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Class 12 · Mathematics · Part 1

Chapter 5: Continuity and Differentiability

Revision notes, worked problems, a question bank, assertion and reason questions and answers to the NCERT exercises. Tap a group to open it; opening one closes the others.

5.2ContinuityAt a point, one-sided limits and algebra of continuous functions

Definition

A function f is continuous at x = c if

limx→c f(x) = f(c)

Three things must hold: f(c) is defined, the limit exists, and the two are equal. In terms of one-sided limits: LHL = RHL = f(c), where LHL = limx→c⁻ f(x) and RHL = limx→c⁺ f(x). If f is not continuous at c, then f is discontinuous at c.

Example: f(x) = 2x² − 1 at x = 3 has lim f(x) = 17 = f(3), so f is continuous at 3.

How continuity can fail

Continuouslimit = valuef(c)Removablelimit ≠ valueLHLRHLJumpleft ≠ rightInfinitelimit blows up
At x = c, a function is continuous only when the limit exists and equals f(c). Removable, jump and infinite breaks all violate this.
  • Removable: the limit exists but is not equal to f(c) (or f(c) is not defined). Changing the value at c repairs it.
  • Jump: the left and right limits both exist but are different. The greatest integer function [x] jumps at every integer.
  • Infinite: the function blows up near c, as 1/x does at 0.

Continuous on an interval

A function is continuous on an interval if it is continuous at every point of the interval. A function that is continuous on its whole domain is simply called continuous. Its graph can be drawn without lifting the pencil, which is the idea that the definition makes precise.

Example: f(x) = 1/x is continuous at every point of its domain (x ≠ 0). We do not call it discontinuous at 0, because 0 is not in the domain.

The greatest integer function

123−112xyf(x) = [x]At each integer n:left limit = n − 1right limit = n = f(n)so a jump at every integer
A filled dot means the value is attained, an open circle means it is not. At every integer the graph jumps up by 1.

For g(x) = x − [x] at an integer n, the left limit is 1 and the right limit is 0, so g is discontinuous at every integer.

Algebra of continuous functions

  • If f and g are continuous at c, then so are f + g, f − g, f · g and kf (k constant).
  • f/g is continuous at c if g(c) ≠ 0.
  • Composite: if g is continuous at c and f is continuous at g(c), then f∘g is continuous at c.

So |cos x|, cos(x²) and sin|x| are continuous, because they are composites of continuous functions.

Standard continuous functions

  • Polynomial functions: continuous at every real number.
  • Rational functions p(x)/q(x): continuous wherever q(x) ≠ 0.
  • sin x, cos x, eˣ: continuous at every real number.
  • tan x, cot x, sec x, cosec x: continuous at every point of their domains.
  • log x: continuous for x > 0. Inverse trigonometric functions are continuous on their domains.

Finding the unknown constant

To make a piecewise function continuous at the join, set the left limit equal to the right limit and to the value at the join, then solve.

Example: f(x) = kx² for x ≤ 2 and 3 for x > 2. LHL = 4k and RHL = 3, so 4k = 3 and k = 3/4.

If the join conditions involve two unknowns, write one equation per join. For f = 5 (x ≤ 2), ax + b (2 < x < 10), 21 (x ≥ 10): 2a + b = 5 and 10a + b = 21 give a = 2, b = 1.

5.3DifferentiabilityDerivative, chain rule, implicit functions and inverse trig functions

Definition

The derivative of f at c is

f′(c) = limh→0 [f(c + h) − f(c)]/h

provided the limit exists. The left hand derivative uses h → 0⁻ and the right hand derivative uses h → 0⁺. f is differentiable at c when the two are equal.

Differentiable implies continuous

Every differentiable function is continuous, because f(x) − f(c) = [(f(x) − f(c))/(x − c)] · (x − c) tends to f′(c) · 0 = 0. The converse is false.

xy−22slope −1slope +1corner at x = 0y = |x|continuous at 0LHD = −1, RHD = 1not differentiable
The two one-sided derivatives at the corner are −1 and 1, so no single tangent exists. The graph has no break, so it is still continuous.

f(x) = |x| is continuous at 0 but its left and right derivatives are −1 and 1, so it is not differentiable at 0.

