What is a Derivative?

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

The Definition

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

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

Let's break this down:

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

What This Fraction Means

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

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

As $h \to 0$, 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 $f(x) = x^2$ at $x = 3$ using the definition.

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

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

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

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

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

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

So $f'(3) = 6$ — the slope at $x = 3$ 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 $y = f(x)$ is our function, then $y' = f'(x)$ 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

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

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

Step 2: Find $f(1)$

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

Step 3: Substitute

$f'(1) = \lim_{h \to 0} \frac{(3 + 4h + h^2) - 3}{h}$ $= \lim_{h \to 0} \frac{4h + h^2}{h}$

Step 4: Factor and simplify

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

Step 5: Evaluate

$= 4 + 0 = 4$

Answer: $f'(1) = 4$