Calculus 1: The Formal Definition of the Derivative and the Limit Process

Differential calculus is primarily concerned with the study of how functions change. The core tool used to measure this change is the derivative.
For a simple linear function (a straight line) defined by $y = mx + b$ , the rate of change is constant. This rate of change is the slope ( $m$ ), which can be determined by taking any two points $(x_1, y_1)$ and $(x_2, y_2)$ on the line and using the formula: $m = \frac{y_2 - y_1}{x_2 - x_1}$ .
In contrast, for non-linear functions (curves), the slope is not constant; it varies continuously depending on the value of $x$ . The derivative provides a mathematical way to calculate the "instantaneous" slope at any specific point on such a curve.

To understand the derivative, we examine the relationship between secant lines and tangent lines on a graph.
Consider a point $P$ on a curve at high coordinates $(x, f(x))$ . Now consider a second point $Q$ that is a small horizontal distance $h$ away from $P$ . The coordinates of $Q$ are $(x + h, f(x + h))$ .
A secant line is a line that passes through these two distinct points, $P$ and $Q$ . The slope of this secant line, denoted as $m_{sec}$ , represents the average rate of change between those two points and is calculated using the difference quotient: $m_{sec} = \frac{f(x + h) - f(x)}{(x + h) - x} = \frac{f(x + h) - f(x)}{h}$
As we decrease the value of $h$ , the point $Q$ slides along the curve toward the point $P$ . The secant line begins to rotate and align more closely with the curve's direction at point $P$ .
When we take the limit as $h$ approaches zero ( $h \rightarrow 0$ ), the distance between the two points becomes infinitely small. At this limit, the secant line transforms into a tangent line.
The tangent line touches the curve at exactly one point ( $P$ ). Its slope represents the instantaneous rate of change of the function at that specific coordinate.

The derivative of a function $f(x)$ , written as $f'(x)$ , is defined mathematically as the limit of the difference quotient as $h$ approaches zero: $f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}$
This definition is often referred to as differentiation "from first principles."
Identifying the derivative allows us to create a new function that describes the slope of the original function at any given value of $x$ .
Terminology Note: The process of calculating a derivative is called differentiation. If the limit exists at a point, the function is said to be differentiable at that point.

Calculating the derivative using the limit definition involves a specific five-step algebraic process: 1. Find $f(x + h)$ : Substitute the expression $(x + h)$ into every instance of $x$ in the original function. 2. Set up the Numerator: Subtract the original function $f(x)$ from the expression found in step 1 ( $f(x + h) - f(x)$ ). 3. Divide by $h$ : Place the resulting expression over the denominator $h$ . 4. Algebraic Simplification: Expand any polynomials, combine like terms, and factor out an $h$ from the numerator. This allows the $h$ in the numerator and denominator to cancel, removing the division-by-zero problem. 5. Apply the Limit: Substitute $h = 0$ into the simplified expression to find the final derivative function $f'(x)$ .

Objective: Find the general derivative formula for $f(x) = x^2$ .
Step 1: Calculate $f(x + h)$ . $f(x + h) = (x + h)^2 = x^2 + 2xh + h^2$
Step 2: Subtract $f(x)$ . $f(x + h) - f(x) = (x^2 + 2xh + h^2) - x^2 = 2xh + h^2$
Step 3: Divide by $h$ . $\frac{2xh + h^2}{h}$
Step 4: Factor and cancel. $\frac{h(2x + h)}{h} = 2x + h$
Step 5: Evaluate the limit. $f'(x) = \lim_{h \rightarrow 0} (2x + h) = 2x + 0 = 2x$
Result: The derivative of $x^2$ is $2x$ . This means for any point on the parabola $y = x^2$ , the slope is exactly twice the x-coordinate.

The derivative can be used to find the specific linear equation for a line tangent to a curve at a designated point. For example, find the tangent line for $f(x) = x^2$ at $x = 3$ .
Step 1: Find the y-coordinate. Plug $x = 3$ into the original function: $f(3) = 3^2 = 9$ . The point of tangency is $(3, 9)$ .
Step 2: Find the slope ( $m$ ). Plug $x = 3$ into the derivative function: $f'(3) = 2(3) = 6$ . The slope of the tangent line is $6$ .
Step 3: Point-Slope Formula. Use the formula $y - y_1 = m(x - x_1)$ . $y - 9 = 6(x - 3)$
Step 4: Solve for Slope-Intercept Form ( $y = mx + b$ ). $y - 9 = 6x - 18$ $y = 6x - 9$
Conclusion: The line $y = 6x - 9$ is perfectly tangent to the curve $y = x^2$ at the coordinate $(3, 9)$ .

Calculus is extensively used in physics to describe motion.
If a function $s(t)$ represents the position of an object over time ( $t$ ), the average velocity over an interval is the slope of the secant line ( $\frac{\Delta s}{\Delta t}$ ).
The instantaneous velocity is the derivative of the position function, $s'(t)$ . This corresponds to the speedometer reading at a specific moment in time.
This concept applies to any rate: the rate at which water fills a tank, the rate of investment growth, or the rate of a chemical reaction.

Different notations are used depending on the context of the problem: * Lagrange's Notation: $f'(x)$ , which reads as "f prime of x." * Leibniz's Notation: $\frac{dy}{dx}$ , which reads as "the derivative of y with respect to x." This notation is helpful because it reminds us that the derivative is fundamentally a ratio of changes ( $dy$ over $dx$ ). * Operator Notation: $\frac{d}{dx} [f(x)]$ , which indicates the action of taking the derivative of the expression inside the brackets.

Question: Why must we use a limit? Why can't we just set $h = 0$ at the very beginning?
Answer: If you set $h = 0$ immediately, the difference quotient becomes $\frac{f(x+0) - f(x)}{0}$ , which results in $\frac{0}{0}$ . This is an indeterminate form, and division by zero is undefined. The limit allows us to explore what happens as $h$ gets infinitely close to zero without actually reaching the undefined state.
Question: Does every function have a derivative?
Answer: No. For a derivative to exist at a point, the function must be continuous and smooth (no sharp corners, vertical tangents, or breaks) at that point. If the limit of the difference quotient does not exist, the function is not differentiable there.