Continuity and Differentiability — Class 12 Maths Notes & Practice

Meet Your Tutor

Continuity and Differentiability rewards a fixed order: inspect the domain, compare the two one-sided limits with the function value, and only then differentiate. I will help you keep these tests separate, recognise standard forms and write each condition explicitly so a correct idea also becomes a full-mark solution.

Take a breath before you start. If the words continuity and differentiability have been sitting in your syllabus looking scary, that is completely normal — almost every Class 12 student feels the same on day one. Here is the reassuring truth: underneath the formal language, this chapter is built on two very ordinary ideas. Continuity just asks, “can I draw this graph without lifting my pen?” Differentiability just asks, “does this curve have a smooth, well-defined direction at every point, or does it turn a sharp corner somewhere?” Everything else — the limits, the chain rule, the substitutions, the second derivatives — grows out of those two questions.

This chapter also happens to be the engine room of the whole Calculus unit. Unit III (Calculus) carries 35 marks in the CBSE Class 12 Maths paper, shared across Continuity and Differentiability, Applications of Derivatives, Integrals, Applications of Integrals and Differential Equations. Every one of those later chapters leans on the differentiation skills you build right here. So the time you invest now pays you back four more times before the paper is over.

We will go slowly. Every rule gets explained in plain words first, then shown on a worked example, then repeated on a harder one. Do not rush, and do not move on from a section until it genuinely feels comfortable. Sit with me, keep a rough notebook open, and let us take it one small step at a time.

What You’ll Learn

Your Game Plan

Here is the order I would work through this chapter in if I were sitting next to you. Resist the urge to jump straight to the formula lists — the first six sections are what make the rest make sense.

1. Days 1–2: Get continuity into your hands. Read the first four sections, then physically sketch five or six piecewise functions on paper and test each one at its joining point. The three-step test should become automatic.
2. Day 3: Discontinuity and unknown constants. These are guaranteed marks. Practise until finding k in a piecewise function takes you under two minutes.
3. Day 4: Differentiability. Understand the corner at x = 0 on y = |x| deeply. If you understand that one picture, you understand the whole idea.
4. Days 5–6: Build your derivative toolkit. Memorise the standard derivatives table, then drill the chain rule until it is muscle memory. Everything after this point is the chain rule wearing different hats.
5. Days 7–8: Inverse trig and the substitution tricks. Slow, careful work here. Learn the substitutions as a small family, not as twenty unrelated tricks.
6. Day 9: Implicit, exponential, logarithmic and logarithmic differentiation. Four techniques, one afternoon, lots of practice.
7. Day 10: Parametric forms and second order derivatives. Short sections, high exam frequency. Do not skip them because they come last.
8. Day 11: The worksheet at the bottom of this page. Attempt every question with the answers hidden first. Only then reveal.

Study Notes

What Continuity Really Means

Picture yourself drawing the graph of a function on paper. You put your pencil down on the left end of the curve and start moving right. If you can reach the other end without ever lifting the pencil, the function is continuous. If at some point you have to lift the pencil, jump across a gap and put it down again, the function is discontinuous at that point. That is the entire idea. Everything formal we do below is just a precise way of saying “no pencil-lifting”.

Now, why do we need the formal version at all? Because “without lifting the pencil” is a description of a picture, and in an exam you are usually handed a formula, not a picture. So we translate the picture into limits. Here is the translation.

Read condition (iii) again, because it is the one students under-appreciate. It says the value the function is heading towards as you approach a must equal the value it actually takes at a. If the function heads towards 5 but is defined to be 7 at that exact point, there is a hole in the graph with a dot floating above it — and your pencil has to jump. Discontinuous.

One more piece of vocabulary before the examples. The left-hand limit (LHL), written lim_x→a− f(x), is the value f approaches when you creep up to a from the left, using x-values slightly smaller than a. The right-hand limit (RHL), lim_x→a+ f(x), is what happens when you creep down from the right. A very useful practical trick: to compute the LHL substitute x = a − h and let h → 0 with h > 0; for the RHL substitute x = a + h and let h → 0.

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The Three-Step Test For Continuity At A Point

Follow the flow chart above exactly, in that order, every time. It costs you thirty seconds and it removes almost all the ways of going wrong. Let us walk through what each box means in practice.

