AP Calculus BC Quiz: Differentiating Inverse Functions
20 questions · exam conditions
0:00
Differentiating Inverse FunctionsQuestion 1 of 20

For f(x)=x5f(x)=x^5 (invertible) with f(2)=32f(2)=32, what is (f−1)′(32)(f^{-1})'(32)?

180\dfrac{1}{80}
1f′(32)\dfrac{1}{f'(32)}
f′(2)f'(2)
8080
1f′(2)\dfrac{1}{f'(2)}
← Back to quizzes

AP Calculus BC Quiz

AP Calculus BC Quiz: Differentiating Inverse Functions

Practice Differentiating Inverse Functions in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Differentiating Inverse Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

For f(x)=x5f(x)=x^5 (invertible) with f(2)=32f(2)=32, what is (f−1)′(32)(f^{-1})'(32)?

  1. 180\dfrac{1}{80}
  2. 1f′(32)\dfrac{1}{f'(32)}
  3. f′(2)f'(2)
  4. 8080
  5. 1f′(2)\dfrac{1}{f'(2)} (correct answer)

Explanation: This problem tests the skill of differentiating inverse functions. To find the derivative of the inverse function at a point b, use the formula (f−1)′(b)=1f′(a)(f^{-1})'(b) = \dfrac{1}{f'(a)}, where a is the value such that f(a)=bf(a) = b. In this case, f(2)=32f(2) = 32, so a = 2 and b = 32. Therefore, (f−1)′(32)=1f′(2)(f^{-1})'(32) = \dfrac{1}{f'(2)}. A tempting distractor is choice B, 1f′(32)\dfrac{1}{f'(32)}, which incorrectly uses the derivative at b instead of at a. A transferable strategy for inverse derivatives is to identify the point a where f(a)=bf(a) = b and then compute the reciprocal of f′(a)f'(a).

Question 2

Let f(x)=x3+1f(x)=x^3+1 be invertible and f(0)=1f(0)=1; what is (f−1)′(1)(f^{-1})'(1)?

  1. 1f′(1)\dfrac{1}{f'(1)}
  2. 13\dfrac{1}{3}
  3. f′(0)f'(0)
  4. 1f′(0)\dfrac{1}{f'(0)} (correct answer)
  5. f′(1)f'(1)

Explanation: This problem tests the skill of differentiating inverse functions. To find the derivative of the inverse function at a point b, use the formula (f−1)′(b)=1/f′(a)(f^{-1})'(b) = 1 / f'(a), where a is the value such that f(a)=bf(a) = b. In this case, f(0)=1f(0) = 1, so a = 0 and b = 1. Therefore, (f−1)′(1)=1/f′(0)(f^{-1})'(1) = 1 / f'(0). A tempting distractor is choice A, 1/f′(1)1/f'(1), which incorrectly uses the derivative at b instead of at a. A transferable strategy for inverse derivatives is to identify the point a where f(a)=bf(a) = b and then compute the reciprocal of f′(a)f'(a).

Question 3

Let f(x)=x+1f(x)=\sqrt{x+1} for x≥−1x\ge-1 and f(3)=2f(3)=2. What is (f−1)′(2)(f^{-1})'(2)?

  1. 14\dfrac{1}{4}
  2. 44 (correct answer)
  3. 12\dfrac{1}{2}
  4. 22
  5. 18\dfrac{1}{8}

Explanation: To find the derivative of an inverse function, we apply (f^{-1})'(b) = rac{1}{f'(a)} where f(a)=bf(a) = b. Given f(3)=2f(3) = 2, we need (f^{-1})'(2) = rac{1}{f'(3)}. For f(x)=sqrtx+1f(x) = sqrt{x+1}, we have f'(x) = rac{1}{2sqrt{x+1}}. Evaluating at x=3x = 3: f'(3) = rac{1}{2sqrt{4}} = rac{1}{4}. Therefore, (f−1)′(2)=4(f^{-1})'(2) = 4. Students might forget the chain rule when differentiating sqrtx+1sqrt{x+1} or evaluate at the wrong point. The key is recognizing that inverse derivatives reciprocate the original function's rate of change.

Question 4

Let f(x)=sin⁡x+xf(x)=\sin x+x on [−π/2,π/2][-\pi/2,\pi/2] with f(0)=0f(0)=0. Find (f−1)′(0)(f^{-1})'(0).

