Continuity and Differentiability

A function f is said to be continuous at a point x = a in its domain if the value of the function at that point equals the limit of the function as x approaches a. Formally, f is continuous at x = a if:

lim_x→a f(x) = f(a)

This single condition packs three separate requirements: (i) f(a) must be defined, (ii) the limit lim_x→a f(x) must exist, which means the left-hand limit (LHL) and right-hand limit (RHL) at a must be equal, and (iii) this common limit value must equal f(a). If even one of these fails, f is discontinuous at x = a.

A function is continuous on an interval if it is continuous at every point of that interval. Polynomial functions, sin x, cos x, e^x, and ln x (wherever defined) are continuous everywhere on their domains. Functions such as the greatest integer function [x] and 1/x are examples of functions that fail to be continuous at certain points (integers, and x = 0, respectively).

Algebra of continuous functions: If f and g are both continuous at x = a, then so are f + g, f − g, f·g, and (provided g(a) ≠ 0) f/g. Also, if f is continuous at a and g is continuous at f(a), the composite function g∘f is continuous at a.

Worked Example: Examine whether the function f defined by f(x) = (x² − 9)/(x − 3) for x ≠ 3, and f(3) = 6, is continuous at x = 3.

For x ≠ 3, (x² − 9)/(x − 3) = (x − 3)(x + 3)/(x − 3) = x + 3. So lim_x→3 f(x) = lim_x→3 (x + 3) = 6. Since f(3) is also given as 6, we have lim_x→3 f(x) = f(3) = 6. Hence f is continuous at x = 3.

Worked Example (discontinuity): Let f(x) = x/|x| for x ≠ 0. As x → 0⁻, f(x) = x/(−x) = −1, so LHL = −1. As x → 0⁺, f(x) = x/x = 1, so RHL = 1. Since LHL ≠ RHL, the limit at x = 0 does not exist, so however f(0) is defined, f is discontinuous at x = 0.

A function f is differentiable at x = a if the following limit exists:

f′(a) = lim_h→0 [f(a + h) − f(a)] / h

This limit is called the derivative of f at a. For the derivative to exist, the left-hand derivative Lf′(a) = lim_h→0⁻ [f(a+h)−f(a)]/h and the right-hand derivative Rf′(a) = lim_h→0⁺ [f(a+h)−f(a)]/h must both exist and be equal.

An important theorem states: every differentiable function is continuous, but the converse is not true — a function can be continuous at a point without being differentiable there. This happens whenever a function's graph has a sharp corner, cusp, or vertical tangent.

Worked Example: Show that f(x) = |x| is continuous at x = 0 but not differentiable there.

Continuity: lim_x→0|x| = 0 = f(0), so f is continuous at 0.

Differentiability: Rf′(0) = lim_h→0⁺ [|h| − 0]/h = lim_h→0⁺ h/h = 1. Lf′(0) = lim_h→0⁻ [|h| − 0]/h = lim_h→0⁻ (−h)/h = −1. Since Lf′(0) ≠ Rf′(0), f is not differentiable at x = 0, even though it is continuous there.

Worked Example (derivative from first principles): Find the derivative of f(x) = x² using the definition.

f′(x) = lim_h→0 [(x+h)² − x²]/h = lim_h→0 [2xh + h²]/h = lim_h→0 (2x + h) = 2x.

Many functions we differentiate are composite functions — a function of a function, such as sin(x²) or √(1 − x²). The chain rule lets us differentiate these systematically: if y = f(u) and u = g(x), then

dy/dx = (dy/du) · (du/dx)

In words: differentiate the "outer" function with respect to the inner expression, then multiply by the derivative of the inner expression.

Worked Example: Differentiate y = sin(x² + 5) with respect to x. Let u = x² + 5, so y = sin u. Then dy/du = cos u and du/dx = 2x. By the chain rule, dy/dx = cos(x² + 5) · 2x = 2x cos(x² + 5).

The chain rule also gives us the derivatives of the inverse trigonometric functions, which are used constantly in calculus:

• 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

Worked Example: Differentiate y = tan⁻¹(√x) with respect to x. Using the chain rule with u = √x: dy/dx = [1/(1 + u²)] · du/dx = [1/(1 + x)] · [1/(2√x)] = 1 / [2√x (1 + x)].

Not every relation between x and y is written explicitly as y = f(x); sometimes x and y are mixed together, as in x² + y² = 25. Such a relation defines y implicitly as a function of x. To find dy/dx, we differentiate both sides of the equation with respect to x, treating y as a function of x and applying the chain rule whenever a term contains y (so d/dx(y²) = 2y · dy/dx, and so on).

