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Continuity and Differentiability

When a function has no breaks (continuity) and when it has a well-defined slope (differentiability), plus rules for differentiating implicit, inverse trig, exponential, log, and parametric functions.

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Reading time~9 min
Revision time~3 min
Last updated2026-07-17
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🎯 Key Points

  • Continuity at a point needs THREE things to match: f(a) defined, limit exists, and limit=f(a) — missing any one breaks continuity
  • Differentiable ⟹ Continuous, but NOT the reverse — f(x)=|x| is the standard counterexample (continuous everywhere, not differentiable at x=0)
  • Logarithmic differentiation is required whenever the variable is in BOTH the base and exponent (y=xˣ, y=f(x)^g(x)) — taking ln first turns an otherwise impossible direct differentiation into a standard product
  • Parametric derivative: dy/dx = (dy/dt)/(dx/dt) — never differentiate y directly with respect to x when both x and y depend on a third parameter t
f(x) = |x|: Continuous but NOT Differentiable at 0x=0 (sharp "kink")LHD = -1RHD = +1LHD ≠ RHD at the kink → not differentiable, even though the curve has no gap (continuous)

f(x)=|x| is perfectly continuous at x=0 (no break, no jump), but the curve has a sharp corner there: approaching from the left gives slope -1 and from the right gives slope +1 — since these one-sided derivatives disagree, the function is not differentiable at that point.

Continuity and Differentiability

This chapter builds the foundation for all of calculus by making precise the ideas of "no breaks" (continuity) and "smooth, well-defined slope" (differentiability), then develops powerful rules for differentiating complicated functions.

Continuity at a Point

A function f is continuous at x = a if three conditions hold together: f(a) is defined, the limit of f(x) as x approaches a exists, and that limit equals f(a). Formally:

  • lim(x→a⁻) f(x) = lim(x→a⁺) f(x) = f(a)
  • If the left-hand limit (LHL) and right-hand limit (RHL) are not equal, the function is discontinuous at a.
  • A function is continuous on an interval if it is continuous at every point of that interval.

Types of Discontinuity

  • Removable discontinuity: LHL = RHL but this common value is not equal to f(a), or f(a) is undefined. The "gap" can be fixed by redefining f(a).
  • Jump (first kind) discontinuity: LHL and RHL both exist but are unequal, for example the greatest integer function f(x) = [x] at integer points.
  • Infinite/essential discontinuity: the function grows without bound near the point, such as f(x) = 1/x at x = 0.

Algebra of Continuous Functions

If f and g are continuous at x = a, then so are: f + g, f - g, f times g, and f/g (provided g(a) ≠ 0). Polynomial functions are continuous everywhere; sin x, cos x, eˣ, and ln x (on their domains) are continuous wherever defined. A composite of continuous functions is continuous.

Differentiability

The derivative of f at x = a is defined as the limit of the difference quotient:

f'(a) = lim(h→0) [f(a + h) - f(a)] / h

f is differentiable at a if this limit exists and is the same whether h approaches 0 from the left or the right (left-hand derivative = right-hand derivative). Geometrically, f'(a) is the slope of the tangent line at x = a.

Continuity vs Differentiability

  • Every differentiable function is continuous at that point. Differentiability is a stronger condition than continuity.
  • The converse is false: a function can be continuous but not differentiable. The classic example is f(x) = |x|, which is continuous at x = 0 but has different left-hand and right-hand derivatives there (-1 and +1), so it is not differentiable at x = 0.

Chain Rule and Implicit Differentiation

  • Chain rule: if y = f(u) and u = g(x), then dy/dx = (dy/du) × (du/dx).
  • Implicit differentiation: when y is not isolated (e.g. x² + y² = 25), differentiate both sides with respect to x, treating y as a function of x and applying the chain rule to every y term, then solve algebraically for dy/dx.

