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AP Calculus AB/BC Unit 2 Visual ReviewA topic-by-topic visual walkthrough of Unit 2: Differentiation — the definition of the derivative, the power/product/quotient rules, and derivatives of key functions.
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TOPIC 2.1
Average & Instantaneous Rate
Average = secant slope
[f(b) − f(a)] / (b − a)
The slope of the line through two points
on the curve.
f(x)=x², [1,3]: (9−1)/2 = 4
Average change over the interval.
Instantaneous = tangent slope
lim(h→0) [f(a+h) − f(a)] / h
The limit of average rates as the interval
shrinks to zero.
This is the slope of the tangent line —
the derivative at that point.
Secant → tangent
As the second point slides toward the first, the secant line rotates into the tangent line at that point.
Average = secant slope ; instantaneous = tangent slope (a limit).
The Review Hub · AP Calculus AB/BC Unit 2
TOPIC 2.2
Defining the Derivative
f′(x) = lim(h→0) [f(x+h) − f(x)] / h
f′(a) = lim(x→a) [f(x) − f(a)] / (x − a)
Notation for the derivative
f′(x), dy/dx, y′, d/dx[f(x)]
All mean the same: the instantaneous
rate of change of f.
f′(a) = value of the derivative at x = a
The derivative IS a function
f′(x) gives the tangent slope at every x
where f is differentiable.
Evaluate f′ at a point to get one slope;
the whole f′ describes all slopes.
f′(x) = lim(h→0) [f(x+h)−f(x)]/h — the derivative is itself a function.The Review Hub · AP Calculus AB/BC Unit 2
TOPIC 2.3
Estimating Derivatives at a Point
// estimate f′(2) from a table of values
x 1.9 2 2.1
f(x) 7.61 8.00 8.41
f′(2) ≈ [f(2.1) − f(1.9)] / (2.1 − 1.9) = (8.41 − 7.61)/0.2 = 4.0
Use a symmetric difference quotient
The average rate over a small interval CENTERED at the point gives the best estimate of the tangent slope.
Estimate f′(a) with a difference quotient over a small interval around a.
The Review Hub · AP Calculus AB/BC Unit 2
TOPIC 2.4
Differentiability & Continuity
differentiable at a ⟹ continuous at a (but NOT the reverse)
Where a derivative fails
Even a continuous function may NOT be
differentiable at:
• a CORNER or CUSP (e.g. |x| at 0)
• a VERTICAL TANGENT (infinite slope)
• a DISCONTINUITY (jump, hole)
The one-way implication
If f is differentiable at a, it MUST be
continuous there.
Continuity does NOT guarantee
differentiability (|x| is a counterexample).
Smooth (no corners) + no breaks = differentiable.
Differentiable ⟹ continuous , but corners/cusps/vertical tangents break differentiability.The Review Hub · AP Calculus AB/BC Unit 2
TOPIC 2.5
Applying the Power Rule
d/dx [ xⁿ ] = n · xⁿ⁻¹
Worked examples
d/dx[x⁵] = 5x⁴
d/dx[x] = 1
d/dx[√x] = d/dx[x^½] = ½x^(−½)
d/dx[1/x] = d/dx[x⁻¹] = −x⁻²
Works for any real exponent n.
Rewrite before differentiating
Convert radicals and fractions to powers:
³√x → x^(1/3)
5/x² → 5x⁻²
Then apply the power rule term by term.
Subtract 1 from the exponent; multiply by the old one.
Power rule: d/dx[xⁿ] = n·xⁿ⁻¹ — rewrite radicals/fractions as powers first.The Review Hub · AP Calculus AB/BC Unit 2
TOPIC 2.6
Constant, Sum, Difference & Multiple
d/dx[c] = 0 (constant rule)
d/dx[c·f] = c·f′ (constant multiple)
d/dx[f ± g] = f′ ± g′ (sum / difference)
Differentiate a polynomial TERM BY TERM.
Worked example
f(x) = 4x³ − 7x² + 2x − 9
f′(x) = 12x² − 14x + 2 (the constant −9 → 0)
Constant → 0; pull out multiples; differentiate sums term by term .The Review Hub · AP Calculus AB/BC Unit 2
TOPIC 2.7
Derivatives of sin, cos, eˣ, ln x
// memorize these building blocks
d/dx[sin x] = cos x
d/dx[cos x] = −sin x
d/dx[eˣ] = eˣ
d/dx[ln x] = 1/x
d/dx[aˣ] = aˣ·ln a
d/dx[log_a x] = 1/(x·ln a)
eˣ is its own derivative
The slope of eˣ at any point equals its height — that's why it appears in exponential growth models.
Memorize: (sin)′=cos, (cos)′=−sin, (eˣ)′=eˣ, (ln x)′=1/x .
The Review Hub · AP Calculus AB/BC Unit 2
TOPIC 2.8
The Product Rule
(f · g)′ = f′·g + f·g′
Worked example
y = x²·sin x
f = x² (f′=2x), g = sin x (g′=cos x)
y′ = 2x·sin x + x²·cos x
"Derivative of first times second, plus
first times derivative of second."
Common mistake
The derivative of a product is NOT the
product of the derivatives:
(f·g)′ ≠ f′·g′
You must use BOTH terms. When possible,
expanding first can avoid the rule.
✖(f·g)′ = f′g + fg′ — not the product of the derivatives.
The Review Hub · AP Calculus AB/BC Unit 2
TOPIC 2.9
The Quotient Rule
(f/g)′ = (f′·g − f·g′) / g²
Worked example
y = x / (x² + 1)
f=x (f′=1), g=x²+1 (g′=2x)
y′ = [1·(x²+1) − x·2x] / (x²+1)²
= (1 − x²) / (x²+1)²
"Low d-high minus high d-low, over low²."
Order matters here
The numerator is f′g − fg′ — subtraction,
so the ORDER cannot be swapped.
Divide the whole thing by the
denominator SQUARED.
Sometimes rewriting as a product is easier.
(f/g)′ = (f′g − fg′)/g² — mind the order and square the denominator.The Review Hub · AP Calculus AB/BC Unit 2
TOPIC 2.10
Derivatives of tan, cot, sec, csc
// derived from sin/cos with the quotient rule
d/dx[tan x] = sec²x
d/dx[cot x] = −csc²x
d/dx[sec x] = sec x·tan x
d/dx[csc x] = −csc x·cot x
The three "co-" functions all get a NEGATIVE sign.
Derivation example: tan x = sin x / cos x
(cos x·cos x − sin x·(−sin x)) / cos²x = (cos²x + sin²x)/cos²x = 1/cos²x = sec²x.
(tan)′=sec²x, (sec)′=sec x tan x ; the co-functions carry a minus sign.The Review Hub · AP Calculus AB/BC Unit 2
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How to use the visual review
Spend 30 seconds per slide before clicking next. Look at the diagram, then ask yourself: "Could I state this rule, or differentiate this function, from memory?"
Use the fullscreen button () on desktop for the best experience. Use arrow keys to navigate. Tap "Show all slides" to jump around.
This is great for review the night before the exam — fast, visual, and covers every idea you need to recognize in Unit 2.