Difficulty: Foundational to Intermediate | Prerequisites: Algebra II, basic function notation, limits
Derivatives measure instantaneous rate of change. Every AP Calculus AB exam leans heavily on your ability to differentiate quickly and accurately, whether the function is a simple polynomial, a composition of functions, a product, a quotient, or defined implicitly. The questions in this set test the core rules you will use on roughly 40–60% of the entire exam. If you can differentiate fluently, the applications sections become far more manageable.
The chain rule, product rule, and quotient rule are the workhorses of AP Calculus AB. Most exam questions combine two or more of these rules in a single problem, and errors almost always come from forgetting an inner derivative or misapplying the quotient rule sign.
Derivative
The instantaneous rate of change of a function at a point; geometrically, the slope of the tangent line. In simple terms, it tells you how fast y is changing as x changes.
Chain rule
If y = f(g(x)), then dy/dx = f'(g(x)) · g'(x). In simple terms, differentiate the outer function, leave the inner function alone, then multiply by the derivative of the inner function.
Product rule
If y = f(x) · g(x), then dy/dx = f'(x) · g(x) + f(x) · g'(x). Think of it as: "derivative of the first times the second, plus the first times the derivative of the second."
Quotient rule
If y = f(x)/g(x), then dy/dx = [f'(x) · g(x) – f(x) · g'(x)] / [g(x)]². Think of it as: "low d-high minus high d-low, over low squared."
Implicit differentiation
A technique for finding dy/dx when y is not isolated. Differentiate both sides with respect to x, apply the chain rule to any term containing y (tacking on dy/dx), then solve for dy/dx.
Derivative of an inverse function
If g(x) = f⁻¹(x), then g'(x) = 1 / f'(g(x)). In simple terms, the slope of the inverse at a point is the reciprocal of the slope of the original at the corresponding point.
The chain rule appears whenever a function is composed with another function. On the AP exam, watch for:
Powers of expressions: y = (x⁴ + 2)² requires treating (x⁴ + 2) as the inner function
Trigonometric compositions: y = sin(3x) has inner function 3x
Exponential compositions: f(x) = xe²ˣ requires both the product rule and chain rule
Logarithmic compositions: f(x) = ln(e⁻⁴ˣ + 5x + 4) has a complex inner function
Worked example (from Q2): y = (x⁴ + 2)²
Outer function: ( )², derivative = 2( )
Inner function: x⁴ + 2, derivative = 4x³
Result: dy/dx = 2(x⁴ + 2) · 4x³ = 8x³(x⁴ + 2)
The answer is A) 8x³(x⁴ + 2). Option C, 2(4x³ + 2), is a common error where the student differentiates the inside but forgets to leave the inner function intact in the outer derivative.
The product rule is needed whenever two variable expressions are multiplied together.
Worked example (from Q8): y = x³ sin(3x)
f(x) = x³, so f'(x) = 3x²
g(x) = sin(3x), so g'(x) = cos(3x) · 3 = 3cos(3x) (chain rule on the inner 3x)
dy/dx = 3x² · sin(3x) + x³ · 3cos(3x) = 3x²(sin(3x) + x cos(3x))
The answer is E) 3x²(sin(3x) + x cos(3x)). The most common mistake is forgetting the chain rule factor of 3 when differentiating sin(3x).
Worked example (from Q16): y = (3x + 4)/(4x + 3)
f(x) = 3x + 4, f'(x) = 3
g(x) = 4x + 3, g'(x) = 4
dy/dx = [3(4x + 3) – (3x + 4)(4)] / (4x + 3)²
= [12x + 9 – 12x – 16] / (4x + 3)²
= –7 / (4x + 3)²
The answer is C) –7/(4x + 3)².
When a curve is defined by an equation involving both x and y (not solved for y), differentiate every term with respect to x. Any time you differentiate a y-term, multiply by dy/dx.
Worked example (from Q23): 2x²y – 3 = y² + 3x² at the point (2, 3)
Differentiate both sides:
d/dx[2x²y] = 4xy + 2x²(dy/dx) (product rule)
d/dx[–3] = 0
d/dx[y²] = 2y(dy/dx)
d/dx[3x²] = 6x
So: 4xy + 2x²(dy/dx) = 2y(dy/dx) + 6x
Collect dy/dx terms: 2x²(dy/dx) – 2y(dy/dx) = 6x – 4xy
dy/dx = (6x – 4xy) / (2x² – 2y)
At (2, 3): dy/dx = (12 – 24) / (8 – 6) = –12/2 = –6
The answer is E) –6.
