Source: AP Calculus BC Mock 3 (College Board 2008 PT)
Tags: derivative, quotient rule, chain rule, implicit differentiation, related rates, inverse functions, critical points, inflection points, Fundamental Theorem of Calculus, AP Calculus BC
Difficulty: Intermediate to Advanced Prerequisites: AP Calculus AB or strong understanding of limits, basic derivatives, and the Fundamental Theorem of Calculus.
Differentiation is the backbone of the entire AP Calculus BC exam, appearing in roughly 40% of questions across both multiple-choice and free-response sections. This set of notes covers the techniques and applications you need: the quotient rule, chain rule in nested and implicit settings, derivatives of inverse functions, related rates, and using derivatives to analyse curves (critical points, inflection, concavity). If you are comfortable taking basic derivatives of polynomials and trigonometric functions but get tripped up by compositions, implicit equations, or applied rate problems, start here.
Know your differentiation rules cold, especially quotient rule and chain rule, because the exam layers them together. Implicit differentiation and inverse-function derivatives both appear regularly and follow directly from the chain rule. Related rates and curve analysis (finding extrema, concavity, inflection) are the main application areas tested.
Quotient rule
The derivative of a ratio of two functions: if f(x) = g(x)/h(x), then f'(x) = [h(x)g'(x) - g(x)h'(x)] / [h(x)]^2. Think of it as "low d-high minus high d-low, over the square of what's below."
Chain rule
The derivative of a composite function: d/dx [f(g(x))] = f'(g(x)) * g'(x). In simple terms, differentiate the outer function first (leaving the inner function alone), then multiply by the derivative of the inner function.
Implicit differentiation
A technique for finding dy/dx when y is not isolated on one side of the equation. Differentiate both sides with respect to x, applying the chain rule to every y term (each picks up a dy/dx factor), then solve for dy/dx. Think of it as treating y as a hidden function of x.
Related rates
Problems where two or more quantities change with time, linked by a geometric or physical equation. You differentiate the linking equation with respect to t and substitute known rates to find an unknown rate. In simple terms, if two things are connected and one is changing, you can work out how fast the other changes.
Critical point
A value of x where f'(x) = 0 or f'(x) does not exist. These are the only candidates for local maxima or minima. Think of it as a place where the curve's slope is flat or has a sharp corner.
Inflection point
A point where the concavity of f changes sign, i.e. f''(x) changes from positive to negative or vice versa. In simple terms, it is where the curve switches from bending upward to bending downward (or the reverse).
Derivative of an inverse function
If g is the inverse of f, then g'(y) = 1 / f'(g(y)). In simple terms, the slope of the inverse at a point is the reciprocal of the slope of the original function at the corresponding point.
Definition of the derivative (limit form)
f'(x) = lim(h->0) [f(x+h) - f(x)] / h. This limit definition appears on the exam in disguise: any expression that looks like a difference quotient with h approaching 0 is asking you to recognise a derivative.
For f(x) = (3x - 2)/(2x + 3), apply the quotient rule directly: f'(x) = [(2x+3)(3) - (3x-2)(2)] / (2x+3)^2 = (6x+9 - 6x+4) / (2x+3)^2 = 13/(2x+3)^2.
The numerator of the derivative often simplifies to a constant when the original is a linear-over-linear rational function. Watch for this shortcut.
Common error: mixing up the order of subtraction. It is always (denominator * derivative of numerator) minus (numerator * derivative of denominator).
The expression [tan(x+h) - tan(x)] / h as h approaches 0 is simply the definition of the derivative of tan(x). The answer is sec^2(x), not some exotic limit.
Any time you see [f(x+h) - f(x)] / h with h -> 0, identify which function's derivative is being asked for. This pattern appears frequently in Part A.
For f(x) = arccos(x^2), the chain rule gives f'(x) = -1/sqrt(1 - x^4) * 2x = -2x/sqrt(1 - x^4).
The outer function is arccos(u), whose derivative is -1/sqrt(1 - u^2). The inner function is u = x^2, whose derivative is 2x.
