Introduction to the Derivative

Discover how to measure instantaneous rates of change

Imagine you’re driving on a highway and glance at your speedometer. It reads 65 miles per hour. But what does that really mean? You haven’t been driving for an hour, and you certainly haven’t traveled 65 miles since you looked. The speedometer is telling you something more subtle and more powerful: your instantaneous speed—how fast you’re going at this very moment.

This simple question—“How fast am I going right now?"—leads us to one of the most profound ideas in all of mathematics: the derivative. The derivative gives us a way to measure rates of change at a single instant, and it turns out this concept underlies nearly everything in physics, engineering, economics, and countless other fields.

Don’t worry if this sounds abstract at first. We’re going to build up to it step by step, starting from ideas you already understand intuitively. By the end of this lesson, you’ll not only understand what a derivative is, but you’ll be able to compute them yourself.

Core Concepts

Average Rate of Change: The Starting Point

Before we can talk about instantaneous rates of change, let’s make sure we’re comfortable with average rates of change—something you’ve actually been working with since you first learned about slope.

Suppose you’re on a road trip. You note that at 2:00 PM you’re at mile marker 100, and at 4:00 PM you’re at mile marker 220. What was your average speed during that time?

You’d calculate:

$$\text{Average speed} = \frac{\text{Change in position}}{\text{Change in time}} = \frac{220 - 100}{4 - 2} = \frac{120 \text{ miles}}{2 \text{ hours}} = 60 \text{ mph}$$

This is exactly the same as calculating slope! If we let $s(t)$ represent your position at time $t$, then your average speed between times $t_1$ and $t_2$ is:

$$\text{Average rate of change} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}$$

Graphically, this is the slope of the line connecting the two points $(t_1, s(t_1))$ and $(t_2, s(t_2))$ on the graph of $s(t)$. This line is called a secant line—a line that cuts through a curve at two points.

For any function $f(x)$, the average rate of change from $x = a$ to $x = b$ is:

$$\frac{f(b) - f(a)}{b - a}$$

This tells us, on average, how much $f$ changes for each unit increase in $x$ over that interval.

The Problem: Capturing an Instant

Here’s where things get interesting. Your average speed of 60 mph tells us something, but it doesn’t capture everything. Maybe you were stuck in traffic for 30 minutes going only 10 mph, then sped along at 80 mph for the rest. The average smooths over all those details.

What if we want to know your exact speed at 3:00 PM?

Your first instinct might be to look at a smaller interval—say, from 2:55 PM to 3:05 PM. That would give you a better approximation. An even smaller interval—from 2:59 PM to 3:01 PM—would be better still.

You can see where this is going. To get the instantaneous speed, we want to shrink the interval down to… zero? But wait—if the interval has length zero, we’d be dividing by zero! We’d have:

$$\frac{s(3) - s(3)}{3 - 3} = \frac{0}{0}$$

This is undefined. We seem to be stuck.

This is the fundamental problem that calculus solves. We can’t actually use a zero-length interval, but we can ask: What happens to the average rate of change as the interval gets closer and closer to zero?

The Solution: Taking a Limit

The breakthrough insight is to use a limit. Instead of actually dividing by zero, we examine what value the average rate of change approaches as the interval width shrinks toward zero.

Let’s set this up carefully. Suppose we want to find the instantaneous rate of change of $f(x)$ at $x = a$. We’ll compute the average rate of change from $x = a$ to $x = a + h$, where $h$ is a small number representing the width of our interval:

$$\text{Average rate of change} = \frac{f(a + h) - f(a)}{(a + h) - a} = \frac{f(a + h) - f(a)}{h}$$

Now, the instantaneous rate of change is what this expression approaches as $h$ gets closer and closer to zero:

$$\text{Instantaneous rate of change at } x = a = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$$

This limit, when it exists, is called the derivative of $f$ at $x = a$.

The Definition of the Derivative

We’re now ready for the formal definition. The derivative of a function $f(x)$ is a new function, denoted $f’(x)$ (read “f prime of x”), defined by:

$$f’(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

This definition tells us the instantaneous rate of change of $f$ at any point $x$ where the limit exists.

There’s an equivalent form that’s sometimes useful. If we let $a$ be a specific point and use a different variable for nearby points, we can write:

$$f’(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$$

Both forms say the same thing: the derivative is the limit of average rates of change as the interval shrinks to zero.

When this limit exists at a point, we say $f$ is differentiable at that point. If $f$ is differentiable at every point in some interval, we say $f$ is differentiable on that interval.

