Continuity and Differentiability

In this chapter: continuity at a point and on an interval, algebra of continuous functions, differentiability and its link with continuity, chain rule, implicit differentiation, derivatives of inverse trigonometric, exponential and logarithmic functions, logarithmic differentiation, parametric forms, second order derivatives, and Rolle's and mean value theorems.

Continuity

A real function f is continuous at x = c (c in the domain of f) if

All three must exist and be equal. f is called a continuous function if it is continuous at every point of its domain. Note the words "of its domain": f(x) = 1/x is a continuous function, because x = 0 is not in its domain and the question of continuity there does not arise.

• Polynomials, sin x, cos x, e^x and |x| are continuous everywhere. log x is continuous for x > 0; tan x is continuous at every point except odd multiples of π/2 (which are outside its domain).
• The greatest integer function [x] is discontinuous at every integer: at x = 2, the left limit is 1 and the right limit is 2.
• If f and g are continuous at c, so are f + g, f − g, fg, and f/g provided g(c) ≠ 0.
• If g is continuous at c and f is continuous at g(c), then f∘g is continuous at c. This is why |sin x| or cos(x²) is continuous everywhere.

Worked example: Find k so that f(x) = kx + 1 for x ≤ 5 and f(x) = 3x − 5 for x > 5 is continuous at x = 5.
Solution: f(5) = LHL = 5k + 1. RHL = lim_x→5⁺(3x − 5) = 10. Continuity requires 5k + 1 = 10, so k = 9/5.

Differentiability

f is differentiable at c if the limit

exists, which means the left hand derivative (h → 0⁻) and the right hand derivative (h → 0⁺) are finite and equal. Every differentiable function is continuous, but the converse is false. The standard counterexample is |x| at 0.

Other standard cases: [x] is not differentiable at integers (it is not even continuous there), and f(x) = x|x| is differentiable at 0 with f′(0) = 0. For a piecewise function, check continuity first, then compare LHD and RHD.

Rules of differentiation

Chain rule and implicit functions

If y = f(t) and t = g(x), then dy/dx = (dy/dt)(dt/dx). For example, d/dx sin(x²) = cos(x²) · 2x. When x and y are mixed in an equation, differentiate every term with respect to x, treating y as a function of x, and collect dy/dx. From x² + y² = 25: 2x + 2y(dy/dx) = 0, so dy/dx = −x/y.

Standard derivatives

Here log means the natural logarithm (base e), as in NCERT. The derivatives of the co-functions are simply the negatives, which follows from sin⁻¹x + cos⁻¹x = π/2 and the similar identities.

A trick worth knowing: substitute to simplify before differentiating. For y = sin⁻¹(2x/(1 + x²)) with |x| ≤ 1, put x = tan θ; then y = sin⁻¹(sin 2θ) = 2θ = 2 tan⁻¹x, so dy/dx = 2/(1 + x²).

Logarithmic differentiation

Use it for a variable raised to a variable power, y = [u(x)]^v(x), and for long products and quotients. Take log of both sides, then differentiate.

Worked example: Find dy/dx if y = x^x, x > 0.
Solution: log y = x log x. Differentiating, (1/y)(dy/dx) = x · (1/x) + log x · 1 = 1 + log x. So dy/dx = x^x(1 + log x). Neither the power rule (which would give x · x^{x − 1}) nor the exponential rule (x^x log x) alone is correct; the true answer is their sum.

Parametric form

If x = f(t) and y = g(t), then dy/dx = (dy/dt) ÷ (dx/dt), provided dx/dt ≠ 0. For the circle x = a cos θ, y = a sin θ, dy/dx = (a cos θ)/(−a sin θ) = −cot θ.

Second order derivative

d²y/dx² = d/dx (dy/dx), also written y″ or y₂. In parametric form, do not divide second derivatives. Differentiate dy/dx with respect to the parameter and then divide by dx/dt once more.

Worked example: If x = at² and y = 2at, find d²y/dx².
Solution: dx/dt = 2at, dy/dt = 2a, so dy/dx = 1/t. Then d²y/dx² = (d/dt of 1/t) ÷ (dx/dt) = (−1/t²)/(2at) = −1/(2at³). The wrong method, (d²y/dt²)/(d²x/dt²) = 0/(2a) = 0, gives a completely different answer.

A common question type: if y = A sin x + B cos x, then y″ = −A sin x − B cos x = −y, so y″ + y = 0.

Rolle's theorem and the mean value theorem

These two theorems were removed from the rationalised NCERT book, but they are still in the MHT‑CET and JEE Advanced syllabus, so learn the statements and the method of finding c.

• 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 in (a, b). Geometrically, the graph has a horizontal tangent somewhere between a and b.
• Mean value theorem (Lagrange): if f is continuous on [a, b] and differentiable on (a, b), then f′(c) = (f(b) − f(a))/(b − a) for some c in (a, b). The tangent at c is parallel to the chord joining the end points.

For f(x) = x² on [2, 4], the chord slope is (16 − 4)/2 = 6, and f′(c) = 2c = 6 gives c = 3, which lies in (2, 4).

Common mistakes: (1) Checking only one side of a limit when testing continuity of a piecewise function. (2) Assuming a continuous function must be differentiable. (3) Differentiating x^x by the power rule. (4) Computing the parametric second derivative as (d²y/dt²)/(d²x/dt²). (5) Forgetting the inner derivative in the chain rule, for example writing d/dx tan⁻¹(x²) = 1/(1 + x⁴) and dropping the 2x.

JEE and MHT‑CET focus

• Finding constants that make a piecewise function continuous or differentiable at a given point.
• Points of discontinuity and non-differentiability of functions built from |x|, [x] and max or min of two functions.
• Derivatives of inverse trigonometric functions after a smart substitution (x = tan θ, sin θ, cos θ).
• Logarithmic differentiation of u^v and of long products.
• Parametric and implicit derivatives, including second derivatives, and proving relations like y″ + y = 0.
• Finding c in Rolle's theorem and the mean value theorem.

Practice questions