AP Precalculus: Introduction to Derivatives
Master rates of change, the derivative concept, and tangent lines
📚 The Big Picture
Derivatives measure how fast things change. The average rate of change tells us the overall rate between two points, while the instantaneous rate tells us the exact rate at a specific moment. This is the foundation of calculus — understanding how limits connect these two ideas through the concept of the derivative.
1 Average Rate of Change
The average rate of change of a function over an interval measures the overall rate at which output values change relative to input values.
Function: \(f(x) = x^2\), find average rate of change on [1, 4]
Calculate: \(\frac{f(4) - f(1)}{4 - 1} = \frac{16 - 1}{3} = \frac{15}{3} = 5\)
Interpretation: On average, f increases by 5 units for each 1-unit increase in x
2 Instantaneous Rate of Change (The Derivative)
The instantaneous rate of change at a point is the limit of average rates as the interval shrinks to zero. This is the derivative, denoted f'(a).
What It Measures
The exact rate of change at a single point x = a, not over an interval
Graphical Meaning
The slope of the tangent line to the curve at x = a
Alternative Form
Function: \(f(x) = x^2\), find f'(3)
Set up: \(f'(3) = \lim_{h \to 0} \frac{(3+h)^2 - 9}{h}\)
Expand: \(= \lim_{h \to 0} \frac{9 + 6h + h^2 - 9}{h} = \lim_{h \to 0} \frac{6h + h^2}{h}\)
Simplify: \(= \lim_{h \to 0} (6 + h) = 6\)
Answer: The slope of the tangent at x = 3 is 6
3 Secant Line vs. Tangent Line
The derivative connects secant lines (cutting through two points) to tangent lines (touching at one point) through the limit process.
As the second point of the secant line approaches the first, the secant line becomes the tangent line. This is exactly what taking the limit h → 0 does!
4 Velocity as a Rate of Change
If s(t) represents position at time t, the velocity is the rate of change of position with respect to time.
Position: \(s(t) = 16t^2\) feet (dropped object)
Average velocity from t = 1 to t = 3:
\(v_{\text{avg}} = \frac{s(3) - s(1)}{3 - 1} = \frac{144 - 16}{2} = \frac{128}{2} = 64\) ft/s
Instantaneous velocity at t = 2:
\(v(2) = s'(2) = \lim_{h \to 0} \frac{16(2+h)^2 - 64}{h} = 64\) ft/s
5 Equation of the Tangent Line
Once you find the derivative f'(a), you can write the equation of the tangent line at x = a using point-slope form.
- Point: The line passes through (a, f(a))
- Slope: The slope is f'(a)
- Form: This is point-slope form: y - y₁ = m(x - x₁)
Function: \(f(x) = x^2\), find tangent line at x = 3
Find point: f(3) = 9, so point is (3, 9)
Find slope: f'(3) = 6 (from earlier example)
Write equation: \(y = 9 + 6(x - 3)\)
Simplify: \(y = 6x - 9\)
Don't forget to find BOTH the point (a, f(a)) AND the slope f'(a). You need both to write the tangent line equation.
📋 Quick Reference
Average Rate
\(\frac{f(b) - f(a)}{b - a}\)
Derivative (Limit Def)
\(\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\)
Alternative Def
\(\lim_{x \to a} \frac{f(x) - f(a)}{x - a}\)
Instantaneous Velocity
\(v(a) = s'(a)\)
Tangent Line
\(y = f(a) + f'(a)(x-a)\)
Secant → Tangent
As h → 0, secant becomes tangent