Rules of differentiation

  • Sum: (u + v)′ = u′ + v′. Product: (uv)′ = u′v + uv′. Quotient: (u/v)′ = (u′v − uv′)/v².
  • (xⁿ)′ = n xⁿ⁻¹, (sin x)′ = cos x, (cos x)′ = −sin x, (tan x)′ = sec²x, (sec x)′ = sec x tan x.
  • These sum, product and quotient rules apply wherever the functions involved are differentiable.

Chain rule

xinputu = g(x)inner stepy = f(u)outer stepdu/dxdy/dudy/dx = (dy/du) · (du/dx)differentiate the outside, keep the inside, then multiply by the derivative of the insideExample: y = sin(x²): (cos x²)(2x)
For a composite function the derivative is a product, one factor per link in the chain.

If y = f(u) and u = g(x), then dy/dx = (dy/du)(du/dx). For three links, y = f(u), u = g(v), v = h(x): dy/dx = (dy/du)(du/dv)(dv/dx).

Example: y = cos(√x). Outer: −sin(√x). Inner: 1/(2√x). So dy/dx = −sin(√x)/(2√x).

Implicit functions

When y cannot easily be solved for, differentiate both sides with respect to x, treating y as a function of x (so d/dx y² = 2y dy/dx), and then solve for dy/dx.

Example: x² + y² = 25. 2x + 2y dy/dx = 0, so dy/dx = −x/y.

Derivatives of inverse trigonometric functions

f(x)f′(x)Domain of f′
sin⁻¹x1/√(1 − x²)−1 < x < 1
cos⁻¹x−1/√(1 − x²)−1 < x < 1
tan⁻¹x1/(1 + x²)all real x

Reason for sin⁻¹: if y = sin⁻¹x then x = sin y, so 1 = cos y · dy/dx and dy/dx = 1/cos y = 1/√(1 − x²), because cos y > 0 on the principal branch.

Simplify first by a substitution

Inverse trigonometric expressions often become simple after a substitution, and then no heavy differentiation is needed.

ExpressionSubstituteSimplifies to
sin⁻¹(2x/(1 + x²))x = tan θ2 tan⁻¹x (|x| < 1)
cos⁻¹((1 − x²)/(1 + x²))x = tan θ2 tan⁻¹x (0 ≤ x < 1)
tan⁻¹((3x − x³)/(1 − 3x²))x = tan θ3 tan⁻¹x
sin⁻¹(2x√(1 − x²))x = sin θ2 sin⁻¹x (|x| < 1/√2)

Always check the range of θ. The result sin⁻¹(sin 2θ) = 2θ is true only when 2θ lies in [−π/2, π/2].

5.4Exponential and logarithmic functionsDerivatives of eˣ, aˣ and log x

Standard results

xy−1123123(0, 1)(1, 0)y = eˣslope at (0, 1) is 1y = log xslope at (1, 0) is 1y = x
log x is the inverse of eˣ, so the graphs reflect in y = x. Since (eˣ)′ = eˣ, the slope of log x at x is 1/x.
f(x)f′(x)
eˣeˣ
aˣ (a > 0)aˣ log a
log x (x > 0)1/x
loga x1/(x log a)

With the chain rule: d/dx e^(g(x)) = e^(g(x)) g′(x) and d/dx log g(x) = g′(x)/g(x).

Examples

  • d/dx e⁻ˣ = −e⁻ˣ.
  • d/dx log(cos eˣ) = (−sin eˣ · eˣ)/cos eˣ = −eˣ tan eˣ.
  • d/dx log(log x) = 1/(x log x), for x > 1.
  • d/dx e^(sin⁻¹x) = e^(sin⁻¹x)/√(1 − x²).
5.5Logarithmic differentiationVariable powers, long products and quotients

When and how

Use it for functions of the form y = [u(x)]^v(x) (variable base and variable exponent) and for long products or quotients. Both u and y must be positive for the logarithm to make sense.