Tangent Lines: A Geometric View

There’s a beautiful geometric interpretation of the derivative. Remember that the average rate of change from $x$ to $x + h$ is the slope of the secant line through those two points.

As $h \to 0$, the second point slides closer and closer to the first point. The secant line pivots, approaching a limiting position. This limiting line is called the tangent line to the curve at that point.

The tangent line is the line that “just touches” the curve at that point—it has exactly the same slope as the curve itself at that instant. And that slope is precisely the derivative!

So we can say: The derivative $f’(a)$ is the slope of the tangent line to the graph of $f$ at the point $(a, f(a))$.

This gives us a way to visualize derivatives. If you look at a graph and imagine a line perfectly hugging the curve at a point, the steepness of that line is the derivative at that point.

Computing Derivatives from the Definition

Let’s see how to actually find derivatives using the limit definition. The process follows a consistent pattern:

  1. Write out $\frac{f(x+h) - f(x)}{h}$
  2. Simplify the expression algebraically
  3. Take the limit as $h \to 0$

The key step is usually the algebraic simplification. You need to manipulate the expression so that you can cancel the $h$ in the denominator (otherwise you’d get $\frac{\text{something}}{0}$, which doesn’t help).

This process can feel tedious, and later you’ll learn shortcut rules that make finding derivatives much faster. But working through the definition a few times helps you understand why those shortcuts work and what the derivative really means.

Notation and Terminology

Mathematicians and scientists have developed several notations for derivatives over the centuries. Each has its advantages, and you’ll encounter all of them.

Term Meaning Example or Notes
Secant line Line through two points on a curve Slope gives average rate of change
Tangent line Line that touches curve at exactly one point Slope gives instantaneous rate of change
$f’(x)$ Derivative of $f$ at $x$ (Lagrange notation) Pronounced “f prime of x”
$\frac{dy}{dx}$ Derivative of $y$ with respect to $x$ (Leibniz notation) Emphasizes “rate of change of $y$ per unit change in $x$”
$\frac{d}{dx}[f(x)]$ Derivative operator applied to $f(x)$ Useful when $f$ is a formula without a name
$\dot{y}$ Derivative with respect to time (Newton notation) Common in physics
Differentiable The derivative exists at that point Function is smooth, with no corners or jumps
Difference quotient $\frac{f(x+h) - f(x)}{h}$ before taking the limit The “before” picture of the derivative

The Leibniz notation $\frac{dy}{dx}$ is particularly useful because it reminds us that the derivative is a ratio of changes. While it’s not literally a fraction (it’s a limit!), it often behaves like a fraction, which makes certain manipulations intuitive.

When $y = f(x)$, all of these mean the same thing:

$$f’(x) = \frac{dy}{dx} = \frac{d}{dx}[f(x)] = y’$$

Examples

Let’s work through several examples, starting simple and building to more challenging cases.

Example 1: Average Rate of Change

Problem: Find the average rate of change of $f(x) = x^2$ from $x = 1$ to $x = 3$.

Solution:

We use the formula for average rate of change:

$$\frac{f(b) - f(a)}{b - a} = \frac{f(3) - f(1)}{3 - 1}$$

First, let’s calculate $f(3)$ and $f(1)$:

  • $f(3) = 3^2 = 9$
  • $f(1) = 1^2 = 1$

Now we can compute:

$$\frac{f(3) - f(1)}{3 - 1} = \frac{9 - 1}{3 - 1} = \frac{8}{2} = 4$$

The average rate of change is 4.

This means that, on average, $f(x) = x^2$ increases by 4 units for each 1-unit increase in $x$ over the interval from 1 to 3.

Example 2: Derivative of a Linear Function

Problem: Find $f’(x)$ for $f(x) = 3x$ using the definition of the derivative.

Solution:

We apply the limit definition:

$$f’(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

First, let’s find $f(x + h)$:

$$f(x + h) = 3(x + h) = 3x + 3h$$

Now we set up the difference quotient:

$$\frac{f(x+h) - f(x)}{h} = \frac{(3x + 3h) - 3x}{h} = \frac{3h}{h} = 3$$

Taking the limit:

$$f’(x) = \lim_{h \to 0} 3 = 3$$

Therefore, $f’(x) = 3$.

This makes perfect sense! The function $f(x) = 3x$ is a line with slope 3. Since a line has the same steepness everywhere, the derivative (which measures steepness) is constant and equals the slope of the line.

