Historical Chronicle / Epigraph: Albert Einstein
Historical Chronicle / Epigraph: Sir Isaac Newton
This chapter introduces the fundamental concepts of continuity and differentiability of real-valued functions, establishing the mathematical framework for calculus.
Suppose $f$ is a real function on a subset of the real numbers and let $c$ be a point in the domain of $f$. Then $f$ is continuous at $c$ if:
$$\lim_{x \to c} f(x) = f(c)$$More elaborately, a function is continuous at $x = c$ if the left hand limit (LHL), right hand limit (RHL), and the value of the function at $x = c$ exist and are all equal to each other:
$$\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$$If $f$ is not continuous at $c$, we say $f$ is discontinuous at $c$, and $c$ is called a point of discontinuity of $f$.
A real function $f$ is said to be continuous if it is continuous at every point in its domain.
Mathematical Problem Statement: Discuss the continuity of the function $f$ given by $f(x) = |x| at x = 0$.
Step-by-Step Algorithmic Solution Trace:
- By definition, the modulus function is:$$f(x) = \begin{cases} -x, & \text{if } x < 0 \\ x, & \text{if } x \ge 0 \end{cases}$$
- Find the Left Hand Limit (LHL) at $x = 0$:$$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x) = 0$$
- Find the Right Hand Limit (RHL) at $x = 0$:$$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x) = 0$$
- Find the value of the function at $x = 0$:$$f(0) = 0$$
- Since LHL = RHL = $f(0) = 0$, the function is continuous at $x = 0$.
Final Evaluation Output: $f(x) = |x|$ is continuous at $x = 0$.
Mathematical Problem Statement: Show that the greatest integer function $f(x) = [x]$ is discontinuous at every integral point.
Step-by-Step Algorithmic Solution Trace:
- Let $c$ be any integer. By definition, $[x]$ denotes the greatest integer less than or equal to $x$.
- Evaluate the Left Hand Limit (LHL) as $x \to c^-$:$$\lim_{x \to c^-} f(x) = \lim_{x \to c^-} [x] = c - 1$$
- Evaluate the Right Hand Limit (RHL) as $x \to c^+$:$$\lim_{x \to c^+} f(x) = \lim_{x \to c^+} [x] = c$$
- Since LHL $\neq$ RHL, the limit of $f(x)$ as $x \to c$ does not exist.
Final Evaluation Output: $f(x) = [x]$ is discontinuous at every integer point.
The algebraic combinations of continuous functions preserve continuity under specific conditions.
Suppose $f$ and $g$ are two real functions continuous at a real number $c$. Then:
| Function Type | Domain of Continuity | Continuity Status |
|---|---|---|
| Polynomial Function $p(x) = a_0 + a_1 x + \dots + a_n x^n$ | $\mathbb{R}$ | Continuous everywhere |
| Rational Function $f(x) = \frac{p(x)}{q(x)}$ | $\{x \in \mathbb{R} : q(x) \neq 0\}$ | Continuous in its domain |
| Sine Function $f(x) = \sin x$ | $\mathbb{R}$ | Continuous everywhere |
| Cosine Function $f(x) = \cos x$ | $\mathbb{R}$ | Continuous everywhere |
| Tangent Function $f(x) = \tan x$ | $\{x \in \mathbb{R} : x \neq (2n+1)\frac{\pi}{2}, n \in \mathbb{Z}\}$ | Continuous in its domain |
Suppose $f$ and $g$ are real-valued functions such that $(f \circ g)$ is defined at $c$. If $g$ is continuous at $c$ and $f$ is continuous at $g(c)$, then $(f \circ g)$ is continuous at $c$.
Mathematical Problem Statement: Show that $f(x) = \sin(x^2)$ is a continuous function.
Step-by-Step Algorithmic Solution Trace:
- Express $f(x)$ as a composition of two functions $g$ and $h$, i.e., $f = g \circ h$.
- Let $h(x) = x^2$ (a polynomial function, hence continuous everywhere).
- Let $g(x) = \sin x$ (a trigonometric function, hence continuous everywhere).
- Since both $g$ and $h$ are continuous everywhere, their composition $(g \circ h)(x) = g(h(x)) = \sin(x^2)$ is also continuous everywhere.
Final Evaluation Output: $f(x) = \sin(x^2)$ is continuous everywhere.
Differentiability measures the rate of change of a function and is a stronger condition than continuity.
Let $f$ be a real function and $c$ be a point in its domain. The derivative of $f$ at $c$ is defined by:
$$f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}$$provided this limit exists. A function is differentiable in an interval $[a, b]$ if it is differentiable at every point in $(a, b)$, and has one-sided derivatives at the endpoints.
If a function $f$ is differentiable at a point $c$, then it is also continuous at that point.
Note: The converse is not true. For example, $f(x) = |x|$ is continuous at $x = 0$ but not differentiable at $x = 0$ because the left-hand derivative ($-1$) and right-hand derivative ($1$) do not coincide.
Let $f$ be a composite of two functions $u$ and $v$, i.e., $f = v \circ u$. Let $t = u(x)$. If both $\frac{dt}{dx}$ and $\frac{dv}{dt}$ exist, then:
$$\frac{df}{dx} = \frac{dv}{dt} \cdot \frac{dt}{dx}$$Mathematical Problem Statement: Find the derivative of $f(x) = \sin(x^2)$ with respect to $x$.
Step-by-Step Algorithmic Solution Trace:
- Let $t = u(x) = x^2$ and $y = v(t) = \sin t$.
