12.4Introduction to Derivatives
Define the derivative as the limit of the difference quotient, compute f'(x) from the definition, write tangent line equations, and interpret derivatives as instantaneous rates of change.
The derivative is the central object of differential calculus. Understanding it as a limit of average rates of change — and connecting it to tangent lines and instantaneous velocity — gives you the conceptual foundation that makes all of Calculus AB and BC accessible.
Essential Question
How can we capture the exact rate of change of a function at a single instant — and why does that idea unlock all of calculus?
Overview
The derivative of a function f at a point x is the instantaneous rate of change of f at that point. It is defined as the limit of the difference quotient (average rate of change) as the interval shrinks to zero:
- Average rate of change on [a, b]: [f(b) − f(a)] / (b − a) — slope of the secant line
- Instantaneous rate of change at x = a: f'(a) — slope of the tangent line
- Difference quotient: [f(x+h) − f(x)] / h — the expression inside the limit
- Tangent line equation: y − f(a) = f'(a)(x − a)
- Velocity interpretation: if s(t) is position, then s'(t) is instantaneous velocity
Worked Examples
Find the average rate of change of f(x) = x² on the interval [1, 3].
Average rate of change = [f(3) − f(1)] / (3 − 1)
f(3) = 3² = 9
f(1) = 1² = 1
[9 − 1] / (3 − 1) = 8 / 2 = 4
This is the slope of the secant line connecting (1, 1) and (3, 9).
Use the limit definition to find the instantaneous rate of change f'(2) for f(x) = x².
Write the difference quotient: [f(2+h) − f(2)] / h
f(2+h) = (2+h)² = 4 + 4h + h²
f(2) = 4
[4 + 4h + h² − 4] / h = [4h + h²] / h
Factor: h(4 + h) / h = 4 + h
Take the limit: lim(h→0) (4 + h) = 4
Find f'(x) using the limit definition for f(x) = 3x² − 2x.
Write: f'(x) = lim(h→0) [f(x+h) − f(x)] / h
f(x+h) = 3(x+h)² − 2(x+h) = 3(x² + 2xh + h²) − 2x − 2h
f(x+h) = 3x² + 6xh + 3h² − 2x − 2h
f(x+h) − f(x) = 6xh + 3h² − 2h
Divide by h: (6xh + 3h² − 2h) / h = 6x + 3h − 2
lim(h→0) (6x + 3h − 2) = 6x − 2
Write the equation of the tangent line to f(x) = x² at x = 1.
Find the slope: f'(x) = 2x (from the limit definition), so f'(1) = 2.
Find the point: f(1) = 1² = 1, so the point is (1, 1).
Use point-slope form: y − y₁ = m(x − x₁)
y − 1 = 2(x − 1)
y = 2x − 2 + 1 = 2x − 1
A ball is thrown upward. Its position is s(t) = −16t² + 64t (feet). Find the instantaneous velocity at t = 1 second.
Instantaneous velocity = s'(t) = lim(h→0) [s(t+h) − s(t)] / h
s(t+h) = −16(t+h)² + 64(t+h) = −16t² − 32th − 16h² + 64t + 64h
s(t+h) − s(t) = −32th − 16h² + 64h
Divide by h: −32t − 16h + 64
lim(h→0): s'(t) = −32t + 64
At t = 1: s'(1) = −32(1) + 64 = 32 ft/s
Guided Practice
Find the average rate of change of f(x) = x³ on the interval [1, 2].
Hint: Use [f(2) − f(1)] / (2 − 1). Compute f(2) = 8 and f(1) = 1, then subtract and divide.
Use the limit definition to find f'(x) for f(x) = 2x + 5.
Hint: Write [f(x+h) − f(x)] / h. Substitute f(x+h) = 2(x+h) + 5 = 2x + 2h + 5. Subtract f(x), simplify, then take the limit as h→0.
Find f'(3) for f(x) = x² − 4x using the limit definition.
Hint: First find f'(x) using the limit definition (expand (x+h)², simplify, divide by h, take limit). Then substitute x = 3.
Write the equation of the tangent line to f(x) = x³ at x = 2.
Hint: Find f'(x) = 3x² (you may use the power rule preview). Evaluate f'(2) for slope and f(2) for the y-coordinate. Use point-slope form.
A particle's position is s(t) = t² − 3t. Find the instantaneous velocity at t = 2.
Hint: Find s'(t) using the limit definition. Expand (t+h)² − 3(t+h), subtract s(t), divide by h, then take the limit. Substitute t = 2.
Common Mistakes
Confusing average rate of change with instantaneous rate of change.
Average rate of change uses two points: [f(b)−f(a)]/(b−a). Instantaneous rate of change is the limit as the interval shrinks to zero: f'(a) = lim(h→0)[f(a+h)−f(a)]/h.
Forgetting to subtract f(x) in the numerator of the difference quotient.
The difference quotient is [f(x+h) − f(x)] / h. Both f(x+h) AND f(x) must appear in the numerator.
Canceling h before simplifying the numerator fully.
Expand and combine all terms in the numerator first so that every remaining term has a factor of h, then cancel h.
Using the wrong point when writing the tangent line equation.
The tangent line at x = a passes through (a, f(a)). Compute f(a) — do not use f'(a) as the y-coordinate.
Math Tips
The difference quotient [f(x+h)−f(x)]/h is just the slope formula (rise/run) with run = h.
After expanding f(x+h), every term without h should cancel with f(x) in the numerator.
The power rule f'(xⁿ) = nxⁿ⁻¹ is a shortcut derived from the limit definition — it always gives the same answer.
A positive derivative means the function is increasing; a negative derivative means it is decreasing.
Velocity is the derivative of position; acceleration is the derivative of velocity.
Derivative
The instantaneous rate of change of a function at a point, defined as f'(x) = lim(h→0) [f(x+h)−f(x)]/h.
Example: For f(x) = x², f'(x) = 2x, so f'(3) = 6.
Difference Quotient
The expression [f(x+h)−f(x)]/h, which gives the average rate of change over an interval of width h.
Example: For f(x) = x², the difference quotient simplifies to 2x + h.
Tangent Line
The line that touches a curve at exactly one point and has slope equal to the derivative at that point.
Example: The tangent to f(x) = x² at x = 1 is y = 2x − 1.
Instantaneous Rate of Change
The rate of change of a function at a single instant, equal to the derivative f'(a) at that point.
Example: The instantaneous velocity of s(t) = t² at t = 3 is s'(3) = 6 ft/s.
Average Rate of Change
The slope of the secant line between two points: [f(b)−f(a)]/(b−a).
Example: For f(x) = x² on [1, 3]: (9−1)/(3−1) = 4.
Interactive Practice — 5 Questions
What is the average rate of change of f(x) = x² on [2, 5]?
Which expression correctly defines the derivative f'(x)?
Using the limit definition, what is f'(x) for f(x) = 5x?
The slope of the tangent line to f(x) = x² at x = 4 is:
If s(t) = t² + 2t represents position, what is the instantaneous velocity at t = 3?