Box 1 — Does f(a) exist? Look at the definition of the function at x = a specifically. For a piecewise function, check carefully which branch owns the point a. A branch written “for x ≤ 2” owns x = 2; a branch written “for x < 2” does not. This single reading error causes more lost marks than any actual mathematics in this chapter.

Boxes 2 and 3 — The two one-sided limits. For a piecewise function, the LHL uses the branch valid just to the left of a, and the RHL uses the branch valid just to the right. Note that neither of them uses the value at a itself; limits do not care what happens exactly at the point, only what happens nearby.

Box 4 — Compare all three numbers. Not two of them. All three. Then write your conclusion sentence.

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Continuity On An Interval

So far we have tested one point at a time. Usually we want to describe a whole stretch of the graph at once, and for that we say a function is continuous on an interval.

The endpoint rule sounds fussy but it is only common sense. At the left endpoint a there is nothing to the left of it inside the interval, so we cannot demand a left-hand limit. We only ask that the function arrives correctly from inside the interval. That is called one-sided continuity.

It saves an enormous amount of time to know which standard functions are continuous everywhere, so that you never have to prove them from scratch. Learn this list.

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The Algebra Of Continuous Functions

Here is a set of results that will save you hours. Once you know a few functions are continuous, you get all their combinations for free.

In everyday language: adding, subtracting, multiplying, dividing (carefully) and nesting continuous functions can never manufacture a break out of nowhere. A break can only appear where something was already undefined — a zero denominator, a negative number under a square root, a non-positive number inside a logarithm.

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Types Of Discontinuity

Not all breaks are the same size or shape. Being able to name the type of discontinuity shows the examiner you understand what is happening, and it often earns an extra mark in “discuss the continuity” questions. There are three types you should recognise.

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Finding Unknown Constants Using Continuity

This is one of the most reliably examined question types in the whole chapter, and the good news is that it follows a fixed recipe. You are given a piecewise function containing an unknown letter (usually k, or a and b) and told the function is continuous. Your job is to run the continuity condition backwards and solve for the letter.

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What Differentiability Really Means

Continuity asked, “is the curve unbroken?” Differentiability asks a more demanding question: “is the curve smooth?” Imagine driving a tiny car along the graph. Continuity means the road never has a gap you would fall through. Differentiability means the road never has a sudden corner that would snap your steering wheel. A road can be perfectly connected and still have a nasty corner — and that is the single most important idea in this section.

What is that fraction actually measuring? The numerator f(a + h) − f(a) is the rise in the function as you move h units to the right of a. Dividing by h gives the slope of the straight line joining the two points (a, f(a)) and (a + h, f(a + h)) — a chord. Letting h → 0 slides the second point closer and closer to the first, so the chord swings round until it becomes the tangent. The derivative is the slope of that tangent.

Now you can see why a corner is fatal. Approaching from the left, the chords settle on one slope. Approaching from the right, they settle on a different slope. There is no single tangent line, so there is no derivative. Look again at the graph of y = |x| above: the left arm arrives with slope −1, the right arm leaves with slope +1, and at x = 0 the curve simply cannot decide which direction it is pointing.

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Differentiability Implies Continuity, But Not The Reverse

This is the relationship between the chapter’s two big ideas, and it is a one-way street. Let us state it carefully, because the direction matters enormously.

Here is the short reason why differentiability forces continuity. Write f(a + h) − f(a) = h × [f(a + h) − f(a)]/h. As h → 0 the bracket approaches the finite number f′(a), and h itself approaches 0. A finite number multiplied by something going to zero goes to zero. So f(a + h) − f(a) → 0, which says exactly that f(a + h) → f(a), which is continuity at a. Notice the argument needed f′(a) to be finite — that is why the vertical-tangent case in Example 23 does not count as differentiable.

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Standard Derivatives You Must Know By Heart

From here on we stop computing limits and start using results. Every derivative in the table below can be proved from first principles, but in the exam you are expected to quote them. Learn them until you can write the whole table from memory in three minutes — that single skill is worth more marks in Unit III than any other piece of memorisation.