  1. 00
  2. 22
  3. 12\dfrac{1}{2} (correct answer)
  4. 11
  5. −12-\dfrac{1}{2}

Explanation: This problem requires finding the derivative of an inverse function at a point. Using (f^{-1})'(b) = rac{1}{f'(a)} where f(a)=bf(a) = b, and knowing f(0)=0f(0) = 0, we need (f^{-1})'(0) = rac{1}{f'(0)}. Taking the derivative: f′(x)=cosx+1f'(x) = cos x + 1, we get f′(0)=cos(0)+1=1+1=2f'(0) = cos(0) + 1 = 1 + 1 = 2. Thus (f^{-1})'(0) = rac{1}{2}. A common mistake is computing f′f' at the wrong point or forgetting that cos(0)=1cos(0) = 1, not 0. Remember that the inverse function derivative formula creates a reciprocal relationship between the rates of change.

Question 5

Let f(x)=x5f(x)=x^5 with inverse f−1(x)=x1/5f^{-1}(x)=x^{1/5}. Since f(2)=32f(2)=32, what is (f−1)′(32)(f^{-1})'(32)?

  1. 8080
  2. 180\dfrac{1}{80} (correct answer)
  3. 132\dfrac{1}{32}
  4. 15\dfrac{1}{5}
  5. 55

Explanation: This problem tests the skill of differentiating inverse functions. To find the derivative of the inverse at a point, use the formula (f−1f^{-1})'(b) = 1 / f'(a), where f(a) = b. Here, a = 2 and b = 32, so compute f'(x) = 5x^4, and f'(2) = 80. Thus, (f−1f^{-1})'(32) = 1/80. A tempting distractor is 80, which is f'(2) instead of its reciprocal. Always remember to take the reciprocal of the original function's derivative at the input point to find the inverse's derivative at the output point.

Question 6

Let f(x)=x3−xf(x)=x^3-x be one-to-one on [1,∞)[1,\infty) with inverse f−1f^{-1}. If f(2)=6f(2)=6, find (f−1)′(6)(f^{-1})'(6).

  1. 111\dfrac{1}{11} (correct answer)
  2. 1111
  3. 16\dfrac{1}{6}
  4. 13\dfrac{1}{3}
  5. 66

Explanation: This problem tests the skill of differentiating inverse functions. To find the derivative of the inverse at a point, use the formula (f−1f^{-1})'(b) = 1 / f'(a), where f(a) = b. Here, a = 2 and b = 6, so compute f'(x) = 3x^2 - 1, and f'(2) = 11. Thus, (f−1f^{-1})'(6) = 1/11. A tempting distractor is 11, which is f'(2) instead of its reciprocal. Always remember to take the reciprocal of the original function's derivative at the input point to find the inverse's derivative at the output point.

Question 7

Let f(x)=x+4f(x)=\sqrt{x+4} with inverse f−1f^{-1}. If f(5)=3f(5)=3, what is (f−1)′(3)(f^{-1})'(3)?

  1. 16\dfrac{1}{6}
  2. 66 (correct answer)
  3. 13\dfrac{1}{3}
  4. 33
  5. 12\dfrac{1}{2}

Explanation: This problem tests the skill of differentiating inverse functions. To find the derivative of the inverse at a point, use the formula (f−1f^{-1})'(b) = 1 / f'(a), where f(a) = b. Here, a = 5 and b = 3, so compute f'(x) = 1/(2√(x+4)), and f'(5) = 1/6. Thus, (f−1f^{-1})'(3) = 6. A tempting distractor is 1/6, which is f'(5) instead of its reciprocal. Always remember to take the reciprocal of the original function's derivative at the input point to find the inverse's derivative at the output point.

Question 8

If f(x)=x5+xf(x)=x^5+x and f(0)=0f(0)=0, what is (f−1)′(0)(f^{-1})'(0)?

  1. 00
  2. 55
  3. 15\dfrac{1}{5}
  4. 11 (correct answer)
  5. 16\dfrac{1}{6}

Explanation: This problem requires finding the derivative of an inverse function at a specific point. Using (f^{-1})'(b) = rac{1}{f'(a)} where f(a)=bf(a) = b, and given f(0)=0f(0) = 0, we need (f^{-1})'(0) = rac{1}{f'(0)}. Computing f′(x)=5x4+1f'(x) = 5x^4 + 1, we get f′(0)=5(0)4+1=1f'(0) = 5(0)^4 + 1 = 1. Thus (f^{-1})'(0) = rac{1}{1} = 1. A tempting error is to think that since f(0)=0f(0) = 0, the derivative must also be 0, but this confuses function values with derivative values. Remember that the inverse derivative formula always involves taking a reciprocal of the original function's derivative.

Question 9

Let f(x)=x3−2x+5f(x)=x^3-2x+5 be invertible and f(1)=4f(1)=4; what is (f−1)′(4)(f^{-1})'(4)?

