cos x. Example 2: Find dy/dx if x³+y³=3xy. Differentiate implicitly: 3x²+3y²·(dy/dx)=3y+3x·(dy/dx). Rearrange: 3y²·(dy/dx)−3x·(dy/dx)=3y−3x². Factor: dy/dx·(y²−x)=y−x². So dy/dx=(y−x²)/(y²−x). Example 3: Given x=2t, y=t². Find dy/dx. dy/dt=2t, dx/dt=2. Hence dy/dx=(2t)/2=t. These three examples cover chain rule, implicit differentiation, and parametric forms—core techniques tested every year in CBSE board exams.

One-Glance Last-Minute Revision Box

Continuity at a: lim(x→a)f(x)=f(a). Differentiable ⇒ continuous. Chain rule: d/dx[f(g(x))]=f'(g(x))·g'(x). Product: (uv)'=u'v+uv'. Quotient: (u/v)'=(u'v−uv')/v². d/dx(sin x)=cos x, d/dx(cos x)=−sin x, d/dx(tan x)=sec²x, d/dx(eˣ)=eˣ, d/dx(ln x)=1/x. Inverse trig: d/dx(sin⁻¹x)=1/√(1−x²), d/dx(tan⁻¹x)=1/(1+x²); co-functions negative. Logarithmic diff: ln both sides, expand, differentiate, multiply by y. Parametric: dy/dx=(dy/dt)/(dx/dt). Implicit: differentiate both sides, each y term →·dy/dx. Rolle's: f(a)=f(b), continuous [a,b], differentiable (a,b) ⇒ ∃c: f'(c)=0. MVT: f'(c)=[f(b)−f(a)]/(b−a). Check domains, signs, chain-rule multipliers. Practise past 5 years' board papers for pattern recognition and speed.

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Frequently asked questions

What is the difference between continuity and differentiability in Class 12 Mathematics Chapter 5?+

Continuity means the graph has no breaks or jumps at a point: lim(x→a)f(x)=f(a). Differentiability means the graph has a unique non-vertical tangent at that point, i.e., f'(a) exists. Every differentiable function is continuous, but a continuous function may have a sharp corner (like |x| at 0) and fail to be differentiable there.

How many marks does Continuity and Differentiability carry in the CBSE Class 12 board exam?+

This chapter typically contributes 10–13 marks in the board paper. Expect one 4-mark long-answer question on differentiability of piecewise functions or logarithmic differentiation, one 2-mark question on derivatives of inverse trig functions, and short conceptual questions on Rolle's or Mean Value Theorem totalling around 4 marks.

Which derivative formulas from Chapter 5 must I memorize for the board exam?+

Memorize all six basic trig derivatives (sin, cos, tan, cot, sec, csc), all six inverse trig derivatives (sin⁻¹, cos⁻¹, tan⁻¹, cot⁻¹, sec⁻¹, csc⁻¹), exponential (eˣ, aˣ), logarithmic (ln x, logₐx), and power rule xⁿ. Pay special attention to signs: cos, cot, csc and their inverses cos⁻¹, cot⁻¹, csc⁻¹ all carry a negative sign in the derivative.

What is the chain rule and why is it so important in this chapter?+

The chain rule states d/dx[f(g(x))]=f'(g(x))·g'(x). It lets you differentiate composite functions layer by layer. Most board exam differentiation problems—such as sin(x²), e^(tan x), or (3x+1)⁵—require the chain rule. Forgetting the multiplier g'(x) is the single most common mistake that costs students marks.

How do I apply logarithmic differentiation, and when should I use it?+

Use logarithmic differentiation when the function is a product, quotient, or has a variable exponent, e.g., y=xˣ or y=(x²·sin x)/eˣ. Take ln of both sides, apply log properties (ln(ab)=ln a+ln b, ln(aⁿ)=n·ln a), differentiate implicitly to get (1/y)(dy/dx), then multiply both sides by y to isolate dy/dx.

What are the three conditions for Rolle's Theorem?+

Rolle's Theorem requires: (1) f is continuous on the closed interval [a,b], (2) f is differentiable on the open interval (a,b), and (3) f(a)=f(b). If all three hold, there exists at least one c in (a,b) where f'(c)=0, meaning the tangent is horizontal at c.

How do I find dy/dx for parametric equations like x=f(t), y=g(t)?+

Use the formula dy/dx = (dy/dt)/(dx/dt). Differentiate y with respect to t to get dy/dt, differentiate x with respect to t to get dx/dt, then divide the first by the second. Ensure dx/dt ≠ 0 at the point of interest. This method is tested in 2–3 mark short questions every year.

What is the correct derivative of tan⁻¹x and how do I avoid sign mistakes?+

d/dx(tan⁻¹x) = 1/(1+x²), which is always positive. Its co-function cot⁻¹x has derivative −1/(1+x²), negative. Remember 'co-functions are negative': cos, cot, csc, and their inverses cos⁻¹, cot⁻¹, csc⁻¹ all have a minus sign in front.

Can a function be continuous but not differentiable? Give an example.+

Yes. The classic example is f(x)=|x| at x=0. It is continuous because lim(x→0)|x|=0=f(0). However, the left-hand derivative at 0 is −1 and the right-hand derivative is +1, so they do not match. The graph has a sharp corner, hence no unique tangent, and the function is not differentiable at 0.

How does CBSETUTOR.AI help with difficult differentiation problems in Chapter 5?+

CBSETUTOR.AI provides a 24×7 AI tutor that accepts photo uploads of your work. Snap a picture of your logarithmic differentiation or implicit differentiation problem, and the AI identifies where you forgot the chain rule multiplier or made a sign error, then walks you through the correct steps. It costs ₹999/month for all classes (6–12) with a 3-day free trial, cheaper than a single private tutor session yet available anytime.

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