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The Chain Rule And Derivatives Of Composite Functions

The chain rule is the single most important tool in this chapter. Almost every derivative you meet from now on is the chain rule in disguise, so it is worth investing real practice time here. The idea is the onion picture above: a composite function has layers, and to differentiate it you peel one layer at a time, differentiate each layer, and multiply everything you collected.

The phrase “keeping the inside untouched” is the part to hold on to. When you differentiate sin(x²), the derivative of the outer sine is cosine, and the thing sitting inside it stays exactly as it was: cos(x²), not cos(2x). Only then do you multiply by the derivative of the inside, 2x.

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Derivatives Of Inverse Trigonometric Functions

The syllabus names sin⁻¹x, cos⁻¹x and tan⁻¹x specifically, but the whole family is worth knowing. Two things make these questions manageable: the derivatives are algebraic (no trig functions in the answers), and they always arrive paired with the chain rule.

Combined with the chain rule, if u is a function of x then d/dx (sin⁻¹u) = u′/√(1 − u²), d/dx (tan⁻¹u) = u′/(1 + u²), and so on. The pattern is always “standard formula with u in place of x, multiplied by u′”.

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Smart Substitutions For Inverse Trigonometric Derivatives

Some inverse-trig derivatives look terrifying — expressions like tan⁻¹(2x/(1 − x²)) or sin⁻¹(2x√(1 − x²)). Attacking them head-on with the chain rule produces a page of algebra. There is a much better way: substitute a trigonometric variable so the messy inside collapses into a standard identity. This is a favourite board exam topic, and once you see the pattern it becomes almost mechanical.

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Derivatives Of Implicit Functions

So far every function has been given explicitly: y written alone on the left, a formula in x on the right. But plenty of curves refuse to cooperate. The circle x² + y² = 25 defines y in terms of x, yet you cannot write a single formula for y without splitting into two halves. Such a relation is called implicit, and there is a neat way to find dy/dx without ever solving for y.

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Exponential And Logarithmic Functions

Before we differentiate these, the syllabus asks you to be comfortable with the concept. Let us take a minute on that, because a shaky grasp here makes the next two sections much harder than they need to be.

An exponential function is f(x) = a^x where a > 0 and a ≠ 1. The base a stays fixed and the exponent varies. Its graph passes through (0, 1) for every base, rises steeply for a > 1, falls for 0 < a < 1, is defined for every real x, and is always positive — so it never touches the x-axis. The special base is e ≈ 2.71828, and f(x) = e^x is called the natural exponential function. Its defining property is that it is the one exponential curve whose slope at every point equals its own height.

A logarithmic function is the inverse of the exponential. Saying log_a N = x means exactly the same as a^x = N. Because the exponential only ever produces positive outputs, the logarithm only accepts positive inputs: the domain of log_ax is x > 0. The graphs of a^x and log_ax are mirror images of each other in the line y = x. Throughout this chapter, “log x” with no base written means the natural logarithm, log_ex, often written ln x.

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Derivatives Of Exponential And Logarithmic Functions

That middle result, d/dx (log u) = u′/u, is worth its own box in your memory. “Derivative of the inside over the inside” handles an enormous number of exam questions in a single line.

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Logarithmic Differentiation

Two kinds of function defeat every rule you have learned so far. The first is a variable raised to a variable power, like x^x — it is neither a power function (the exponent is not constant) nor an exponential function (the base is not constant), so neither formula applies. The second is a long product or quotient of many factors, where the product and quotient rules would take half a page. Logarithmic differentiation handles both.

Step 4 is the one to internalise: because y is a function of x, d/dx (log y) = (1/y)(dy/dx). Everything else is ordinary differentiation.

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Derivatives Of Functions In Parametric Form

Sometimes x and y are both expressed in terms of a third variable, called a parameter and usually written t or θ. For example x = a cos t, y = a sin t traces a circle as t runs from 0 to 2π. You could eliminate t to get x² + y² = a², but that is often messy or impossible — and completely unnecessary.

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Second Order Derivatives

Differentiating once gives you the slope. Differentiating the slope gives you the second derivative — the rate at which the slope itself is changing, which is what makes a curve bend. It is written in several equivalent ways, and all of them appear in exam papers:

That parametric warning is the single biggest source of lost marks in this section, so read it twice. And a practical tip: always simplify dy/dx before differentiating again. A messy first derivative turns into a nightmare second derivative.