Example 3: Derivative of a Quadratic Function

Problem: Find $f’(x)$ for $f(x) = x^2$ using the definition of the derivative.

Solution:

We start with the limit definition:

$$f’(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

First, we need to expand $f(x + h)$:

$$f(x + h) = (x + h)^2 = x^2 + 2xh + h^2$$

Now we form the difference quotient:

$$\frac{f(x+h) - f(x)}{h} = \frac{(x^2 + 2xh + h^2) - x^2}{h}$$

Simplify the numerator:

$$= \frac{2xh + h^2}{h}$$

Factor out $h$ from the numerator:

$$= \frac{h(2x + h)}{h}$$

Cancel the common factor of $h$ (valid since $h \neq 0$ as we approach the limit):

$$= 2x + h$$

Now take the limit as $h \to 0$:

$$f’(x) = \lim_{h \to 0} (2x + h) = 2x + 0 = 2x$$

Therefore, $f’(x) = 2x$.

This tells us that the instantaneous rate of change of $x^2$ depends on where you are. At $x = 1$, the derivative is $2(1) = 2$. At $x = 3$, the derivative is $2(3) = 6$. The parabola gets steeper as you move away from the origin.

Example 4: Finding the Equation of a Tangent Line

Problem: Find the equation of the tangent line to $f(x) = x^2$ at the point where $x = 2$.

Solution:

To write the equation of a line, we need two things: a point on the line and the slope.

Step 1: Find the point.

When $x = 2$: $$f(2) = 2^2 = 4$$

So the point of tangency is $(2, 4)$.

Step 2: Find the slope.

The slope of the tangent line is the derivative evaluated at $x = 2$. From Example 3, we know that $f’(x) = 2x$, so:

$$f’(2) = 2(2) = 4$$

The slope of the tangent line is $4$.

Step 3: Write the equation.

Using point-slope form with point $(2, 4)$ and slope $m = 4$:

$$y - y_1 = m(x - x_1)$$ $$y - 4 = 4(x - 2)$$ $$y - 4 = 4x - 8$$ $$y = 4x - 4$$

The equation of the tangent line is $y = 4x - 4$.

You can verify this makes sense: the line passes through $(2, 4)$ (check: $4(2) - 4 = 4$ ) and has slope 4.

Example 5: Derivative of a Reciprocal Function

Problem: Find $f’(x)$ for $f(x) = \frac{1}{x}$ using the definition of the derivative.

Solution:

This one requires more careful algebra. Let’s apply the definition:

$$f’(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

First, write $f(x + h)$:

$$f(x + h) = \frac{1}{x + h}$$

Now form the difference quotient:

$$\frac{f(x+h) - f(x)}{h} = \frac{\frac{1}{x+h} - \frac{1}{x}}{h}$$

To subtract these fractions in the numerator, we need a common denominator:

$$\frac{1}{x+h} - \frac{1}{x} = \frac{x}{x(x+h)} - \frac{x+h}{x(x+h)} = \frac{x - (x+h)}{x(x+h)} = \frac{-h}{x(x+h)}$$

Substituting back into the difference quotient:

$$\frac{\frac{-h}{x(x+h)}}{h} = \frac{-h}{x(x+h)} \cdot \frac{1}{h} = \frac{-1}{x(x+h)}$$

Now we can take the limit:

$$f’(x) = \lim_{h \to 0} \frac{-1}{x(x+h)} = \frac{-1}{x(x+0)} = \frac{-1}{x^2}$$

Therefore, $f’(x) = -\frac{1}{x^2}$.

The negative sign tells us that $\frac{1}{x}$ is decreasing wherever $x > 0$. This matches what we see on the graph: as $x$ increases (for positive $x$), the function values get smaller.

Example 6: Derivative of a Square Root Function

Problem: Find $f’(x)$ for $f(x) = \sqrt{x}$ using the definition of the derivative.

Solution:

This example requires a special algebraic trick: multiplying by the conjugate.

We start with:

$$f’(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$$

The trouble is that directly substituting $h = 0$ gives us $\frac{0}{0}$. We need to manipulate the expression.

The key is to multiply by the conjugate of the numerator:

$$\frac{\sqrt{x+h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}$$

The numerator becomes a difference of squares:

$$= \frac{(\sqrt{x+h})^2 - (\sqrt{x})^2}{h(\sqrt{x+h} + \sqrt{x})} = \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})} = \frac{h}{h(\sqrt{x+h} + \sqrt{x})}$$

Cancel the $h$:

$$= \frac{1}{\sqrt{x+h} + \sqrt{x}}$$

Now take the limit as $h \to 0$:

$$f’(x) = \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}} = \frac{1}{\sqrt{x} + \sqrt{x}} = \frac{1}{2\sqrt{x}}$$

Therefore, $f’(x) = \frac{1}{2\sqrt{x}}$.