Derivatives of Inverse Trigonometric Functions

  • d/dx (sin⁻¹x) = 1/√(1-x²), for -1 < x < 1
  • d/dx (cos⁻¹x) = -1/√(1-x²), for -1 < x < 1
  • d/dx (tan⁻¹x) = 1/(1+x²)
  • d/dx (cot⁻¹x) = -1/(1+x²)
  • d/dx (sec⁻¹x) = 1/(|x|√(x²-1)), for |x| > 1
  • d/dx (cosec⁻¹x) = -1/(|x|√(x²-1)), for |x| > 1

Exponential, Logarithmic, and Logarithmic Differentiation

  • d/dx (eˣ) = eˣ; d/dx (aˣ) = aˣ ln a; d/dx (ln x) = 1/x; d/dx (logₐx) = 1/(x ln a)
  • Logarithmic differentiation is used when the variable appears in both the base and the exponent, such as y = xˣ or y = f(x)^g(x). Take ln of both sides first: ln y = g(x) ln f(x), then differentiate implicitly with respect to x and finally solve for dy/dx by multiplying back by y.
  • It is also a convenient shortcut for differentiating products and quotients of many factors.

Parametric Form and Second Order Derivatives

  • If x = f(t) and y = g(t) are both given in terms of a parameter t, then dy/dx = (dy/dt) / (dx/dt), provided dx/dt ≠ 0.
  • The second order derivative d²y/dx² is the derivative of dy/dx with respect to x; it measures concavity (d²y/dx² > 0 means concave up, < 0 means concave down).

Rolle's Theorem and Mean Value Theorem

  • Rolle's Theorem: if f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then there exists some c in (a, b) such that f'(c) = 0.
  • Lagrange's Mean Value Theorem (MVT): if f is continuous on [a, b] and differentiable on (a, b), then there exists c in (a, b) such that f'(c) = [f(b) - f(a)] / (b - a). This says some tangent is parallel to the chord joining the endpoints.

Differentiation from First Principles (ab initio)

Finding the derivative straight from the definition f'(x) = lim(h→0) [f(x+h) - f(x)]/h is called differentiation from first principles. Example: for f(x) = x², f'(x) = lim(h→0) [(x+h)² - x²]/h = lim(h→0) [2xh + h²]/h = lim(h→0)(2x + h) = 2x. Similarly d/dx(sin x) = lim(h→0) [sin(x+h) - sin x]/h = cos x, using sin(x+h) - sin x = 2 cos(x + h/2) sin(h/2) together with lim(h→0) sin(h/2)/(h/2) = 1.

Standard Derivatives (Reference Table)

f(x)f'(x)
xⁿn xⁿ⁻¹
sin x / cos xcos x / -sin x
tan x / cot xsec²x / -cosec²x
sec x / cosec xsec x tan x / -cosec x cot x
eˣ / aˣeˣ / aˣ ln a
ln x / logₐx1/x / 1/(x ln a)

Derivative of One Function with Respect to Another

To differentiate u = f(x) with respect to v = g(x), use du/dv = (du/dx)/(dv/dx). Example: differentiate sin²x with respect to cos²x. Here du/dx = 2 sin x cos x and dv/dx = -2 cos x sin x, so du/dv = (2 sin x cos x)/(-2 sin x cos x) = -1.

Differentiability of Standard Functions

  • Polynomials, sin x, cos x, eˣ, and constant functions are differentiable for every real x.
  • |x| is not differentiable at x=0 (sharp corner); [x] (greatest integer) is neither continuous nor differentiable at integers.
  • √x is not differentiable at x=0 (vertical tangent); ln x is differentiable only for x>0.
  • A function differentiable on an interval is automatically continuous there, so any point of discontinuity is automatically a point of non-differentiability.

🚀 JEE Advanced Edge

Checking differentiability at a piecewise-junction point: For a piecewise function, continuity at the junction is necessary but not sufficient for differentiability — you must separately compute the left-hand derivative (LHD) and right-hand derivative (RHD) at that point and confirm LHD=RHD. A function can be perfectly continuous at a join (no visible "jump") yet have a "kink" (like |x| at 0) where the two one-sided derivatives disagree.

Why composite functions need care with differentiability, not just continuity: If f is differentiable and g is differentiable, f∘g is differentiable (chain rule applies cleanly) — but if either f or g is merely continuous (not differentiable) at the relevant point, the composite's differentiability must be checked directly from the limit definition, since no shortcut rule guarantees it.

Worked problem: Check whether f(x) = |x-2| is differentiable at x=2. Approach: LHD = lim(h→0⁻)[f(2+h)-f(2)]/h = lim(h→0⁻)[|h|-0]/h = lim(h→0⁻)(-h)/h = -1 (since h<0, |h|=-h). RHD = lim(h→0⁺)[|h|-0]/h = lim(h→0⁺)h/h = 1. Since LHD(-1) ≠ RHD(1), f is continuous at x=2 (both one-sided limits of f itself agree, equal to 0) but NOT differentiable there.