Worked example (from Q25): f(x) = x³ – 2x, g(x) = f⁻¹(x), g(4) = 2. Find g'(4).
Since g = f⁻¹, the formula gives g'(4) = 1/f'(g(4)) = 1/f'(2).
f'(x) = 3x² – 2, so f'(2) = 12 – 2 = 10.
g'(4) = 1/10
The answer is A) 1/10.
Worked example (from Q26): f(x) = ln(e⁻⁴ˣ + 5x + 4). Find f'(0).
f'(x) = (–4e⁻⁴ˣ + 5) / (e⁻⁴ˣ + 5x + 4)
At x = 0: f'(0) = (–4e⁰ + 5) / (e⁰ + 0 + 4) = (–4 + 5) / (1 + 4) = 1/5
The answer is A) 1/5.
Core derivative rules to memorise:
d/dx[xⁿ] = nxⁿ⁻¹
d/dx[sin x] = cos x
d/dx[cos x] = –sin x
d/dx[eˣ] = eˣ
d/dx[ln x] = 1/x
d/dx[eᵘ] = eᵘ · u' (chain rule form)
d/dx[ln u] = u'/u (chain rule form)
Product rule: (fg)' = f'g + fg'
Quotient rule: (f/g)' = (f'g – fg') / g²
Inverse function: (f⁻¹)'(a) = 1 / f'(f⁻¹(a))
Derivatives describe rates of change in every quantitative field: velocity in physics, marginal cost in economics, reaction rates in chemistry. When you differentiate a position function you get velocity; when you differentiate a cost function you get the cost of producing one more unit.
Students often forget the inner derivative when applying the chain rule. If the inner function is just "x", the chain rule still applies; its factor is simply 1.
When using the quotient rule, the numerator is "low d-high MINUS high d-low". Swapping the order flips the sign of the answer.
Implicit differentiation requires the chain rule on every y-term. Students sometimes differentiate y² as 2y instead of 2y(dy/dx).
The derivative of the inverse is 1/f'(g(x)), not 1/f'(x). You must evaluate f' at the corresponding point on the original function.
⚠️ The chain rule is tested on nearly every AP exam, often buried inside product or quotient rule problems. Build the habit of identifying the inner function before writing anything.
⚠️ Implicit differentiation appears in both multiple choice and free response. Practice isolating dy/dx cleanly.
⚠️ Inverse function derivatives appear roughly once per exam. The formula is short but the conceptual step of finding the corresponding point trips students up.
True or false: The derivative of (sin x)³ is 3(sin x)² · cos x.
Fill in the blank: The quotient rule numerator is f'g ______ fg'.
True or false: When differentiating implicitly, d/dx[y³] = 3y².
True or false: If g = f⁻¹ and f(2) = 5, then g'(5) = 1/f'(2).
Fill in the blank: d/dx[ln(u)] = ______.
Answers: 1. True. 2. minus (–). 3. False; it is 3y²(dy/dx). 4. True. 5. u'/u.
Q: y = (x⁴ + 2)². What is dy/dx?
A: 8x³(x⁴ + 2). Apply the chain rule: bring down the 2, keep the inner function, multiply by the derivative of the inner function (4x³).
Q: y = x³ sin(3x). What is dy/dx?
A: 3x²(sin(3x) + x cos(3x)). Product rule, with a chain rule on sin(3x) producing the factor of 3.
Q: What is the slope of the tangent to 2x²y – 3 = y² + 3x² at (2, 3)?
A: –6. Implicit differentiation, then substitute the point.
Q: f(x) = x³ – 2x, g = f⁻¹, g(4) = 2. What is g'(4)?
A: 1/10. Use the inverse derivative formula: g'(4) = 1/f'(2) = 1/10.
Q: y = (3x + 4)/(4x + 3). What is dy/dx?
A: –7/(4x + 3)². Quotient rule; the numerator simplifies to –7.
These derivative techniques feed directly into every applications topic: finding tangent lines, determining where a function is increasing or decreasing, solving related rates problems, and analysing motion along a line. Mastering the mechanics here is a prerequisite for the applications sections of the course.
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