  1. 13\dfrac{1}{3}
  2. 1f′(4)\dfrac{1}{f'(4)}
  3. f′(1)f'(1)
  4. 1f′(1)\dfrac{1}{f'(1)} (correct answer)
  5. f′(4)f'(4)

Explanation: This problem tests the skill of differentiating inverse functions. To find the derivative of the inverse function at a point b, use the formula (f−1f^{-1})'(b) = 1 / f'(a), where a is the value such that f(a) = b. In this case, f(1) = 4, so a = 1 and b = 4. Therefore, (f−1f^{-1})'(4) = 1 / f'(1). A tempting distractor is choice B, 1/f'(4), which incorrectly uses the derivative at b instead of at a. A transferable strategy for inverse derivatives is to identify the point a where f(a) = b and then compute the reciprocal of f'(a).

Question 10

Let f(x)=ex+2f(x)=e^x+2 be invertible and f(0)=3f(0)=3; what is (f−1)′(3)(f^{-1})'(3)?

  1. e3e^3
  2. 1e3\dfrac{1}{e^3}
  3. 1f′(0)\dfrac{1}{f'(0)} (correct answer)
  4. f′(0)f'(0)
  5. 1f′(3)\dfrac{1}{f'(3)}

Explanation: This problem tests the skill of differentiating inverse functions. To find the derivative of the inverse function at a point bb, use the formula (f−1)′(b)=1f′(a)(f^{-1})'(b) = \frac{1}{f'(a)}, where aa is the value such that f(a)=bf(a) = b. In this case, f(0)=3f(0) = 3, so a=0a = 0 and b=3b = 3. Therefore, (f−1)′(3)=1f′(0)(f^{-1})'(3) = \frac{1}{f'(0)}. A tempting distractor is choice E, 1f′(3)\frac{1}{f'(3)}, which incorrectly uses the derivative at bb instead of at aa. A transferable strategy for inverse derivatives is to identify the point aa where f(a)=bf(a) = b and then compute the reciprocal of f′(a)f'(a).

Question 11

Given f(x)=x−1x+1f(x)=\dfrac{x-1}{x+1} (invertible on x>−1x>-1) and f(1)=0f(1)=0, find (f−1)′(0)(f^{-1})'(0).

  1. 1f′(0)\dfrac{1}{f'(0)}
  2. 1f′(1)\dfrac{1}{f'(1)} (correct answer)
  3. f′(1)f'(1)
  4. 12\dfrac{1}{2}
  5. f′(0)f'(0)

Explanation: This problem tests the skill of differentiating inverse functions. To find the derivative of the inverse function at a point b, use the formula (f−1f^{-1})'(b) = 1 / f'(a), where a is the value such that f(a) = b. Here, f(1) = 0, so a = 1 and b = 0. Thus, (f−1f^{-1})'(0) = 1 / f'(1). A tempting distractor is choice A, 1/f'(0), which mistakenly evaluates the derivative at b rather than at a. A transferable strategy for inverse derivatives is to identify the point a where f(a) = b and then compute the reciprocal of f'(a).

Question 12

If f(x)=ln⁡(x−1)f(x)=\ln(x-1) for x>1x>1 and f(e+1)=1f(e+1)=1, what is (f−1)′(1)(f^{-1})'(1)?

  1. 1f′(1)\dfrac{1}{f'(1)}
  2. e+1e+1
  3. 1e\dfrac{1}{e}
  4. 1f′(e+1)\dfrac{1}{f'(e+1)} (correct answer)
  5. f′(e+1)f'(e+1)

Explanation: This problem tests the skill of differentiating inverse functions. To find the derivative of the inverse function at a point b, use the formula (f−1)′(b)=1f′(a)(f^{-1})'(b) = \frac{1}{f'(a)}, where a is the value such that f(a)=bf(a) = b. Here, f(e+1)=1f(e+1) = 1, so a = e+1 and b = 1. Thus, (f−1)′(1)=1f′(e+1)(f^{-1})'(1) = \frac{1}{f'(e+1)}. A tempting distractor is choice A, 1/f′(1)1/f'(1), which mistakenly evaluates the derivative at b rather than at a. A transferable strategy for inverse derivatives is to identify the point a where f(a)=bf(a) = b and then compute the reciprocal of f′(a)f'(a).

Question 13

If f(x)=x+1f(x)=\sqrt{x+1} for x≥−1x\ge -1 and f(3)=2f(3)=2, what is (f−1)′(2)(f^{-1})'(2)?

  1. 22
  2. 1f′(2)\dfrac{1}{f'(2)}
  3. 14\dfrac{1}{4}
  4. 1f′(3)\dfrac{1}{f'(3)} (correct answer)
  5. f′(3)f'(3)

Explanation: This problem tests the skill of differentiating inverse functions. To find the derivative of the inverse function at a point b, use the formula (f−1f^{-1})'(b) = 1 / f'(a), where a is the value such that f(a) = b. In this case, f(3) = 2, so a = 3 and b = 2. Therefore, (f−1f^{-1})'(2) = 1 / f'(3). A tempting distractor is choice B, 1/f'(2), which incorrectly uses the derivative at b instead of at a. A transferable strategy for inverse derivatives is to identify the point a where f(a) = b and then compute the reciprocal of f'(a).