  • Take logarithms: log y = v(x) log u(x).
  • Differentiate both sides: (1/y) dy/dx = v′ log u + v u′/u.
  • Multiply by y: dy/dx = [u]^v ( v′ log u + v u′/u ).

Example

y = xˣ. Then log y = x log x, so (1/y) dy/dx = log x + 1 and dy/dx = xˣ(1 + log x).

Compare: d/dx xⁿ = n xⁿ⁻¹ needs a constant exponent, and d/dx aˣ = aˣ log a needs a constant base. xˣ needs logarithmic differentiation.

Long products and quotients

For y = (x + 3)²(x + 4)³(x + 5)⁴, take logs: log y = 2 log(x + 3) + 3 log(x + 4) + 4 log(x + 5), so dy/dx = y [2/(x + 3) + 3/(x + 4) + 4/(x + 5)].

Two such terms added

If y = u + v and each term has a variable power, find u′ and v′ separately by logarithmic differentiation, then add. Taking the log of the sum does not help.

5.6Parametric formsx and y both given in terms of a third variable

Rule

If x = f(t) and y = g(t), then

dy/dx = (dy/dt)/(dx/dt), dx/dt ≠ 0

This follows from the chain rule. There is no need to eliminate t.

Examples

  • x = a cos θ, y = a sin θ: dy/dx = (a cos θ)/(−a sin θ) = −cot θ.
  • x = at², y = 2at: dy/dx = 2a/(2at) = 1/t.
  • x = a(θ − sin θ), y = a(1 + cos θ): dy/dx = −sin θ/(1 − cos θ) = −cot(θ/2).

Common simplifications

Use 1 − cos θ = 2 sin²(θ/2), 1 + cos θ = 2 cos²(θ/2) and sin θ = 2 sin(θ/2) cos(θ/2) to get half-angle results.

5.7Second order derivativeDifferentiating twice

Definition

If y = f(x), then dy/dx = f′(x). Differentiating again gives the second order derivative d²y/dx² = f″(x), also written y″ or y₂.

Example: y = x³ log x. y′ = 3x² log x + x², and y″ = 6x log x + 5x.

Proving a differential relation

To show a relation like (1 − x²)y″ − xy′ = 0 for y = sin⁻¹x, differentiate once, rearrange to remove the square root if needed, and differentiate again.

  • From y₁ = 1/√(1 − x²): (1 − x²)y₁² = 1.
  • Differentiate: (1 − x²) · 2y₁y₂ + y₁²(−2x) = 0, and dividing by 2y₁: (1 − x²)y₂ − xy₁ = 0.

Second derivative in parametric form

If x = f(t), y = g(t): first find dy/dx = (dy/dt)/(dx/dt). Then d²y/dx² = [d/dt (dy/dx)]/(dx/dt). Do not differentiate dy/dx with respect to x directly.

Rolle's theorem and the mean value theorem are not part of this reprint (2026-27) of the chapter. The exercises here follow the reprint.

RefQuick referenceFormulas and common slips

Formulas

ResultCondition
f continuous at c ⇔ LHL = RHL = f(c)c in the domain
differentiable ⇒ continuousconverse is false
(sin⁻¹x)′ = 1/√(1 − x²)−1 < x < 1
(cos⁻¹x)′ = −1/√(1 − x²)−1 < x < 1
(tan⁻¹x)′ = 1/(1 + x²)all real x
(eˣ)′ = eˣ, (aˣ)′ = aˣ log aa > 0
(log x)′ = 1/xx > 0
dy/dx = (dy/dt)/(dx/dt)dx/dt ≠ 0
(xˣ)′ = xˣ(1 + log x)x > 0

Common slips

  • Writing d/dx (xˣ) = x xˣ⁻¹ (the exponent is not a constant).
  • Forgetting the chain-rule factor g′(x) on e^(g(x)), sin g(x) and log g(x).
  • Saying |x| is differentiable at 0 because it is continuous there.
  • Forgetting dy/dx when differentiating y² in an implicit relation.
  • Using a substitution like x = tan θ without checking the range of θ.
  • Taking the log of a function that is negative or zero.
  • Computing d²y/dx² in parametric form as d/dt of dy/dx without dividing by dx/dt.