01 · Explore
The Concept of Continuity
Continuity is a fundamental property of functions that describes their behavior without sudden jumps or breaks. Naively, a function is continuous if its graph can be drawn without lifting the pen from the paper.
Mathematically, a real function f is continuous at a point c in its domain if the limit of the function as x approaches c is equal to the value of the function at c. This is expressed as lim (x → c) f(x) = f(c). For this to hold, three conditions must be met: the function must be defined at c, the limit of f(x) as x approaches c must exist, and this limit must equal f(c).
More elaborately, if the left hand limit (LHL), right hand limit (RHL), and the value of the function at x = c exist and are equal to each other, then f is said to be continuous at x = c. If any of these conditions fail, the function is called discontinuous at c, and c is termed a point of discontinuity.
A function is said to be a continuous function if it is continuous at every point in its domain. For functions defined on a closed interval [a, b], continuity at the endpoints is defined using one-sided limits: the right hand limit at 'a' must equal f(a) and the left hand limit at 'b' must equal f(b). If a function's domain is a singleton (a single point), it is considered continuous there by definition.
Checking Continuity of a Linear Function
f(x) = 2x + 3 at x = 1
1. Find the value at x = 1: f(1) = 2(1) + 3 = 5. 2. Find the limit: lim (x → 1) (2x + 3) = 2(1) + 3 = 5. 3. Since lim (x → 1) f(x) = f(1), the function is continuous at x = 1.
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Is the function f(x) = x² continuous at x = 0?
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02 · Explore
Algebra of Continuous Functions
Continuous functions follow specific algebraic rules, allowing us to determine the continuity of complex functions by breaking them down into simpler components.
Theorem 1 states that if f and g are two real functions continuous at a real number c, then their sum (f + g), difference (f - g), and product (f . g) are also continuous at x = c. The quotient (f/g) is also continuous at x = c, provided that g(c) is not zero.
As a consequence of these rules, several standard functions are always continuous within their domains. Every polynomial function is continuous at every real number. Every rational function (the ratio of two polynomials) is continuous wherever its denominator is non-zero.
Trigonometric functions like sine and cosine are continuous for all real numbers. Functions like tangent, cotangent, secant, and cosecant are continuous wherever they are defined (i.e., where their respective denominators are not zero).
| Function Type | Continuity Domain | Reasoning |
|---|---|---|
| Polynomial | All Real Numbers (R) | Defined by sum/product of identity and constant functions |
| Rational (p/q) | R except where q(x) = 0 | Quotient of two continuous polynomial functions |
| Sine / Cosine | All Real Numbers (R) | Fundamental trigonometric properties |
| Tangent | R except (2n+1)π/2 | Quotient of sin x / cos x; cos x must not be zero |
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If f(x) is continuous and k is a constant, is k . f(x) continuous?
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03 · Explore
Continuity of Composite Functions
The composition of functions preserves continuity under specific conditions.
Theorem 2 states that if g is continuous at c and f is continuous at g(c), then the composite function (f o g) is continuous at c. This allows us to prove the continuity of complex expressions like sin(x²) or |1 - x + |x||.
For example, to show f(x) = sin(x²) is continuous, we identify it as a composition of g(x) = sin x and h(x) = x². Since both g and h are continuous everywhere, their composition is also continuous.
Continuity of Absolute Value Composition
f(x) = |1 - x + |x||
Let g(x) = 1 - x + |x| and h(x) = |x|. Then f(x) = (h o g)(x). Since g(x) is the sum of a polynomial and a modulus function (both continuous), and h(x) is a modulus function (continuous), the composition f(x) is continuous.
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Question
Is f(x) = cos(x²) a continuous function?
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04 · Explore
Differentiability and its Relation to Continuity
Differentiability refers to the existence of a derivative at a point, representing the rate of change of the function.