Worked Example: Chain Rule

Differentiate y = sin(x²).

Let u = x², so y = sin u. By chain rule: dy/dx = (dy/du)(du/dx) = cos(u) · 2x = 2x cos(x²). Chain rule: differentiate the outer function (leaving the inner unchanged) and multiply by the derivative of the inner function.

Worked Example: Implicit Differentiation

If x² + y² = 25, find dy/dx.

Differentiate both sides with respect to x: 2x + 2y(dy/dx) = 0 → dy/dx = −x/y. For implicit equations, differentiate term by term; any term containing y picks up a dy/dx factor via the chain rule.

2 Revise ~3 min before the exam

📐 Formula Sheet

  • Continuity at x = a: limx→a⁻ f(x) = limx→a⁺ f(x) = f(a) — all three must exist and agree
  • Differentiability at a: the left-hand and right-hand derivatives must be equal
  • Key implication: differentiable ⇒ continuous, but continuous ⇏ differentiable (e.g. |x| at 0)
  • Chain rule: dy/dx = (dy/du)(du/dx)
  • Implicit differentiation: differentiate both sides with respect to x, then solve for dy/dx
  • Logarithmic differentiation: for y = [f(x)]g(x), take ln of both sides first
  • Parametric: dy/dx = (dy/dt)/(dx/dt)
  • Rolle's theorem: if f is continuous on [a,b], differentiable on (a,b) and f(a) = f(b), then f'(c) = 0 for some c
  • Mean value theorem: f'(c) = [f(b) − f(a)]/(b − a) for some c in (a, b)
3 Practice apply it

✍️ Worked Examples

Example 1 — Continuity of a piecewise function
Q: Find k so that f(x) = kx² for x ≤ 2 and f(x) = 3 for x > 2 is continuous at x = 2.
Step 1 — Left-hand limit: limx→2⁻ kx² = 4k, and this also equals f(2).
Step 2 — Right-hand limit: limx→2⁺ 3 = 3.
Step 3 — Continuity requires them equal: 4k = 3.
Answer: k = 3/4. Note: continuity only needs the values to meet; the slopes may still differ, so the function need not be differentiable there.

Example 2 — Continuous but not differentiable
Q: Show f(x) = |x| is continuous but not differentiable at x = 0.
Step 1 — Continuity: both one-sided limits give 0, and f(0) = 0, so it is continuous.
Step 2 — Left-hand derivative: for x < 0, f(x) = −x, so the slope is −1.
Step 3 — Right-hand derivative: for x > 0, f(x) = x, so the slope is +1.
Step 4 — Since −1 ≠ 1, the derivative does not exist at 0.
Answer: continuous everywhere, not differentiable at 0 — the graph has a sharp corner there.

Example 3 — Logarithmic differentiation
Q: Differentiate y = xx.
Step 1 — Neither the power rule nor the exponential rule applies, since both base and exponent vary. Take logs: ln y = x·ln x.
Step 2 — Differentiate both sides: (1/y)(dy/dx) = ln x + x(1/x) = ln x + 1.
Step 3 — Multiply through by y: dy/dx = y(ln x + 1).
Step 4 — Substitute back: dy/dx = xx(ln x + 1).
Answer: xx(1 + ln x). Trap: writing x·xx−1 treats the exponent as constant and is wrong.

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Frequently Asked Questions — Continuity and Differentiability

What are the key concepts in Continuity and Differentiability?
When a function has no breaks (continuity) and when it has a well-defined slope (differentiability), plus rules for differentiating implicit, inverse trig, exponential, log, and parametric functions.
Is Continuity and Differentiability important for JEE?
Yes. Continuity and Differentiability is part of the Mathematics Class 12 NCERT syllabus and is directly tested in JEE examinations. StudyHub provides structured notes, diagrams, and practice questions covering all exam-level subtopics.
How can I practice Continuity and Differentiability questions on StudyHub?
Open StudyHub and select Mathematics → Continuity and Differentiability. Choose Easy, Medium, or Hard difficulty. Hard-tier questions are at JEE level with full step-by-step explanations.

References

  1. NCERT Class 12 Mathematics Textbook — Chapter: Continuity and Differentiability
  2. CBSE Curriculum — Mathematics (Class 12)
  3. NTA JEE Main Official Syllabus — subject-wise topic list