Source: AP Practice Questions, Units 2 & 3
Tags: derivatives, product rule, quotient rule, chain rule, trig derivatives, logarithmic differentiation, exponential derivatives, AP Calculus AB
Units 2 and 3 cover the core toolkit for taking derivatives: the product rule, quotient rule, and chain rule, plus the specific derivative formulas for trig, exponential, and logarithmic functions. Nearly every AP question in this section combines two or more of these rules in a single problem, so fluency with each rule individually and in combination is essential.
Derivative
The instantaneous rate of change of a function at a point. Written f'(x) or dy/dx.
Product rule
For two functions multiplied together: d/dx[f(x)g(x)] = f'(x)g(x) + f(x)g'(x). Think "derivative of the first times the second, plus the first times the derivative of the second."
Quotient rule
For a fraction of two functions: d/dx[f(x)/g(x)] = [f'(x)g(x) - f(x)g'(x)] / [g(x)]². Think "low d-high minus high d-low, over low squared."
Chain rule
For composite functions: d/dx[f(g(x))] = f'(g(x)) · g'(x). Differentiate the outer function, keep the inner function inside, then multiply by the derivative of the inner function.
Composite function
A function inside another function, written f(g(x)). The chain rule is the tool for differentiating these.
The chain rule is the single most tested skill across these units. It appears on its own and layered inside product and quotient rule problems.
If y = (x³ + 1)², then dy/dx = 2(x³ + 1) · 3x² = 6x²(x³ + 1).
Outer function: ( )², derivative is 2( ).
Inner function: x³ + 1, derivative is 3x².
Answer: (E)
If f(x) = cos(3x), then f'(x) = -3sin(3x).
At x = π/9: f'(π/9) = -3 sin(3 · π/9) = -3 sin(π/3) = -3 · (√3/2) = -3√3/2.
Answer: (E)
When a function is a product and one (or both) factors require the chain rule, apply the product rule first, then chain-rule each factor as needed.
If f(x) = (x - 1)(x² + 2)³, then:
f'(x) = (1)(x² + 2)³ + (x - 1) · 3(x² + 2)² · (2x)
Factor out (x² + 2)²: = (x² + 2)²[(x² + 2) + 6x(x - 1)]
Simplify the bracket: = (x² + 2)²[7x² - 6x + 2]
Answer: (D)
If y = x² sin(2x), then:
dy/dx = 2x · sin(2x) + x² · 2cos(2x)
Factor: = 2x[sin(2x) + x cos(2x)]
Answer: (E)
Applied when the function is a ratio. Watch the sign in the numerator: it is always "low d-high minus high d-low."
If y = (2x + 3)/(3x + 2), then:
dy/dx = [(2)(3x + 2) - (3)(2x + 3)] / (3x + 2)²
Numerator: 6x + 4 - 6x - 9 = -5
Result: dy/dx = -5 / (3x + 2)²
Answer: (D)
The derivative of ln(u) is (1/u) · u', combining the basic ln derivative with the chain rule.
If f(x) = ln(x + 4 + e^(-3x)), then:
f'(x) = (1 - 3e^(-3x)) / (x + 4 + e^(-3x))
At x = 0: f'(0) = (1 - 3·1) / (0 + 4 + 1) = -2/5
Answer: (A)
When asked for d/dx[f(g(x))], apply the chain rule using the given formula for f.
If f(x) = x² + 2x, find d/dx[f(ln x)]:
f(ln x) = (ln x)² + 2 ln x
d/dx = 2(ln x) · (1/x) + 2 · (1/x) = (2 ln x + 2) / x
Answer: (A)
For y = e^(u), the first derivative is e^(u) · u'. The second derivative requires the product rule on that result.
If y = e^(x³), then:
y' = 3x² · e^(x³)
y'' = 6x · e^(x³) + 3x² · 3x² · e^(x³) = (6x + 9x⁴) · e^(x³)
Answer: (D)
Core derivative formulas to memorise:
d/dx[xⁿ] = nxⁿ⁻¹
d/dx[sin x] = cos x
d/dx[cos x] = -sin x
d/dx[tan x] = sec²x
d/dx[eˣ] = eˣ
d/dx[ln x] = 1/x
d/dx[arctan x] = 1/(1 + x²)
Combined rules template:
Product rule: (fg)' = f'g + fg'
Quotient rule: (f/g)' = (f'g - fg') / g²
Chain rule: [f(g(x))]' = f'(g(x)) · g'(x)
⚠️ The chain rule appears in almost every derivative problem on the AP exam. If there is a function inside another function, you need it.
⚠️ Common mistake: forgetting to multiply by the inner derivative when applying the chain rule. For cos(3x), the derivative is -3sin(3x), not just -sin(3x).
⚠️ In quotient rule problems, the most frequent error is getting the sign wrong in the numerator. It is f'g - fg', not fg' - f'g.
⚠️ When a problem gives f(x) and asks for d/dx[f(g(x))], this is a chain rule problem in disguise. Substitute g(x) into f, then differentiate.
⚠️ For second derivatives of exponentials like e^(u), you must use the product rule on y' = u' · e^(u). Do not skip this step.
Q: If y = (x³ + 1)², what is dy/dx?
A: 6x²(x³ + 1). Apply chain rule: bring down the 2, keep the inner function, multiply by the derivative of x³ + 1, which is 3x².
Q: If f(x) = cos(3x), what is f'(π/9)?
A: -3√3/2. The derivative is -3sin(3x). At π/9, you get -3 sin(π/3) = -3(√3/2).
Q: If y = (2x + 3)/(3x + 2), what is dy/dx?
A: -5/(3x + 2)². Quotient rule gives numerator (2)(3x+2) - (3)(2x+3) = -5.
Q: If f(x) = ln(x + 4 + e^(-3x)), what is f'(0)?
A: -2/5. The numerator of the derivative at x = 0 is 1 - 3 = -2, the denominator is 0 + 4 + 1 = 5.
Q: If f(x) = (x - 1)(x² + 2)³, what is f'(x)?
A: (x² + 2)²(7x² - 6x + 2). Product rule first, chain rule on the cubic factor, then factor out (x² + 2)².
Q: If y = x² sin(2x), what is dy/dx?
A: 2x[sin(2x) + x cos(2x)]. Product rule with chain rule on sin(2x).
differentiation, derivative rules, product rule, quotient rule, chain rule, composite functions, trig derivatives, cosine derivative, sine derivative, exponential derivative, logarithmic derivative, ln derivative, AP Calculus AB, Unit 2, Unit 3, second derivative, d/dx