What is a Derivative?

Understanding the fundamental concept of derivatives

Understanding Derivatives

A derivative measures how a function changes. It's the mathematical way to describe rates of change.

The Big Idea

The derivative tells you how fast something is changing at a specific moment.

Think of it like this:

  • Speed is the derivative of position (how fast your location changes)
  • Acceleration is the derivative of speed (how fast your speed changes)
  • The slope of a curve at a point is the derivative

From Slope to Derivative

Remember the slope of a line between two points?

m=y2y1x2x1=ΔyΔxm = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\Delta y}{\Delta x}

The derivative is what happens when we make those points infinitely close together!

The Definition

The derivative of f(x)f(x) at x=ax = a is:

f(a)=limh0f(a+h)f(a)hf'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}

Let's break this down:

  • f(a+h)f(a + h): Function value slightly to the right
  • f(a)f(a): Function value at the point
  • f(a+h)f(a)f(a + h) - f(a): Change in y-values
  • hh: Change in x-values (approaching 0)
  • The limit: Make h infinitesimally small

What This Fraction Means

f(a+h)f(a)h\frac{f(a + h) - f(a)}{h}

This is the average rate of change over the interval from aa to a+ha + h.

As h0h \to 0, we get the instantaneous rate of change (the derivative!)

Geometric Interpretation

The derivative at a point is the slope of the tangent line to the curve at that point.

  • Secant line: Connects two points on the curve
  • Tangent line: Touches the curve at exactly one point
  • As the two points get closer, the secant line → tangent line

Example: Computing from Definition

Find the derivative of f(x)=x2f(x) = x^2 at x=3x = 3 using the definition.

f(3)=limh0f(3+h)f(3)hf'(3) = \lim_{h \to 0} \frac{f(3 + h) - f(3)}{h}

Step 1: Find f(3+h)f(3 + h) and f(3)f(3)

  • f(3+h)=(3+h)2=9+6h+h2f(3 + h) = (3 + h)^2 = 9 + 6h + h^2
  • f(3)=9f(3) = 9

Step 2: Substitute f(3)=limh0(9+6h+h2)9h=limh06h+h2hf'(3) = \lim_{h \to 0} \frac{(9 + 6h + h^2) - 9}{h} = \lim_{h \to 0} \frac{6h + h^2}{h}

Step 3: Factor and simplify =limh0h(6+h)h=limh0(6+h)= \lim_{h \to 0} \frac{h(6 + h)}{h} = \lim_{h \to 0} (6 + h)

Step 4: Evaluate =6+0=6= 6 + 0 = 6

So f(3)=6f'(3) = 6 — the slope at x=3x = 3 is 6!

Why Derivatives Matter

Derivatives help us:

  • Find maximum and minimum values
  • Determine where functions are increasing or decreasing
  • Analyze motion (velocity, acceleration)
  • Optimize real-world problems
  • Understand rates of change in science and economics

The Function vs. Its Derivative

If y=f(x)y = f(x) is our function, then y=f(x)y' = f'(x) is the derivative function that gives the slope at every point.

Original function → tells you the value Derivative function → tells you the rate of change

📚 Practice Problems

1Problem 1easy

Question:

Use the definition of derivative to find f(2)f'(2) if f(x)=3x1f(x) = 3x - 1.

💡 Show Solution

Using the definition:

f(2)=limh0f(2+h)f(2)hf'(2) = \lim_{h \to 0} \frac{f(2 + h) - f(2)}{h}

Step 1: Find the function values

f(2+h)=3(2+h)1=6+3h1=5+3hf(2 + h) = 3(2 + h) - 1 = 6 + 3h - 1 = 5 + 3h f(2)=3(2)1=5f(2) = 3(2) - 1 = 5

Step 2: Substitute into the definition

f(2)=limh0(5+3h)5h=limh03hhf'(2) = \lim_{h \to 0} \frac{(5 + 3h) - 5}{h} = \lim_{h \to 0} \frac{3h}{h}

Step 3: Simplify

=limh03=3= \lim_{h \to 0} 3 = 3

Answer: f(2)=3f'(2) = 3

This makes sense! f(x)=3x1f(x) = 3x - 1 is a line with slope 3, so the derivative (slope) at any point is 3.

2Problem 2medium

Question:

Find the derivative of f(x)=x2+2xf(x) = x^2 + 2x at x=1x = 1 using the limit definition.

💡 Show Solution

Using the definition:

f(1)=limh0f(1+h)f(1)hf'(1) = \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h}

Step 1: Find f(1+h)f(1 + h)

f(1+h)=(1+h)2+2(1+h)f(1 + h) = (1 + h)^2 + 2(1 + h) =1+2h+h2+2+2h= 1 + 2h + h^2 + 2 + 2h =3+4h+h2= 3 + 4h + h^2

Step 2: Find f(1)f(1)

f(1)=12+2(1)=1+2=3f(1) = 1^2 + 2(1) = 1 + 2 = 3

Step 3: Substitute

f(1)=limh0(3+4h+h2)3hf'(1) = \lim_{h \to 0} \frac{(3 + 4h + h^2) - 3}{h} =limh04h+h2h= \lim_{h \to 0} \frac{4h + h^2}{h}

Step 4: Factor and simplify

=limh0h(4+h)h=limh0(4+h)= \lim_{h \to 0} \frac{h(4 + h)}{h} = \lim_{h \to 0} (4 + h)

Step 5: Evaluate

=4+0=4= 4 + 0 = 4

Answer: f(1)=4f'(1) = 4