- Compute the derivatives:$$\frac{dt}{dx} = 2x \quad \text{and} \quad \frac{dy}{dt} = \cos t$$
- Apply the Chain Rule:$$\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \cos t \cdot 2x$$
- Substitute $t = x^2$ back into the expression:$$\frac{dy}{dx} = 2x \cos(x^2)$$
Final Evaluation Output: $\frac{d}{dx}(\sin(x^2)) = 2x \cos(x^2)$
When a relationship between $x$ and $y$ is expressed implicitly (e.g., $y + \sin y = \cos x$), we differentiate both sides directly with respect to $x$ and solve for $\frac{dy}{dx}$.
Mathematical Problem Statement: Find $\frac{dy}{dx}$ if $y + \sin y = \cos x$.
Step-by-Step Algorithmic Solution Trace:
- Differentiate both sides of the equation with respect to $x$:$$\frac{d}{dx}(y) + \frac{d}{dx}(\sin y) = \frac{d}{dx}(\cos x)$$
- Apply the chain rule to the term involving $y$:$$\frac{dy}{dx} + \cos y \cdot \frac{dy}{dx} = -\sin x$$
- Factor out $\frac{dy}{dx}$:$$\frac{dy}{dx}(1 + \cos y) = -\sin x$$
- Solve for $\frac{dy}{dx}$:$$\frac{dy}{dx} = -\frac{\sin x}{1 + \cos y} \quad \text{where } y \neq (2n+1)\pi$$
Final Evaluation Output: $\frac{dy}{dx} = -\frac{\sin x}{1 + \cos y}$
| Function $f(x)$ | Derivative $f'(x)$ | Domain |
|---|---|---|
| $\sin^{-1} x$ | $\frac{1}{\sqrt{1 - x^2}}$ | $(-1, 1)$ |
| $\cos^{-1} x$ | $-\frac{1}{\sqrt{1 - x^2}}$ | $(-1, 1)$ |
| $\tan^{-1} x$ | $\frac{1}{1 + x^2}$ | $\mathbb{R}$ |
Mathematical Problem Statement: Prove that $\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}}$ for $x \in (-1, 1)$.
Step-by-Step Algorithmic Solution Trace:
- Let $y = \sin^{-1} x$. Then $x = \sin y$.
- Differentiate both sides with respect to $x$:$$1 = \cos y \cdot \frac{dy}{dx}$$
- Solve for $\frac{dy}{dx}$:$$\frac{dy}{dx} = \frac{1}{\cos y}$$
- Since $y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, $\cos y$ is positive:$$\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$$
- Substitute back:$$\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$$
Final Evaluation Output: $\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}}$
Core Logarithmic Identities:
Mathematical Problem Statement: Differentiate $y = e^{-x}$ with respect to $x$.
Step-by-Step Algorithmic Solution Trace:
- Let $u = -x$, so $y = e^u$.
- Compute derivatives:$$\frac{dy}{du} = e^u \quad \text{and} \quad \frac{du}{dx} = -1$$
- Apply the chain rule:$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = e^u \cdot (-1) = -e^{-x}$$
Final Evaluation Output: $\frac{d}{dx}(e^{-x}) = -e^{-x}$
This technique is highly useful for differentiating functions of the form $y = [u(x)]^{v(x)}$ or complex products and quotients.
Mathematical Problem Statement: Differentiate $y = x^{\sin x}$ with respect to $x$ for $x > 0$.
Step-by-Step Algorithmic Solution Trace:
- Take the natural logarithm on both sides:$$\log y = \sin x \log x$$
- Differentiate both sides with respect to $x$ using the product rule:$$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\sin x) \cdot \log x + \sin x \cdot \frac{d}{dx}(\log x)$$$$\frac{1}{y} \frac{dy}{dx} = \cos x \log x + \frac{\sin x}{x}$$
- Multiply by $y$:$$\frac{dy}{dx} = y \left( \cos x \log x + \frac{\sin x}{x} \right)$$
- Substitute $y = x^{\sin x}$:$$\frac{dy}{dx} = x^{\sin x} \left( \cos x \log x + \frac{\sin x}{x} \right)$$
Final Evaluation Output: $\frac{dy}{dx} = x^{\sin x} \left( \cos x \log x + \frac{\sin x}{x} \right)$
When $x$ and $y$ are expressed in terms of a third variable $t$ (called the parameter), i.e., $x = f(t)$ and $y = g(t)$, we find $\frac{dy}{dx}$ without eliminating $t$.
Mathematical Problem Statement: Find $\frac{dy}{dx}$ if $x = a \cos \theta$ and $y = a \sin \theta$.
Step-by-Step Algorithmic Solution Trace:
- Differentiate $x$ with respect to $\theta$:$$\frac{dx}{d\theta} = -a \sin \theta$$
- Differentiate $y$ with respect to $\theta$:$$\frac{dy}{d\theta} = a \cos \theta$$
- Apply the parametric derivative formula:$$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{a \cos \theta}{-a \sin \theta} = -\cot \theta$$
Final Evaluation Output: $\frac{dy}{dx} = -\cot \theta$
The derivative of the first derivative with respect to $x$ is called the second order derivative.
If $y = f(x)$, then:
$$\frac{dy}{dx} = f'(x)$$ $$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = f''(x) = y'' = y_2$$Mathematical Problem Statement: If $y = A \sin x + B \cos x$, prove that $\frac{d^2y}{dx^2} + y = 0$.
Step-by-Step Algorithmic Solution Trace:
- Find the first derivative:$$\frac{dy}{dx} = A \cos x - B \sin x$$
- Differentiate again with respect to $x$:$$\frac{d^2y}{dx^2} = -A \sin x - B \cos x$$
- Factor out the negative sign:$$\frac{d^2y}{dx^2} = -(A \sin x + B \cos x) = -y$$
- Rearrange the terms:$$\frac{d^2y}{dx^2} + y = 0$$
Final Evaluation Output: $\frac{d^2y}{dx^2} + y = 0$ is proved.