A function f is differentiable at a point c in its domain if the limit of [f(c + h) - f(c)] / h as h approaches 0 exists. This limit is called the derivative of f at c, denoted as f'(c). For the limit to exist, the Left Hand Derivative (LHD) and Right Hand Derivative (RHD) must both be finite and equal.
A crucial relationship exists between these two concepts: Theorem 3 states that if a function is differentiable at a point c, it must also be continuous at that point. Thus, differentiability is a stronger condition than continuity.
However, the converse is not true. A function can be continuous at a point but not differentiable there. A classic example is the modulus function f(x) = |x| at x = 0. While it is continuous at 0, the LHD is -1 and the RHD is 1. Since they are not equal, the function is not differentiable at x = 0.
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True or False: Every continuous function is differentiable.
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05 · Explore
The Chain Rule for Composite Functions
The chain rule is a powerful technique used to differentiate composite functions.
If a function f is a composite of two functions u and v (f = v o u), and we let t = u(x), then the derivative of f with respect to x is given by df/dx = (dv/dt) . (dt/dx). This allows us to differentiate functions 'layer by layer' from the outside in.
This rule can be extended to compositions of three or more functions. For f = (w o u) o v, where t = v(x) and s = u(t), the derivative is df/dx = (dw/ds) . (ds/dt) . (dt/dx).
Common mistakes often involve forgetting to multiply by the derivative of the inner function (the 'dt/dx' part). Always identify the inner and outer functions clearly before applying the rule.
The Chain Rule Process
- 1
Identify Layers
Break the function into outer function v(t) and inner function t = u(x).
- 2
Differentiate Outer
Find dv/dt (derivative of outer with respect to inner variable).
- 3
Differentiate Inner
Find dt/dx (derivative of inner with respect to x).
- 4
Multiply
Multiply the results: (dv/dt) times (dt/dx).
- 5
Substitute Back
Replace the intermediate variable t with the original expression u(x).
Sequential steps to differentiate a composite function using the Chain Rule.
Differentiating sin(x²)
f(x) = sin(x²)
1. Let t = x² (inner) and v = sin t (outer). 2. dt/dx = 2x. 3. dv/dt = cos t. 4. df/dx = (cos t) . (2x). 5. Substitute t: df/dx = 2x cos(x²).
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Find the derivative of (2x + 1)³ using the chain rule.
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06 · Explore
Implicit Differentiation
Implicit differentiation is used when y is not explicitly isolated as a function of x, but is instead part of an equation involving both variables.
In an explicit function like y = x² + 1, y is clearly isolated. In an implicit relation like x² + y² = 100 or y + sin y = cos x, y is defined implicitly. To find dy/dx, we differentiate every term in the equation with respect to x, treating y as a function of x and applying the chain rule where necessary.
When differentiating a term involving y, such as sin y, we apply the chain rule: d/dx (sin y) = (d/dy sin y) . (dy/dx) = cos y . (dy/dx). After differentiating all terms, we collect all terms containing dy/dx on one side and solve for it.
Differentiating y + sin y = cos x
y + sin y = cos x
1. Differentiate w.r.t x: d/dx(y) + d/dx(sin y) = d/dx(cos x). 2. Apply chain rule: dy/dx + (cos y . dy/dx) = -sin x. 3. Factor out dy/dx: dy/dx (1 + cos y) = -sin x. 4. Solve: dy/dx = -sin x / (1 + cos y).
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Question
Find dy/dx if x - y = π.
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07 · Explore
Derivatives of Inverse Trigonometric Functions
Inverse trigonometric functions are continuous, and their derivatives can be found using implicit differentiation and the chain rule.
To find the derivative of y = sin⁻¹ x, we rewrite it as x = sin y. Differentiating both sides with respect to x gives 1 = cos y . (dy/dx). Thus, dy/dx = 1/cos y. Using the identity cos² y = 1 - sin² y, and knowing sin y = x, we get cos y = √(1 - x²). Therefore, d/dx (sin⁻¹ x) = 1/√(1 - x²).