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Rolle’s And Mean Value Theorems: Removed From The Current Syllabus

If you are working from an older book, a senior’s notes or a random website, you may run into Rolle’s Theorem and Lagrange’s Mean Value Theorem presented as part of this chapter. You should know that both were removed during the rationalisation of the syllabus and are not part of the CBSE Class 12 Mathematics syllabus for 2026–27. They will not be examined.

The prescribed content of this chapter for 2026–27 is exactly: continuity and differentiability; the chain rule and derivatives of composite functions; derivatives of inverse trigonometric functions such as sin⁻¹x, cos⁻¹x and tan⁻¹x; derivatives of implicit functions; the concept of exponential and logarithmic functions and their derivatives; logarithmic differentiation; derivatives of functions expressed in parametric form; and second order derivatives. Everything on this page sits inside that list — nothing more, nothing less.

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Practice Worksheet

Ten questions, in roughly increasing order of difficulty. Work each one on paper with the answer hidden — that is the only way this is useful. Then click Show Answer and compare not just the final result but your working line by line. If your method differed but your answer matched, that is fine; maths often has more than one good route.

Q1. Find the value of k for which f(x) = (x² − 16)/(x − 4) for x ≠ 4, and f(4) = k, is continuous at x = 4. ▸ Show Answer

For x ≠ 4: (x² − 16)/(x − 4) = (x − 4)(x + 4)/(x − 4) = x + 4 LHL = lim (h→0) [(4 − h) + 4] = 8 RHL = lim (h→0) [(4 + h) + 4] = 8 f(4) = k Continuity needs LHL = RHL = f(4): k = 8

Answer: k = 8. The factor (x − 4) cancels legitimately because we never evaluate at x = 4 itself, only near it.

Q2. Show that f(x) = |x − 2| is continuous at x = 2 but is not differentiable there. ▸ Show Answer

CONTINUITY at x = 2: f(2) = |2 − 2| = 0 LHL = lim (h→0) |(2 − h) − 2| = lim h = 0 RHL = lim (h→0) |(2 + h) − 2| = lim h = 0 All three equal 0 ⇒ continuous at x = 2. ✓ DIFFERENTIABILITY at x = 2: LHD = lim (h→0−) [ |2 + h − 2| − 0 ] / h = lim (h→0−) |h|/h , h < 0 so |h| = −h = −1 RHD = lim (h→0+) |h|/h , h > 0 so |h| = h = +1 LHD = −1 ≠ +1 = RHD

Answer: f is continuous at x = 2 but not differentiable there, because the graph has a sharp corner at (2, 0).

Q3. Differentiate y = sin(x³ + 5) with respect to x. ▸ Show Answer

Chain rule, outer sin, inner x³ + 5. u = x³ + 5, u′ = 3x² dy/dx = cos u · u′ = 3x² cos(x³ + 5)

Answer: dy/dx = 3x² cos(x³ + 5). The inside stays untouched inside the cosine; the 3x² multiplies from outside.

Q4. Differentiate y = tan⁻¹[ (1 + x)/(1 − x) ] for x < 1. ▸ Show Answer

Substitute x = tan θ, so θ = tan⁻¹x. (1 + x)/(1 − x) = (1 + tanθ)/(1 − tanθ) Recognise the compound angle with tan(π/4) = 1: = (tan(π/4) + tanθ)/(1 − tan(π/4)tanθ) = tan(π/4 + θ) So y = tan⁻¹[ tan(π/4 + θ) ] = π/4 + θ = π/4 + tan⁻¹x (for θ in the principal range) dy/dx = 0 + 1/(1 + x²) = 1/(1 + x²)

Answer: dy/dx = 1/(1 + x²). The π/4 is a constant, so it contributes nothing — a lovely example of simplifying before differentiating.

