Class 12 Maths Chapter 5: Continuity and Differentiability

Continuity and Differentiability is Chapter 5 of NCERT Class 12 Maths. It explains when a function is continuous or differentiable at a point and then builds a full toolkit of differentiation: the chain rule, inverse trigonometric functions, implicit and parametric functions, exponential and logarithmic functions, and second order derivatives. Watch the video lesson below, then use these notes, formulas and solved examples to practise for your CBSE board exam.

What you will learn

The meaning of continuity at a point and how to test it using left and right hand limits
The link between differentiability and continuity
The chain rule and derivatives of inverse trigonometric functions
Differentiating implicit, exponential, logarithmic and parametric functions
Using logarithmic differentiation and finding second order derivatives

Video Lessons

Continuity and Differentiability - Part 1

Key Concepts and Formulas

Continuity of a Function

A function is continuous at a point if its graph has no break there. Mathematically, the left hand limit, the right hand limit and the value of the function at that point must all be equal.

f is continuous at x = c if
lim(x→c⁻) f(x) = lim(x→c⁺) f(x) = f(c)

A function is continuous if it is continuous at every point of its domain.
Polynomial, sine, cosine, exponential and logarithmic functions are continuous on their domains.
If f and g are continuous, then f + g, f − g, fg and f/g (where g ≠ 0) are also continuous.

Differentiability

f is differentiable at x = c if the left hand derivative and the right hand derivative at c exist and are equal.

f′(c) = lim(h→0) [f(c + h) − f(c)] / h
LHD = lim(h→0⁻) [f(c + h) − f(c)] / h
RHD = lim(h→0⁺) [f(c + h) − f(c)] / h
Differentiable at c ⇔ LHD = RHD

Graph of y equals mod x showing a corner at zero where it is not differentiable, Class 12 Maths — The modulus function has a sharp corner at x = 0

Every differentiable function is continuous, but the converse is not true.
Example: f(x) = |x| is continuous at 0, but its left hand derivative is −1 and right hand derivative is 1, so it is not differentiable at 0.

Chain Rule and Inverse Trigonometric Derivatives

The chain rule handles a function inside another function: differentiate the outer function, keep the inside unchanged, and multiply by the derivative of the inside.

If y = f(t) and t = g(x):
dy/dx = (dy/dt) × (dt/dx)
d/dx (sin⁻¹x) = 1/√(1 − x²),   −1 < x < 1
d/dx (cos⁻¹x) = −1/√(1 − x²),  −1 < x < 1
d/dx (tan⁻¹x) = 1/(1 + x²)
d/dx (cot⁻¹x) = −1/(1 + x²)

Implicit, Exponential and Logarithmic Functions

When y is not written separately as a function of x, differentiate both sides with respect to x, treating y as a function of x, and then collect the dy/dx terms. In NCERT, log x means the natural logarithm.

d/dx (eˣ) = eˣ
d/dx (log x) = 1/x,      x > 0
d/dx (aˣ) = aˣ log a,    a > 0
d/dx (y²) = 2y (dy/dx)   (implicit)
y = u^v ⇒ log y = v log u

Logarithmic differentiation: take log on both sides first. It is useful for products of many factors and for a function raised to another function.

Parametric Functions and Second Order Derivatives

When x and y are both given in terms of a third variable t (a parameter), find each derivative separately and divide.

If x = f(t) and y = g(t):
dy/dx = (dy/dt) / (dx/dt),  dx/dt ≠ 0
d²y/dx² = d/dx (dy/dx)

The second order derivative is also written y″ or y₂.

Solved Examples

Example 1: Find k if f(x) = kx + 1 for x ≤ 2 and f(x) = 3x − 1 for x > 2 is continuous at x = 2.

LHL = lim(x→2⁻) (kx + 1) = 2k + 1
RHL = lim(x→2⁺) (3x − 1) = 5
f(2) = 2k + 1
For continuity: 2k + 1 = 5
∴ k = 2

Graphs comparing a continuous piecewise function and a jump discontinuity at x = 2, Class 12 Continuity — Equal limits give continuity, unequal limits give a jump

Example 2: Differentiate y = sin(x³ + 2x).

Let t = x³ + 2x, so y = sin t
dy/dt = cos t,  dt/dx = 3x² + 2
dy/dx = cos t × (3x² + 2)
∴ dy/dx = (3x² + 2) cos(x³ + 2x)

Example 3: Find dy/dx if x² + xy + y² = 7.

x² + xy + y² = 7
2x + (y + x dy/dx) + 2y dy/dx = 0
(x + 2y) dy/dx = −(2x + y)
∴ dy/dx = −(2x + y)/(x + 2y)

Example 4: Differentiate y = sin⁻¹(x²).

dy/dx = 1/√(1 − (x²)²) × d/dx (x²)
      = 1/√(1 − x⁴) × 2x
∴ dy/dx = 2x/√(1 − x⁴)

Example 5: Differentiate y = (2x + 1)³(x − 4)² using logarithms.

log y = 3 log(2x + 1) + 2 log(x − 4)
(1/y) dy/dx = 3 × 2/(2x + 1) + 2/(x − 4)
            = 6/(2x + 1) + 2/(x − 4)
∴ dy/dx = (2x + 1)³(x − 4)²
          × [6/(2x + 1) + 2/(x − 4)]

Example 6: If x = 2t² and y = t³ + 1, find dy/dx at t = 2.

dx/dt = 4t
dy/dt = 3t²
dy/dx = 3t² / 4t = 3t/4
At t = 2: dy/dx = 3 × 2/4
∴ dy/dx = 3/2

Example 7: Find the second derivative of y = e²ˣ + 3 sin x.

y = e²ˣ + 3 sin x
dy/dx = 2e²ˣ + 3 cos x
∴ d²y/dx² = 4e²ˣ − 3 sin x

Common Mistakes to Avoid

For continuity, check all three values: left hand limit, right hand limit and f(c). Matching only the two limits is not enough.
Do not assume a continuous function is differentiable; sharp corners like |x| at 0 break differentiability.
In implicit differentiation, the derivative of y² is 2y (dy/dx), not just 2y; missing dy/dx is the most common error.
In parametric questions, divide dy/dt by dx/dt in the right order, and simplify before substituting the value of t.

Quick Revision

Continuous at c: LHL = RHL = f(c)
Chain rule: dy/dx = (dy/dt)(dt/dx)
d/dx sin⁻¹x = 1/√(1 − x²)
d/dx tan⁻¹x = 1/(1 + x²)
d/dx eˣ = eˣ,  d/dx log x = 1/x
Parametric: dy/dx = (dy/dt)/(dx/dt)