This can also be written as $\frac{1}{2}x^{-1/2}$, which fits a pattern you’ll learn soon.

Key Properties and Rules

While the limit definition always works in principle, it can be cumbersome. Here are some properties that make computing derivatives easier:

Constant Rule

If $f(x) = c$ (a constant), then $f’(x) = 0$.

This makes intuitive sense: a constant function doesn’t change, so its rate of change is zero.

Constant Multiple Rule

If $g(x) = c \cdot f(x)$, then $g’(x) = c \cdot f’(x)$.

You can “pull out” constant multipliers when taking derivatives.

Sum and Difference Rules

If $h(x) = f(x) + g(x)$, then $h’(x) = f’(x) + g’(x)$.

If $h(x) = f(x) - g(x)$, then $h’(x) = f’(x) - g’(x)$.

The derivative of a sum (or difference) is the sum (or difference) of the derivatives.

The Power Rule (Preview)

From our examples, you might notice a pattern:

  • The derivative of $x^1$ is $1$ (since $\frac{d}{dx}[x] = 1$)
  • The derivative of $x^2$ is $2x$
  • The derivative of $x^{-1} = \frac{1}{x}$ is $-x^{-2} = -\frac{1}{x^2}$
  • The derivative of $x^{1/2} = \sqrt{x}$ is $\frac{1}{2}x^{-1/2}$

The general pattern is the Power Rule:

$$\frac{d}{dx}[x^n] = n \cdot x^{n-1}$$

This works for any real number $n$—positive, negative, fractional, you name it. You’ll prove this rule more rigorously later, but for now, notice how all our examples fit this pattern.

Real-World Applications

The derivative appears everywhere in science, engineering, and economics. Here are some of the most important applications:

Velocity and Acceleration

If $s(t)$ represents the position of an object at time $t$, then:

  • $s’(t) = v(t)$ is the velocity (instantaneous rate of change of position)
  • $v’(t) = s’’(t) = a(t)$ is the acceleration (rate of change of velocity)

When you see a speedometer, you’re looking at a derivative. When you feel yourself pushed back into your seat as a car accelerates, you’re experiencing the derivative of velocity.

Economics: Marginal Cost and Revenue

In economics, if $C(x)$ represents the total cost of producing $x$ units of a product, then $C’(x)$ is called the marginal cost—approximately how much it costs to produce one more unit.

Similarly, if $R(x)$ is the revenue from selling $x$ units, then $R’(x)$ is the marginal revenue—approximately how much additional revenue you get from selling one more unit.

Business decisions often hinge on comparing marginal cost to marginal revenue.

Biology: Population Growth

If $P(t)$ represents the size of a population at time $t$, then $P’(t)$ tells us how fast the population is growing (or shrinking). This rate of change depends on factors like birth rates, death rates, and resource availability—leading to fascinating models like exponential growth and logistic growth.

Physics: Rates of Change Everywhere

Derivatives appear throughout physics:

  • Rate of change of temperature (heat transfer)
  • Rate of change of charge (electric current)
  • Rate of change of work with respect to position (force)
  • Rate of change of momentum (force, again—Newton’s second law!)

In fact, most of physics can be expressed as relationships between quantities and their derivatives. Understanding derivatives opens the door to understanding the physical world.

Self-Test Problems

Test your understanding with these practice problems. Try to solve each one before checking the answer.

Problem 1: Find the average rate of change of $f(x) = x^3$ from $x = 1$ to $x = 2$.

Show Answer

Average rate of change $= \frac{f(2) - f(1)}{2 - 1} = \frac{8 - 1}{1} = 7$

Problem 2: Use the limit definition to find $f’(x)$ for $f(x) = 5x - 2$.

Show Answer

$$f’(x) = \lim_{h \to 0} \frac{[5(x+h) - 2] - [5x - 2]}{h}$$

$$= \lim_{h \to 0} \frac{5x + 5h - 2 - 5x + 2}{h} = \lim_{h \to 0} \frac{5h}{h} = \lim_{h \to 0} 5 = 5$$

This makes sense: $f(x) = 5x - 2$ is a line with slope 5, so its derivative is 5 everywhere.

