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10 multiple-choice questions and 17 flashcards on Differentiation: Composite, Implicit, and Inverse Functions, about 7% of the High School Calculus bank. Every one carries a written rationale.
Differentiation: Composite, Implicit, and Inverse Functions is one of 8 chapters in CoStudy's High School Calculus bank, and it holds 10 of the bank's 150 multiple-choice questions — roughly 7% of the total. That proportion is not arbitrary: chapters follow the certifying body's published exam outline, and the number of questions in each is set by that domain's published weight, so the share of your practice time this chapter takes matches the share of the real exam it accounts for.
Studying by chapter is worth doing once you have a diagnostic score. A single overall percentage tells you whether you are close; it does not tell you which domain is dragging. Working a weak chapter in isolation, and re-testing it in isolation, is the fastest way to move a score that has stalled — and it is why the mock exams in CoStudy report by domain rather than as one number.
5 questions drawn from this chapter, with the full rationale shown — the controlling principle behind the right answer, and why each wrong option tempts and fails.
An object's velocity is v(t) = 3t² + 2t. Its acceleration at t = 2 is:
Answer: D — 14 (a(t) = v'(t) = 6t + 2; a(2) = 14)
Acceleration = derivative of velocity. a(t) = dv/dt = 6t + 2. At t=2: a(2) = 12 + 2 = 14. Foundation of motion problems in calculus.
What is the integral of cos(x)?
Answer: C — sin(x) + C (antiderivative of cos is sin)
Antiderivatives: ∫cos(x)dx = sin(x) + C, ∫sin(x)dx = -cos(x) + C, ∫sec²(x)dx = tan(x) + C. Distinguish from derivatives. Always include +C.
d/dx [sin(3x)] =
Answer: D — 3 cos(3x)
D) Chain rule: derivative of outer · derivative of inner = cos(3x) · 3 = 3cos(3x). A) Forgot to multiply by the derivative of the inner function (3). B) Did not differentiate at all. C) Confused with derivative of cos.
Apply the PRODUCT RULE to find d/dx[x·sin(x)].
Answer: D — sin(x) + x·cos(x) (product rule: (fg)' = f'g + fg' = (1)sin(x) + (x)cos(x))
Product rule: (fg)' = f'g + fg'. With f=x (f'=1) and g=sin(x) (g'=cos(x)): (1)(sin x) + (x)(cos x) = sin(x) + x cos(x). Critical differentiation rule.
What is implicit differentiation used for?
Answer: B — Functions not solvable for y; differentiate both sides w.r.t. x treating y as function of x, then solve for dy/dx
Implicit differentiation: when y can't be isolated (e.g., x² + y² = 25). Differentiate term-by-term w.r.t. x. Each y term gets dy/dx via chain rule. Then solve algebraically for dy/dx. Used for circles, ellipses, related rates.
4 cards from the 17 in this chapter.
What is d/dx[arctan x]?
1/(1 + x²).
What is d/dx[csc x]?
−csc x · cot x.
Find f''(x) for f(x) = x⁴.
f'(x) = 4x³; f''(x) = 12x².
What is d/dx[log_a x]?
1/(x · ln a).
These are a sample. The full Differentiation: Composite, Implicit, and Inverse Functions chapter runs 27 items with per-chapter progress tracking, on the web and in the iOS app.
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