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AP Calculus AB/BC · Topic 1.1 · Anatomy of a lesson

From average change to an instant.

A concept-first, difficulty-layered progression, built into one original AP Calculus learning loop.

10 teaching stagesOriginal contentConcept-first

The calculus teaching flow

  1. 1Concept
  2. 2Intuition
  3. 3Concept Lab
  4. 4Formal Mathematics
  5. 5Worked Example
  6. 6Try It
  7. 7Feedback
  8. 8Deeper Reasoning
  9. 9AP Connection
  10. 10Mastery
01

Stage 1 of 10

Concept

What are we learning?

Key concepts first

The derivative measures change now.

By the end, you can:

  • ✓Distinguish the average rate of change on [a,a+h] from the instantaneous rate at x=a.
  • ✓Interpret f'(a) as both a limit and the slope of the tangent line.
  • ✓Use one-sided derivatives to decide whether a continuous function is differentiable.
02

Stage 2 of 10

Intuition

What should it feel like?

Intuition before formulas

A speedometer does not wait for an interval to finish. It reports speed at one instant. Calculus builds that number from ordinary average speeds: choose a nearby time, calculate the secant slope, then shrink the time gap until the nearby point merges with the fixed point.
03

Stage 3 of 10

Concept Lab

What changes as we zoom in?

See it change dynamically

Secant slopes approaching a tangent slopeA curve with a fixed point P and points Q approaching P. Secant slopes approach the tangent slope.PQ₁Q₂Q₃h → 0
hsecant slope
19
0.58.5
0.18.1
0.018.01
0→ 8
For s(t)=t^2+2t at t=3, the secant slope 8+h approaches the tangent slope 8 from nearby intervals.
04

Stage 4 of 10

Formal Mathematics

How do we say it precisely?

Precise, equivalent definitions

Average rate · secant slope

m_{sec}=\frac{f(a+h)-f(a)}{h},\qquad h\ne0

Instantaneous rate · tangent slope

f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}

Equivalent point form

f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}
05

Stage 5 of 10

Worked Example

How does an expert solve it?

Expert worked example

A particle’s position is s(t)=t^2+2t feet. Use the definition of the derivative to find its instantaneous velocity at t=3.

  1. 1Start with the average velocity from t=3 to t=3+h: \dfrac{s(3+h)-s(3)}{h}.
  2. 2For s(t)=t^2+2t, compute s(3+h)=15+8h+h^2 and s(3)=15.
  3. 3Simplify before taking the limit: \dfrac{8h+h^2}{h}=8+h for h\ne0.
  4. 4Let h\to0. Therefore s'(3)=8 feet per second.
Answer: 8 feet per second.

Watch for: substituting h=0 before cancelling. The difference quotient is undefined at h=0; only its limit is evaluated there.

06

Stage 6 of 10

Try It

Can you take the next step?

Single-correct practice

Try it · answer before continuing

For f(x)=x^2-3x, use the limit definition to find f'(2).

07

Stage 7 of 10

Feedback

What does your answer reveal?

What your answer reveals

A linked follow-up identifies which step failed before giving the correction.

If you answered 4

Your method differentiated only x^2. Carry every term through the difference quotient: the linear term contributes -3 to the slope.

If you answered -2

You found height, not slope. Build a quotient from two function values, simplify it, and only then take the limit.

08

Stage 8 of 10

Deeper Reasoning

Why does the method work?

Concept stress test

Concept check

Which statement must be true if f is differentiable at x=a?

09

Stage 9 of 10

AP Connection

How will the exam ask it?

AP-style application

AP

A table gives s(4.9)=33.81, s(5)=35, and s(5.1)=36.21, where s is measured in feet. Estimate s'(5) and interpret it.

Use the symmetric difference: s'(5)\\approx\\dfrac{36.21-33.81}{0.2}=12. At t=5 seconds, the particle’s position is increasing at approximately 12 feet per second.

1 · Difference quotient1 · Value1 · Meaning + units
10

Stage 10 of 10

Mastery

Can you do it without support?

Mixed cumulative review

Three proofs of mastery

Limit recognition · numerical estimate · differentiability

3/3
1.Recognize and evaluate \displaystyle \lim_{x\to2}\frac{x^3-8}{x-2}.＋

It is f'(2) for f(x)=x^3. Since f'(x)=3x^2, the limit is 12.

2.A table gives f(1.9)=3.61 and f(2.1)=4.41. Estimate f'(2) symmetrically.＋

f'(2)\approx\dfrac{4.41-3.61}{2.1-1.9}=4. The estimate uses values on both sides of 2.

3.Why is g(x)=|x-1| continuous but not differentiable at x=1?＋

Both function limits equal g(1)=0, but the left derivative is -1 and the right derivative is 1. The tangent slope is not unique.