Problem 3: Use the limit definition to find $f’(x)$ for $f(x) = x^2 + 3x$.

Show Answer

$$f’(x) = \lim_{h \to 0} \frac{[(x+h)^2 + 3(x+h)] - [x^2 + 3x]}{h}$$

Expand: $$= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 + 3x + 3h - x^2 - 3x}{h}$$

Simplify: $$= \lim_{h \to 0} \frac{2xh + h^2 + 3h}{h} = \lim_{h \to 0} \frac{h(2x + h + 3)}{h}$$

$$= \lim_{h \to 0} (2x + h + 3) = 2x + 3$$

Therefore, $f’(x) = 2x + 3$.

Problem 4: Find the equation of the tangent line to $f(x) = x^3$ at $x = 1$.

Show Answer

Step 1: Find the point. At $x = 1$: $f(1) = 1^3 = 1$. Point: $(1, 1)$.

Step 2: Find the slope. First, we need $f’(x)$. Using the power rule (or working through the limit definition): $f’(x) = 3x^2$

At $x = 1$: $f’(1) = 3(1)^2 = 3$. Slope: $m = 3$.

Step 3: Write the equation. $y - 1 = 3(x - 1)$ $y = 3x - 3 + 1$ $y = 3x - 2$

Problem 5: If $f(x) = x^2$, find the value(s) of $x$ where the tangent line is horizontal.

Show Answer

A horizontal line has slope 0, so we need to find where $f’(x) = 0$.

We know $f’(x) = 2x$.

Setting $f’(x) = 0$: $2x = 0$ $x = 0$

The tangent line is horizontal at $x = 0$ (the vertex of the parabola).

Problem 6: The position of a particle is given by $s(t) = t^2 - 4t + 3$ meters, where $t$ is in seconds. Find: (a) The velocity at time $t$. (b) The velocity at $t = 3$ seconds. (c) When the particle is at rest.

Show Answer

(a) Velocity is the derivative of position: $v(t) = s’(t) = 2t - 4$ meters per second

(b) At $t = 3$: $v(3) = 2(3) - 4 = 6 - 4 = 2$ m/s

(c) The particle is at rest when $v(t) = 0$: $2t - 4 = 0$ $t = 2$ seconds

Problem 7 (Challenge): Use the limit definition to find the derivative of $f(x) = \frac{1}{x+1}$.

Show Answer

$$f’(x) = \lim_{h \to 0} \frac{\frac{1}{(x+h)+1} - \frac{1}{x+1}}{h} = \lim_{h \to 0} \frac{\frac{1}{x+h+1} - \frac{1}{x+1}}{h}$$

Find common denominator for the numerator: $$= \lim_{h \to 0} \frac{1}{h} \cdot \frac{(x+1) - (x+h+1)}{(x+h+1)(x+1)}$$

$$= \lim_{h \to 0} \frac{1}{h} \cdot \frac{-h}{(x+h+1)(x+1)}$$

$$= \lim_{h \to 0} \frac{-1}{(x+h+1)(x+1)}$$

$$= \frac{-1}{(x+1)(x+1)} = \frac{-1}{(x+1)^2}$$

Therefore, $f’(x) = -\frac{1}{(x+1)^2}$.

Summary

Congratulations on working through this introduction to the derivative. Here are the key ideas to remember:

  • The derivative measures instantaneous rate of change. While average rate of change looks at two points, the derivative captures what’s happening at a single instant.

  • The derivative is defined as a limit: $$f’(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

  • Geometrically, the derivative is the slope of the tangent line to the graph at that point.

  • Multiple notations exist: $f’(x)$, $\frac{dy}{dx}$, and $\frac{d}{dx}[f(x)]$ all mean the same thing.

  • Computing derivatives from the definition involves setting up the difference quotient, simplifying algebraically to cancel the $h$ in the denominator, and then taking the limit.

  • The Power Rule provides a shortcut: $\frac{d}{dx}[x^n] = nx^{n-1}$.

  • Derivatives appear throughout science and everyday life: velocity, acceleration, growth rates, marginal costs, and countless other quantities are all derivatives.

The derivative is just the beginning. In upcoming lessons, you’ll learn more efficient rules for computing derivatives, see how to use derivatives to analyze functions and solve optimization problems, and discover the profound connection between derivatives and integrals. The tools you’ve learned here form the foundation for all of that.

Keep practicing with the limit definition until it feels natural. Even though you’ll soon have faster methods, understanding the definition helps you truly grasp what derivatives mean and why they work.