Similar derivations lead to the derivatives of other inverse functions. Note the restricted domains for these derivatives to ensure the expressions are well-defined (e.g., denominators cannot be zero).
| Function f(x) | Derivative f'(x) | Domain |
|---|---|---|
| sin⁻¹ x | 1 / √(1 - x²) | (-1, 1) |
| cos⁻¹ x | -1 / √(1 - x²) | (-1, 1) |
| tan⁻¹ x | 1 / (1 + x²) | R |
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Question
What is the derivative of tan⁻¹ x?
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08 · Explore
Exponential and Logarithmic Functions
Exponential functions grow faster than any polynomial, and logarithmic functions are their inverses.
An exponential function with base b > 1 is y = bˣ. Its domain is all real numbers, and its range is all positive real numbers. The natural exponential function uses the base 'e' (approximately 2.718), denoted as y = eˣ. A unique property of eˣ is that its derivative is itself: d/dx (eˣ) = eˣ.
The logarithmic function is the inverse of the exponential function. log_b(a) = x if bˣ = a. The natural logarithm (base e) is denoted as log x or ln x. Its domain is (0, ∞) and its range is R. The derivative of log x is 1/x.
Key properties of logarithms include: log(pq) = log p + log q, log(p/q) = log p - log q, and log(pⁿ) = n log p. These properties are essential for logarithmic differentiation.
Differentiating sin(log x)
y = sin(log x)
Using chain rule: dy/dx = cos(log x) . d/dx(log x). Since d/dx(log x) = 1/x, the result is cos(log x) / x.
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What is the derivative of e^(-x)?
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09 · Explore
Logarithmic Differentiation
This technique is used for functions of the form y = [u(x)]^v(x) or for complex products and quotients.
When a function has a variable in both the base and the exponent, standard power rules do not apply. Instead, we take the natural logarithm of both sides: log y = v(x) log[u(x)]. We then differentiate implicitly.
The process involves: 1. Taking log on both sides. 2. Using log properties to simplify. 3. Differentiating both sides w.r.t x (using product and chain rules). 4. Multiplying by y to isolate dy/dx.
- 1
Let y = f(x).
- 2
Take natural log: log y = log [f(x)].
- 3
Differentiate: (1/y) . (dy/dx) = d/dx [log f(x)].
- 4
Solve for dy/dx: dy/dx = y . d/dx [log f(x)].
dy/dx = f(x) . d/dx [log f(x)]
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Question
Differentiate aˣ where a is a constant.
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10 · Explore
Derivatives of Functions in Parametric Form
Sometimes x and y are both expressed in terms of a third variable, called a parameter.
If x = f(t) and y = g(t), then dy/dx can be found without eliminating the parameter t. By the chain rule, dy/dt = (dy/dx) . (dx/dt).
Rearranging this gives the formula: dy/dx = (dy/dt) / (dx/dt), provided dx/dt is not zero. The final result is usually expressed in terms of the parameter t.
Differentiation of a Circle in Parametric Form
x = a cos θ, y = a sin θ
1. dx/dθ = -a sin θ. 2. dy/dθ = a cos θ. 3. dy/dx = (a cos θ) / (-a sin θ) = -cot θ.
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Question
Find dy/dx if x = at², y = 2at.
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11 · Explore
Second Order Derivatives
The second order derivative is the derivative of the first derivative.
If y = f(x), the first derivative is dy/dx = f'(x). If f'(x) is also differentiable, its derivative with respect to x is called the second order derivative of y, denoted as d²y/dx² or f''(x).
Higher order derivatives are defined similarly by repeated differentiation. Second order derivatives are often used in physics to represent acceleration (the rate of change of velocity) and in calculus to determine the concavity of a curve.
Finding Second Order Derivative
y = x³ + tan x
1. First derivative: dy/dx = 3x² + sec² x. 2. Second derivative: d/dx(3x² + sec² x) = 6x + 2 sec x . (sec x tan x) = 6x + 2 sec² x tan x.
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Question
If y = sin x, what is d²y/dx²?
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