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This free AP Calculus AB Unit 3 practice test covers Differentiation: Composite, Implicit, and Inverse Functions only: the chain rule from formulas, tables and graphs, implicit differentiation and tangent lines, derivatives of inverse functions, the derivatives of arcsin, arccos and arctan, choosing the right procedure, and second derivatives, including implicit ones. It has 16 four-choice multiple-choice questions and one 9-point free-response question on an implicitly defined curve, 25 points in 45 minutes with no calculator. Every explanation shows the working and says why each wrong answer is wrong.

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  1. Question 11 mark · Chain rule
    The table gives selected values of the differentiable functions ff and gg and their derivatives. If h(x)=f(g(x))h(x) = f(g(x)), what is the value of h′(2)h^{\prime}(2)?
    xf(x)f′(x)g(x)g′(x)16−40323−34531−2724−1−280\begin{array}{c|c|c|c|c} \hline x & f(x) & f^{\prime}(x) & g(x) & g^{\prime}(x) \\ \hline 1 & 6 & -4 & 0 & 3 \\ 2 & 3 & -3 & 4 & 5 \\ 3 & 1 & -2 & 7 & 2 \\ 4 & -1 & -2 & 8 & 0 \\ \hline \end{array}
    • A.
      −15-15
    • B.
      −2-2
    • C.
      −10-10
    • D.
      −6-6
  2. Question 21 mark · Chain rule
    The graph of the function gg on the interval 0≤x≤60 \leq x \leq 6 consists of two line segments: one from (0,1)(0, 1) to (3,7)(3, 7) and one from (3,7)(3, 7) to (6,4)(6, 4). If h(x)=g(x2)h(x) = g(x^2), what is the value of h′(2)h^{\prime}(2)?
    • A.
      −4-4
    • B.
      −1-1
    • C.
      88
    • D.
      2424
  3. Question 31 mark · Chain rule
    If y=cos⁡3(2x)y = \cos^3(2x), then dydx=\dfrac{dy}{dx} =
    • A.
      3cos⁡2(2x)3\cos^2(2x)
    • B.
      −3cos⁡2(2x)sin⁡(2x)-3\cos^2(2x)\sin(2x)
    • C.
      6cos⁡2(2x)sin⁡(2x)6\cos^2(2x)\sin(2x)
    • D.
      −6cos⁡2(2x)sin⁡(2x)-6\cos^2(2x)\sin(2x)
  4. Question 41 mark · Chain rule
    If f(x)=2x2+1f(x) = \sqrt{2x^2 + 1}, what is the slope of the line tangent to the graph of ff at x=2x = 2?
    • A.
      16\frac{1}{6}
    • B.
      43\frac{4}{3}
    • C.
      83\frac{8}{3}
    • D.
      23\frac{2}{3}
  5. Question 51 mark · Implicit differentiation
    The curve defined by x2y+y3=10x^2 y + y^3 = 10 passes through the point (3,1)(3, 1). What is the value of dydx\dfrac{dy}{dx} at (3,1)(3, 1)?
    • A.
      −2-2
    • B.
      −12-\frac{1}{2}
    • C.
      −1-1
    • D.
      00
  6. Question 61 mark · Implicit differentiation
    At which of the following points on the curve x2+2xy+3y2=6x^2 + 2xy + 3y^2 = 6 is the line tangent to the curve horizontal?
    • A.
      (0,2)(0, \sqrt{2})
    • B.
      (3,−1)(3, -1)
    • C.
      (1,−1)(1, -1)
    • D.
      (3,−3)(\sqrt{3}, -\sqrt{3})
  7. Question 71 mark · Implicit differentiation
    Which of the following is an equation of the line tangent to the curve xey+y=2x e^{y} + y = 2 at the point (2,0)(2, 0)?
    • A.
      y=−13(x−2)y = -\frac{1}{3}(x - 2)
    • B.
      y=−12(x−2)y = -\frac{1}{2}(x - 2)
    • C.
      y=−3(x−2)y = -3(x - 2)
    • D.
      y=3(x−2)y = 3(x - 2)
  8. Question 81 mark · Derivatives of inverse functions
    Let f(x)=x3+2x+1f(x) = x^3 + 2x + 1, and let gg be the inverse function of ff. What is the value of g′(4)g^{\prime}(4)?
    • A.
      150\frac{1}{50}
    • B.
      55
    • C.
      15\frac{1}{5}
    • D.
      11
  9. Question 91 mark · Derivatives of inverse functions
    The function ff is differentiable and increasing, and selected values of ff and its derivative f′f^{\prime} are given in the table. If gg is the inverse function of ff, what is the value of g′(5)g^{\prime}(5)?
    x1235f(x)02515f′(x)1346\begin{array}{c|c|c|c|c} \hline x & 1 & 2 & 3 & 5 \\ \hline f(x) & 0 & 2 & 5 & 15 \\ \hline f^{\prime}(x) & 1 & 3 & 4 & 6 \\ \hline \end{array}
    • A.
      14\frac{1}{4}
    • B.
      16\frac{1}{6}
    • C.
      33
    • D.
      44
  10. Question 101 mark · Derivatives of inverse functions
    The function ff is differentiable and one-to-one, with f(2)=7f(2) = 7 and f′(2)=−3f^{\prime}(2) = -3. Which of the following is an equation of the line tangent to the graph of f−1f^{-1} at x=7x = 7?
    • A.
      y−7=−13(x−2)y - 7 = -\frac{1}{3}(x - 2)
    • B.
      y−2=−3(x−7)y - 2 = -3(x - 7)
    • C.
      y−2=13(x−7)y - 2 = \frac{1}{3}(x - 7)
    • D.
      y−2=−13(x−7)y - 2 = -\frac{1}{3}(x - 7)
  11. Question 111 mark · Inverse trig derivatives
    ddx[arctan⁡(3x)]=\dfrac{d}{dx}\left[\arctan(3x)\right] =
    • A.
      11+9x2\dfrac{1}{1 + 9x^2}
    • B.
      31+9x2\dfrac{3}{1 + 9x^2}
    • C.
      31+3x2\dfrac{3}{1 + 3x^2}
    • D.
      3sec⁡2(3x)3\sec^2(3x)
  12. Question 121 mark · Inverse trig derivatives
    If f(x)=arccos⁡(2x)f(x) = \arccos(2x), then f′(14)=f^{\prime}\left(\frac{1}{4}\right) =
    • A.
      433\frac{4\sqrt{3}}{3}
    • B.
      −233-\frac{2\sqrt{3}}{3}
    • C.
      −433-\frac{4\sqrt{3}}{3}
    • D.
      −81515-\frac{8\sqrt{15}}{15}
  13. Question 131 mark · Selecting derivative procedures
    If f(x)=ln⁡(x2x+1)f(x) = \ln\left(\dfrac{x^2}{x + 1}\right) for x>0x > 0, then f′(1)=f^{\prime}(1) =
    • A.
      22
    • B.
      34\frac{3}{4}
    • C.
      52\frac{5}{2}
    • D.
      32\frac{3}{2}
  14. Question 141 mark · Selecting derivative procedures
    If h(x)=xarcsin⁡xh(x) = x \arcsin x, then h′(12)=h^{\prime}\left(\frac{1}{2}\right) =
    • A.
      π6+33\frac{\pi}{6} + \frac{\sqrt{3}}{3}
    • B.
      233\frac{2\sqrt{3}}{3}
    • C.
      π6+233\frac{\pi}{6} + \frac{2\sqrt{3}}{3}
    • D.
      π3+33\frac{\pi}{3} + \frac{\sqrt{3}}{3}
  15. Question 151 mark · Higher-order derivatives
    For the curve x2−y2=16x^2 - y^2 = 16, what is the value of d2ydx2\dfrac{d^2y}{dx^2} at the point (5,3)(5, 3)?
    • A.
      13\frac{1}{3}
    • B.
      1627\frac{16}{27}
    • C.
      −1627-\frac{16}{27}
    • D.
      −169-\frac{16}{9}
  16. Question 161 mark · Higher-order derivatives
    If f(x)=ln⁡(1+x2)f(x) = \ln(1 + x^2), then f′′(2)=f^{\prime\prime}(2) =
    • A.
      625\frac{6}{25}
    • B.
      −625-\frac{6}{25}
    • C.
      25\frac{2}{5}
    • D.
      −65-\frac{6}{5}
  17. Question 179 marks · Implicit differentiation & tangent lines
    Implicit differentiation. Consider the curve given by the equation x3−2xy+y2=4x^3 - 2xy + y^2 = 4. The point (1,3)(1, 3) lies on the curve.
    x3−2xy+y2=4x^3 - 2xy + y^2 = 4
    (a)
    Show that dydx=3x2−2y2x−2y\dfrac{dy}{dx} = \dfrac{3x^2 - 2y}{2x - 2y}.
    [2]
    (b)
    Write an equation for the line tangent to the curve at the point (1,3)(1, 3).
    [2]
    (c)
    Find the coordinates of every point on the curve at which the line tangent to the curve is vertical. Justify your answer.
    [3]
    (d)
    Evaluate d2ydx2\dfrac{d^2y}{dx^2} at the point (1,3)(1, 3).
    [2]

