AP Calculus AB: Differentiation
Differentiation is the engine of AP Calculus AB: nearly every later topic, from curve sketching to related rates to motion problems, depends on being fluent with these rules. This free, complete lesson covers the definition of the derivative, the Product, Quotient, and Chain Rules, implicit differentiation, derivatives of inverse functions, differentiability and continuity (including estimating derivatives numerically and graphically), the Mean Value Theorem, and L'Hospital's Rule. Each idea is followed by a fully worked example, and then a set of original practice problems with instant feedback and full explanations. Everything here is free, and you can keep practicing afterward with the full AP Calculus AB Question Bank linked below.
1. The Definition of the Derivative
At any x in its domain, the derivative of y = f(x) is defined as the limit of the difference quotient:
f'(x) = limh→0 f(x + h) − f(x)h
Geometrically, the difference quotient is the slope of a secant line; its limit is the slope of the tangent line, the instantaneous rate of change of f at that point. The second derivative, f''(x), is simply the derivative of f'(x).
Worked Example: Use the limit definition to find f'(x) for f(x) = x² + 3x.
f'(x) = limh→0 [(x+h)² + 3(x+h)] − [x² + 3x]h = limh→0 2xh + h² + 3hh = limh→0 (2x + h + 3) = 2x + 3
1. Use the limit definition to find f'(x) for f(x) = x² + 5x.
Explanation: [(x+h)² + 5(x+h) − (x² + 5x)]/h = (2xh + h² + 5h)/h = 2x + h + 5, which approaches 2x + 5 as h → 0.
2. For f(x) = 4x − 7, find f'(3) using the difference quotient [f(3 + h) − f(3)]/h as h → 0.
Explanation: f(3+h) = 5 + 4h and f(3) = 5, so the difference quotient is 4h/h = 4 for every h ≠ 0, giving a limit of 4.
3. As h → 0, a function's difference quotient [f(a + h) − f(a)]/h approaches 12. What does this represent geometrically?
Explanation: The limit of the difference quotient is, by definition, f'(a), the slope of the tangent line at that point.
4. The second derivative f''(x) is defined as…
Explanation: The second derivative is just the first derivative of f'(x), applying the same limit definition a second time.
2. The Product, Quotient, and Chain Rules
(uv)' = u · v' + v · u' uv' = v · u' − u · v'v²
The Chain Rule differentiates a composite function: find the derivative of the "outside" function first, then multiply by the derivative of the "inside" one. Formally, if y = f(u) and u = g(x), then dydx = dydu · dudx.
Worked Example: If y = (x² + 3x − 1)⁴, find dydx using the Chain Rule.
dydx = 4(x² + 3x − 1)³ · (2x + 3)
1. Find ddx (x³ + 4)⁵.
Explanation: By the Chain Rule: 5(x³ + 4)⁴ · 3x² = 15x²(x³ + 4)⁴.
2. Find ddx √(5x − 3).
Explanation: Writing as (5x − 3)1/2: 12(5x − 3)−1/2 · 5 = 52√(5x − 3).
3. If f(t) = (t² + 1)sin(2t), find f'(0) using the Product Rule.
Explanation: f'(t) = (t²+1) · 2cos(2t) + sin(2t) · 2t. At t = 0: (1)(2)(1) + (0)(0) = 2.
4. If f(v) = 3v1 − v², find f'(v) using the Quotient Rule, as a single fraction.
Explanation: f'(v) = (1 − v²)(3) − 3v(−2v)(1 − v²)² = 3 − 3v² + 6v²(1 − v²)² = 3 + 3v²(1 − v²)².
5. Find ddx [sin(3x² + 1)].
Explanation: By the Chain Rule: cos(3x² + 1) · 6x.
6. If y = ecos x, find dydx.
Explanation: By the Chain Rule: ecos x · (−sin x) = −sin(x) · ecos x.
3. Implicit Differentiation
When y is defined implicitly by an equation F(x, y) = 0 rather than solved explicitly, differentiate both sides with respect to x (treating y as a differentiable function of x and applying the Chain Rule to any y-term), then solve for dydx.
Worked Example: If x² + y² = 25, find dydx.
2x + 2ydydx = 0, so dydx = −xy
1. If x² − 3xy + y² = 7, find dydx.
Explanation: 2x − 3[x(dy/dx) + y] + 2y(dy/dx) = 0, so (dy/dx)(2y − 3x) = 3y − 2x, giving 3y − 2x2y − 3x.