Worked Example: Find dy/dx if x² + y² = 25, and evaluate it at the point (3, 4). Differentiating both sides with respect to x: 2x + 2y (dy/dx) = 0, so dy/dx = −x/y. At (3, 4): dy/dx = −3/4.

Next, the derivatives of exponential and logarithmic functions:

• d/dx (e^x) = e^x
• d/dx (a^x) = a^x ln a, for a > 0
• d/dx (ln x) = 1/x, for x > 0
• d/dx (log_ax) = 1/(x ln a), for x > 0

Worked Example: Differentiate y = e^3x ln x with respect to x. Using the product rule: dy/dx = e^3x · 3 · ln x + e^3x · (1/x) = e^3x(3 ln x + 1/x).

Logarithmic differentiation is a technique used when y is a product/quotient of several factors, or when the variable appears in both the base and the exponent (such as y = x^x), where ordinary power or exponential rules cannot be applied directly. The method: take the natural logarithm of both sides, use log properties to simplify, then differentiate implicitly.

Worked Example: Differentiate y = x^x with respect to x. Taking ln of both sides: ln y = x ln x. Differentiating both sides with respect to x: (1/y)(dy/dx) = ln x + x·(1/x) = ln x + 1. So dy/dx = y(ln x + 1) = x^x(ln x + 1).

Sometimes x and y are both given as functions of a third variable (a parameter), say x = f(t) and y = g(t), rather than y being directly a function of x. To find dy/dx in this case, we use:

dy/dx = (dy/dt) / (dx/dt), provided dx/dt ≠ 0.

Worked Example: If x = a cos θ and y = a sin θ, find dy/dx. Here dx/dθ = −a sin θ and dy/dθ = a cos θ. So dy/dx = (a cos θ)/(−a sin θ) = −cot θ.

The second order derivative d²y/dx² is simply the derivative of the first derivative dy/dx with respect to x: d²y/dx² = d/dx(dy/dx). It measures how the slope of the curve itself is changing, and is used to study concavity.

Worked Example: If y = x³ − 3x, find d²y/dx² at x = 2. First derivative: dy/dx = 3x² − 3. Second derivative: d²y/dx² = 6x. At x = 2, d²y/dx² = 12.

For a curve in parametric form, finding d²y/dx² needs an extra step, because dy/dx is itself a function of the parameter t, not directly of x:

d²y/dx² = [d/dt(dy/dx)] / (dx/dt)

Worked Example: If x = t² and y = t³, find d²y/dx². First, dx/dt = 2t and dy/dt = 3t², so dy/dx = 3t²/2t = 3t/2. Differentiating dy/dx with respect to t: d/dt(3t/2) = 3/2. Dividing by dx/dt: d²y/dx² = (3/2)/(2t) = 3/(4t).

Rolle's Theorem: If a function f is (i) continuous on the closed interval [a, b], (ii) differentiable on the open interval (a, b), and (iii) f(a) = f(b), then there exists at least one point c in (a, b) such that f′(c) = 0.

Geometrically, this means that if a smooth curve starts and ends at the same height, somewhere in between it must have a horizontal tangent (a point where the slope is zero) — for instance, at a local maximum or minimum.

Worked Example: Verify Rolle's Theorem for f(x) = x² − 4x + 3 on [1, 3]. f is a polynomial, so it is continuous on [1, 3] and differentiable on (1, 3). f(1) = 1 − 4 + 3 = 0 and f(3) = 9 − 12 + 3 = 0, so f(1) = f(3). All conditions of Rolle's Theorem are satisfied. Now f′(x) = 2x − 4. Setting f′(c) = 0 gives 2c − 4 = 0, so c = 2, which indeed lies in (1, 3). Rolle's Theorem is verified.

Lagrange's Mean Value Theorem (LMVT): If a function f is (i) continuous on [a, b] and (ii) differentiable on (a, b), then there exists at least one point c in (a, b) such that:

f′(c) = [f(b) − f(a)] / (b − a)

Geometrically, this says that somewhere between a and b, the tangent to the curve is parallel to the chord joining the points (a, f(a)) and (b, f(b)). Rolle's Theorem is simply the special case of LMVT where f(a) = f(b), making the right-hand side zero.

Worked Example: Verify LMVT for f(x) = x² on [1, 4]. f is continuous on [1, 4] and differentiable on (1, 4). [f(4) − f(1)]/(4 − 1) = (16 − 1)/3 = 5. Now f′(x) = 2x, so we need 2c = 5, giving c = 2.5, which lies in (1, 4). LMVT is verified.