Nested chain rule: for functions like e^(sin(x^2)), peel off layers one at a time, outside in.
Given y + 2 = x^2/2 - 2sin(y), differentiate both sides with respect to x: dy/dx = x - 2cos(y) * dy/dx. Collect dy/dx terms and solve.
At the point (2, 0): dy/dx + 0 = 2 - 2cos(0) * dy/dx, so dy/dx + 2(dy/dx) = 2, giving dy/dx = 2/3.
When evaluating at a specific point, substitute the point's coordinates after differentiating, not before.
If f'(x) = 2/x and f(sqrt(e)) = 5, find f(e). Since f'(x) = 2/x, we have f(x) = 2ln(x) + C. Use f(sqrt(e)) = 5 to get C: 2 * (1/2) + C = 5, so C = 4. Then f(e) = 2(1) + 4 = 6.
For a table of values of a function and its derivative, if g is the inverse of f, then g'(b) = 1/f'(a) where f(a) = b. Read the table carefully: find the row where f(x) equals the argument of g'.
Rectangle with sides x and y and diagonal z: x^2 + y^2 = z^2. Differentiate with respect to t: 2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt).
At x = 4, y = 3: z = 5, and dx/dt = dz/dt. If dy/dt = k * dz/dt, substitute and solve for k: 2(4)(dz/dt) + 2(3)(k)(dz/dt) = 2(5)(dz/dt). Divide through by 2(dz/dt): 4 + 3k = 5, so k = 1/3.
Circle area and circumference: A = pir^2, C = 2pir. dA/dt = 2pir(dr/dt), dC/dt = 2pi(dr/dt). If dA/dt = 2 * dC/dt, then 2pir*(dr/dt) = 2 * 2pi(dr/dt), giving r = 2.
For f(x) = 9x^(2/3) + 3x - 6, set f'(x) = 6x^(-1/3) + 3 = 0, giving x^(-1/3) = -1/2, so x = -8. Check the sign of f' around x = -8 and at x = 0 (where f' is undefined) to classify.
If f is twice differentiable with no inflection points, and you know f(6), f'(6), f''(6), then the tangent line at x = 6 gives bounds on f(7). Since f''(6) = -2 < 0 and there are no inflection points, f is concave down throughout, so the tangent line overestimates: f(7) < f(6) + f'(6)(1) = 3 - 1/2 = 2.5.
For f(x) = integral from 4 to 2x of sqrt(t^2 - t) dt, by the chain rule version of FTC: f'(x) = sqrt((2x)^2 - 2x) * 2 = 2*sqrt(4x^2 - 2x).
At x = 2: f'(2) = 2sqrt(16 - 4) = 2sqrt(12) = 4sqrt(3). But check each answer choice format: 2sqrt(12) = 4sqrt(3), which may appear as 2sqrt(12) in the choices.
If the integral of f(x) from 0 to 4 equals 8, and you know f and f' at the endpoints, you can find the integral of xf'(x) dx from 0 to 4 using integration by parts: integral of xf'(x) dx = xf(x) - integral of f(x) dx. Evaluating: [4f(4) - 0*f(0)] - 8 = 4(-3) - 0 - 8 = -20.
If f'(x) = |x - 2|, then f' is always non-negative, so f is always non-decreasing. f' has a minimum of 0 at x = 2 (a cusp in f'). f is piecewise linear with a corner at x = 2. The graph of f has a V-shaped bottom at x = 2.
From the graph of f', inflection points of f correspond to local extrema of f'. Count the peaks and troughs of f' to count inflection points of f.
\frac{d}{dx}\left[\frac{g(x)}{h(x)}\right] = \frac{h(x)\,g'(x) - g(x)\,h'(x)}{[h(x)]^2}\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\text{If } g = f^{-1}, \text{ then } g'(y) = \frac{1}{f'(g(y))}\frac{d}{dx}\left[\int_a^{u(x)} f(t)\,dt\right] = f(u(x)) \cdot u'(x)\frac{d}{dx}[\arccos(u)] = \frac{-u'}{\sqrt{1-u^2}}, \quad \frac{d}{dx}[\arcsin(u)] = \frac{u'}{\sqrt{1-u^2}}, \quad \frac{d}{dx}[\arctan(u)] = \frac{u'}{1+u^2}Students often reverse the subtraction order in the quotient rule, writing gh' - hg' instead of hg' - gh'. The mnemonic "lo d-hi minus hi d-lo" helps, but only if you remember that "lo" (the denominator) comes first.