2. If x³ + y³ = 6xy, find dydx.
Explanation: 3x² + 3y²(dy/dx) = 6[x(dy/dx) + y], so (dy/dx)(3y² − 6x) = 6y − 3x². Dividing by 3: 2y − x²y² − 2x.
3. For the curve x² + 4y² = 16, at what point(s) is the tangent line vertical?
Explanation: Implicitly, dy/dx = −x/(4y), which is undefined (vertical tangent) when y = 0. Substituting y = 0 into the original equation: x² = 16, so x = ±4.
4. Using implicit differentiation on x = cos y, verify the derivative of y = cos⁻¹x (arccos x).
Explanation: Differentiating x = cos y: 1 = −sin y(dy/dx), so dy/dx = −1/sin y. Since sin y ≥ 0 for 0 ≤ y ≤ π, sin y = √(1 − x²), giving −1√(1 − x²).
4. Derivatives of Inverse Functions
If f and g are inverse functions, their graphs are reflections of each other across y = x. The derivative of the inverse of a function at a point is the reciprocal of the derivative of the function at the corresponding (swapped-coordinate) point:
(f⁻¹)'(x) = 1f'(f⁻¹(x))
Worked Example: If f(4) = 10 and f'(4) = 6, what do we know about f⁻¹?
Since f passes through (4, 10), f⁻¹ must pass through (10, 4). And since f has slope 6 at that point, f⁻¹ must have slope 16 at (10, 4).
1. If f(2) = 9 and f'(2) = 3, what is the slope of f⁻¹ at x = 9?
Explanation: The slope of the inverse is the reciprocal: 1/f'(2) = 1/3.
2. A function f satisfies f(1) = 5 with f'(1) = 2, and f(5) = 11 with f'(5) = 4. If g is the inverse of f, find g'(5).
Explanation: To find g'(5), look at the point on f where y = 5, namely (1, 5). Since f'(1) = 2, g'(5) = 1/2.
3. Let y = f(x) = x³ + 2x − 3, and let g be the inverse function. Evaluate g'(0).
Explanation: f'(x) = 3x² + 2. By inspection, f(1) = 1 + 2 − 3 = 0, so x = 1 corresponds to y = 0. g'(0) = 1/f'(1) = 1/(3 + 2) = 1/5.
4. If f is one-to-one, differentiable, and f'(x) ≠ 0 everywhere, what is the general formula for (f⁻¹)'(x)?
Explanation: This is the general inverse-derivative formula: the reciprocal of f' evaluated at the corresponding point, f⁻¹(x).
5. Differentiability, Continuity, and Estimating Derivatives
If f is differentiable at x = c, then f is automatically continuous there, this follows directly from the definition of the derivative. The converse is false: a function can be continuous at a point without being differentiable there, for example, at a corner, a cusp, or a vertical tangent. When only tabular data is available, the symmetric difference quotient gives a numerical estimate:
f'(a) ≈ f(a + h) − f(a − h)2h
1. If a function f is differentiable at x = c, what can we conclude?
Explanation: Differentiability at a point always guarantees continuity at that same point.
2. If a function's graph has a sharp corner at x = c, what does this tell us about differentiability there?
Explanation: At a corner, the slope approaching from the left disagrees with the slope approaching from the right, so no single derivative value exists there.
3. Is it possible for a function to be continuous at a point but not differentiable there?
Explanation: f(x) = |x| is continuous at x = 0 but has a corner there, so it's not differentiable at that point. Differentiability implies continuity, but not the other way around.
4. A vertical tangent line at x = c indicates what about the derivative there?
Explanation: A vertical tangent would require an infinite slope, so the derivative does not exist there, even though the function itself can still be continuous.
5. A cyclist's distance (miles) is recorded every 2 minutes: d(0) = 0, d(2) = 1.5, d(4) = 3.4, d(6) = 5.0. Estimate d'(2) using the symmetric difference quotient.
Explanation: d'(2) ≈ [d(4) − d(0)] / (2 × 2) = (3.4 − 0) / 4 = 0.85 miles per minute.
6. The Mean Value Theorem
The Mean Value Theorem (MVT): if f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one c in (a, b) such that:
f(b) − f(a)b − a = f'(c)
Rolle's Theorem is a special case: if, in addition, f(a) = f(b), then there is a c in (a, b) where f'(c) = 0.
Worked Example: You drive 240 miles in 4 hours. What does the Mean Value Theorem guarantee about your speed along the way?
Your average speed was 240 ÷ 4 = 60 mph. Since your instantaneous speed (the derivative of distance) existed continuously throughout the trip, the MVT guarantees that your speedometer read exactly 60 mph at least once.