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AP Calculus AB exam tips

  • Answer with the calculus, not just a number. A vertical or horizontal tangent needs the reason in words, for example "at (a, b), which is on the curve, the denominator of dy/dx is 0 and the numerator is not". A point found from dy/dx alone, without checking that it satisfies the original equation, loses the answer point. Numerical answers do not need to be simplified, so stop once the value is right.
  • For the chain rule from a table, work from the inside out and write each step. For h(x) = f(g(x)) at x = a: read g(a), look up f' at THAT value, then multiply by g'(a). Evaluating f' at a instead of at g(a), and dropping the factor g'(a), are the two mistakes most distractors are built on.
  • For the derivative of an inverse at x = a, find the input first: solve f(b) = a, from the formula or by reading the f(x) row of a table backwards, then take 1 over f'(b). The classic wrong answer is 1 over f'(a). The graph of the inverse passes through (a, b), so a tangent line to the inverse uses that point and the reciprocal slope, with the same sign.
  • For an implicit second derivative, differentiate dy/dx again with the quotient rule and treat every y as a function of x, so each y term brings its own dy/dx. Then substitute the point and the slope you already found there before you simplify anything; simplifying the general expression first takes longer and invites slips.

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What does Unit 3 cover on the real AP Calculus AB exam?

Unit 3, Differentiation: Composite, Implicit, and Inverse Functions, is weighted at about 9–13 percent of the AP Calculus AB exam in the College Board course and exam description. Its six topics are the chain rule, implicit differentiation, differentiating inverse functions, differentiating inverse trigonometric functions, selecting procedures for calculating derivatives, and calculating higher-order derivatives. The chain rule is then used throughout the later units, and differentiating an implicitly defined curve, writing a tangent line, locating horizontal or vertical tangents and evaluating a second derivative are standard AP free-response skills, which is why this set ends with a free-response question on an implicit curve.

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