When using implicit differentiation, students forget to attach dy/dx every time they differentiate a y term. If y appears inside a trig or exponential function, the chain rule must still apply to y.
In related rates, a common mistake is substituting numerical values for changing quantities before differentiating. Substitute after you differentiate, not before, or the changing quantity vanishes.
Students often confuse the derivative of an inverse function with 1/f(x). The correct formula is g'(y) = 1/f'(g(y)), which requires evaluating f' at the corresponding x-value, not at y.
⚠️ The quotient rule appears almost every year in Part A (no calculator). Be fast and accurate with the algebra.
⚠️ Recognising the limit definition of the derivative in disguise is a near-guaranteed question. If you see [f(a+h) - f(a)]/h, stop and name the derivative.
⚠️ The chain rule version of the Fundamental Theorem of Calculus (d/dx of an integral with a variable upper limit) is tested in both MC and FRQ.
⚠️ Related rates problems appear in the free-response section most years. Draw a diagram, label variables, write the linking equation, then differentiate.
⚠️ Questions about the graph of f' and what it tells you about f (increasing/decreasing, concavity, inflection) are heavily tested in Part B.
True or false: The derivative of arctan(x^3) is 3x^2 / (1 + x^6). (True)
Fill in the blank: If f(2) = 7 and g is the inverse of f, then g(7) = ______. (2)
True or false: If f'(c) = 0, then f has a local extremum at x = c. (False, it could be an inflection point of f, e.g. f(x) = x^3 at x = 0)
Fill in the blank: In the quotient rule, the denominator of the result is the original denominator ______. (squared)
True or false: To find dy/dx implicitly from x^2 + y^2 = 25, you differentiate and get 2x + 2y(dy/dx) = 0. (True)
Q: If f(x) = (3x - 2)/(2x + 3), what is f'(x)?
A: 13/(2x + 3)^2. Apply the quotient rule: numerator = (2x+3)(3) - (3x-2)(2) = 6x+9 - 6x+4 = 13.
Q: For values of h close to 0, which function best approximates [tan(x+h) - tan(x)]/h?
A: sec^2(x). The expression is the limit definition of the derivative of tan(x).
Q: If f'(x) = 2/x and f(sqrt(e)) = 5, what is f(e)?
A: 6. Since f(x) = 2ln(x) + C, use f(sqrt(e)) = 2*(1/2) + C = 5, so C = 4. Then f(e) = 2(1) + 4 = 6.
Q: The sides and diagonal of a rectangle are increasing with time. At the instant when x = 4 and y = 3, dx/dt = dz/dt and dy/dt = k(dz/dt). What is k?
A: 1/3. From x^2 + y^2 = z^2, differentiate and substitute: 8(dz/dt) + 6k(dz/dt) = 10(dz/dt), giving 8 + 6k = 10, so k = 1/3.
Q: The radius of a circle is increasing. At a certain instant, dA/dt = 2(dC/dt). What is r?
A: 2. dA/dt = 2pir*(dr/dt), dC/dt = 2pi(dr/dt). Setting 2pir*(dr/dt) = 2*[2pi(dr/dt)] gives r = 2.
Q: What is the slope of the tangent to y + 2 = x^2/2 - 2sin(y) at (2, 0)?
A: 2/3. Implicit differentiation gives (1 + 2cos(y)) dy/dx = x. At (2,0): 3(dy/dx) = 2.
This material connects directly to integration (the Fundamental Theorem ties differentiation and integration together, and integration by parts relies on the product rule in reverse). Related rates problems link to parametric and polar topics, which also require the chain rule extensively. Curve analysis with first and second derivatives feeds into the free-response questions about particle motion and function behaviour.
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