1. A runner covers 8 miles in 1 hour. By the Mean Value Theorem, what must be true about the runner's speed at some point during the run?
Explanation: The average speed (8 mph) must be matched by the instantaneous speed at least once, by the MVT.
2. What are the two hypotheses required for the Mean Value Theorem to apply to f on [a, b]?
Explanation: The MVT requires continuity on the closed interval and differentiability on the open interval, the endpoints themselves need not be differentiable.
3. For f(x) = x² − 4x on [0, 6], find the value of c guaranteed by the Mean Value Theorem.
Explanation: Average rate: [f(6) − f(0)] / 6 = (12 − 0)/6 = 2. Setting f'(x) = 2x − 4 equal to 2: 2x − 4 = 2, so x = 3.
4. Rolle's Theorem is a special case of the Mean Value Theorem that applies when…
Explanation: When f(a) = f(b), the average rate of change is 0, so Rolle's Theorem guarantees a point where the instantaneous rate (f') is also 0.
7. L'Hospital's Rule and Recognizing a Limit as a Derivative
L'Hospital's Rule: if a limit of f(x)g(x) is indeterminate of the form 00, take the derivative of the numerator and denominator separately and try the limit again.
A limit can also sometimes be recognized directly as the definition of a derivative: limh→0 [f(c + h) − f(c)]/h is exactly f'(c) for some function f and value c.
Worked Example: Find limx→2 x² − 4x − 2 using L'Hospital's Rule.
This is of the form 0/0. Differentiating numerator and denominator separately:
limx→2 2x1 = 2(2) = 4
1. Find limx→0 sin(4x)x using L'Hospital's Rule.
Explanation: This is 0/0. Differentiating: 4cos(4x)1, which at x = 0 gives 4(1) = 4.
2. Find limx→∞ ln xx².
Explanation: This is ∞/∞. Differentiating: 1/x2x = 12x², which approaches 0 as x → ∞.
3. Find limx→0 e³x − 1x.
Explanation: This is 0/0. Differentiating: 3e³x1, which at x = 0 gives 3(1) = 3.
4. Which indeterminate forms are eligible for L'Hospital's Rule as tested on the AP Calculus AB exam?
Explanation: While L'Hospital's Rule can theoretically be extended to other indeterminate forms with algebraic manipulation, the AP exam only tests it directly on 0/0 and ∞/∞.
5. The limit limh→0 (5 + h)³ − 5³h is the derivative of which function, evaluated at which point, and what is its value?
Explanation: This matches the definition f'(c) = limh→0 [f(c+h) − f(c)]/h with f(x) = x³ and c = 5. Since f'(x) = 3x², f'(5) = 3(25) = 75.
Common Mistakes to Avoid
- Forgetting the Chain Rule factor when differentiating a composition. Whenever the "inside" of a function isn't simply x, an extra factor (the inside function's own derivative) must be multiplied in, this is one of the single most common AP scoring deductions.
- Mixing up the order of terms in the Quotient Rule. It's "bottom times derivative of top, minus top times derivative of bottom," reversing this order flips the sign of the answer.
- Forgetting to differentiate y-terms with an extra dy/dx factor during implicit differentiation. Every time you differentiate a term containing y, the Chain Rule requires multiplying by dy/dx, since y is itself a function of x.
- Applying L'Hospital's Rule to a limit that isn't actually indeterminate. Confirm the limit truly evaluates to 0/0 or ∞/∞ by direct substitution first, using the rule on a limit that's already a real number gives a wrong answer.
- Assuming continuity is enough to guarantee differentiability. A function can be perfectly continuous at a point and still fail to have a derivative there, due to a corner, cusp, or vertical tangent.
Frequently Asked Questions
How can I tell whether to use the Product Rule or the Chain Rule?
The Product Rule applies when two separate functions are being multiplied together, like x² · sin x. The Chain Rule applies when one function is nested inside another, like sin(x²). Some problems, like (x²+1)sin(2x), genuinely need both rules working together.
Why does implicit differentiation need the extra dy/dx factor?
Since y is being treated as an unknown function of x rather than an independent variable, every time the Chain Rule is applied to a y-term, it must include the derivative of y with respect to x, exactly as it would for any other composite function.
Is Rolle's Theorem tested by name on the AP exam?
No, but the underlying idea, that a continuous, differentiable function returning to the same value must have a horizontal tangent somewhere in between, is still worth understanding as a special case of the more general Mean Value Theorem.
Where can I practice more problems like these?
The Differentiation quizzes in the AP Calculus AB Question Bank include additional original problems on this topic, along with quizzes covering every other topic tested throughout the course.