Q5. Find dy/dx if x² + xy + y² = 100. ▸ Show Answer

Differentiate both sides w.r.t. x, product rule on xy: 2x + [ y + x · dy/dx ] + 2y · dy/dx = 0 2x + y + (x + 2y) · dy/dx = 0 (x + 2y) · dy/dx = −(2x + y) dy/dx = −(2x + y)/(x + 2y), x + 2y ≠ 0

Answer: dy/dx = −(2x + y)/(x + 2y). Note the pleasing symmetry — a good sign that the algebra went right.

Q6. Differentiate y = (log x)^x with respect to x, for x > 1. ▸ Show Answer

Variable base and variable exponent ⇒ logarithmic differentiation. log y = x · log(log x) Differentiate; product rule on the right, and d/dx log(log x) = (1/log x) · (1/x) by the chain rule: (1/y) · dy/dx = 1 · log(log x) + x · 1/(x log x) = log(log x) + 1/(log x) Multiply by y = (log x)ˣ: dy/dx = (log x)ˣ [ log(log x) + 1/(log x) ]

Answer: dy/dx = (log x)^x [ log(log x) + 1/log x ]. The condition x > 1 is needed so that log x > 0 and log(log x) is defined.

Q7. If x = sin t and y = cos 2t, find dy/dx, and evaluate it at t = π/6. ▸ Show Answer

dx/dt = cos t dy/dt = −2 sin 2t = −2(2 sin t cos t) = −4 sin t cos t dy/dx = (−4 sin t cos t)/(cos t) = −4 sin t, cos t ≠ 0 At t = π/6: sin(π/6) = 1/2 dy/dx = −4 × 1/2 = −2

Answer: dy/dx = −4 sin t, and at t = π/6 it equals −2. Expanding sin 2t as 2 sin t cos t is what allows the cos t to cancel.

Q8. If y = e^x cos x, find d²y/dx². ▸ Show Answer

First derivative (product rule): y₁ = eˣ cos x + eˣ(−sin x) = eˣ( cos x − sin x ) Second derivative (product rule again): y₂ = eˣ( cos x − sin x ) + eˣ( −sin x − cos x ) = eˣ[ cos x − sin x − sin x − cos x ] = eˣ( −2 sin x ) d²y/dx² = −2 eˣ sin x

Answer: d²y/dx² = −2e^x sin x. The cos x terms cancel exactly — a neat check that you differentiated correctly.

Q9. Find k so that f(x) = (1 − cos 2x)/x² for x ≠ 0, and f(0) = k, is continuous at x = 0. ▸ Show Answer

Use the identity 1 − cos 2x = 2 sin²x: (1 − cos 2x)/x² = 2 sin²x / x² = 2 (sin x / x)² As x → 0 (from either side), (sin x)/x → 1, so lim (x→0) f(x) = 2 × 1² = 2 Continuity needs k = 2

Answer: k = 2. The double-angle identity converts an awkward limit into the standard limit (sin x)/x → 1.

Q10. Find a and b so that f(x) = 5 for x ≤ 2, f(x) = ax + b for 2 < x < 10, and f(x) = 21 for x ≥ 10 is continuous on R. ▸ Show Answer

Two joining points give two equations. At x = 2: LHL = 5, RHL = 2a + b, f(2) = 5 ⇒ 2a + b = 5 … (1) At x = 10: LHL = 10a + b, RHL = 21, f(10) = 21 ⇒ 10a + b = 21 … (2) (2) − (1): 8a = 16 ⇒ a = 2 From (1): 2(2) + b = 5 ⇒ b = 1 Check: at x = 2, 2(2) + 1 = 5 ✓ at x = 10, 2(10) + 1 = 21 ✓

Answer: a = 2, b = 1. The middle branch is the line y = 2x + 1, which joins the two flat pieces perfectly at both ends.

If you got seven or more of those right on your first attempt, you are in good shape for this chapter. If you got fewer, that is genuinely fine — it just tells you which section to reread, and the sections are short. Go back to the ones that tripped you up, redo the worked examples with the solutions covered, then come back here tomorrow.

And here is the only piece of advice that matters in the long run: aim for one more correct question tomorrow than you managed today. Not ten more. One. Do that consistently and in a month you will be solving problems that look impossible to you right now, without ever having had a heroic study day. Small, steady, unglamorous progress is what actually gets people through this paper. Be patient with yourself, keep your working neat, and keep showing up.

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