Question 14

Let f(x)=ex+xf(x)=e^x+x with f(0)=1f(0)=1. What is (f−1)′(1)(f^{-1})'(1)?

  1. 22
  2. 12\dfrac{1}{2} (correct answer)
  3. 11
  4. 1e\dfrac{1}{e}
  5. ee

Explanation: To find the derivative of an inverse function, we use the relationship (f^{-1})'(b) = rac{1}{f'(a)} where f(a)=bf(a) = b. Given f(0)=1f(0) = 1, we need (f^{-1})'(1) = rac{1}{f'(0)}. Taking the derivative, f′(x)=ex+1f'(x) = e^x + 1, so f′(0)=e0+1=2f'(0) = e^0 + 1 = 2. Therefore, (f^{-1})'(1) = rac{1}{2}. Students might mistakenly compute f′(1)=e+1f'(1) = e + 1 by plugging in the wrong value, which would give approximately rac{1}{3.718}. Remember to use the input value that produces your target output, not the output value itself.

Question 15

Given f(x)=tan⁡xf(x)=\tan x on (−π/2,π/2)(-\pi/2,\pi/2) and f(0)=0f(0)=0, what is (f−1)′(0)(f^{-1})'(0)?

  1. 00
  2. π2\dfrac{\pi}{2}
  3. 11 (correct answer)
  4. 12\dfrac{1}{2}
  5. 22

Explanation: To find the derivative of an inverse function, we use (f^{-1})'(b) = rac{1}{f'(a)} where f(a)=bf(a) = b. Since f(0)=0f(0) = 0 for the tangent function, we need (f^{-1})'(0) = rac{1}{f'(0)}. The derivative of tangent is f′(x)=sec2xf'(x) = sec^2 x, so f′(0)=sec2(0)=1f'(0) = sec^2(0) = 1. Therefore, (f^{-1})'(0) = rac{1}{1} = 1. Students might confuse this with arctan'(0) = rac{1}{1+0^2} = 1, which happens to give the same answer but uses a different approach. The key insight is that at points where f′(a)=1f'(a) = 1, the function and its inverse have equal rates of change.

Question 16

Let f(x)=x2+1f(x)=x^2+1 restricted to [0,∞)[0,\infty) with inverse f−1f^{-1}. Since f(3)=10f(3)=10, what is (f−1)′(10)(f^{-1})'(10)?

  1. 16\dfrac{1}{6} (correct answer)
  2. 66
  3. 110\dfrac{1}{10}
  4. 13\dfrac{1}{3}
  5. 33

Explanation: This problem tests the skill of differentiating inverse functions. To find the derivative of the inverse at a point, use the formula (f−1)′(b)=1f′(a)(f^{-1})'(b) = \frac{1}{f'(a)}, where f(a)=bf(a) = b. Here, a = 3 and b = 10, so compute f′(x)=2xf'(x) = 2x, and f′(3)=6f'(3) = 6. Thus, (f−1)′(10)=16(f^{-1})'(10) = \frac{1}{6}. A tempting distractor is 6, which is f′(3)f'(3) instead of its reciprocal. Always remember to take the reciprocal of the original function's derivative at the input point to find the inverse's derivative at the output point.

Question 17

Let f(x)=sin⁡xf(x)=\sin x restricted to [−π2,π2][-\tfrac{\pi}{2},\tfrac{\pi}{2}] with inverse f−1(x)=arcsin⁡xf^{-1}(x)=\arcsin x. What is (f−1)′(0)(f^{-1})'(0)?

  1. 00
  2. −1-1
  3. 11 (correct answer)
  4. π2\dfrac{\pi}{2}
  5. 2π\dfrac{2}{\pi}

Explanation: This problem tests the skill of differentiating inverse functions. To find the derivative of the inverse at a point, use the formula (f−1f^{-1})'(b) = 1 / f'(a), where f(a) = b. Here, a = 0 and b = 0, so compute f'(x) = cos x, and f'(0) = 1. Thus, (f−1f^{-1})'(0) = 1. A tempting distractor is 0, which might come from confusing the value of f(0) with the derivative. Always remember to take the reciprocal of the original function's derivative at the input point to find the inverse's derivative at the output point.

Question 18

Let f(x)=x3+2x+1f(x)=x^3+2x+1 with inverse f−1f^{-1}. If f(1)=4f(1)=4, what is (f−1)′(4)(f^{-1})'(4)?

  1. 55
  2. 15\dfrac{1}{5} (correct answer)
  3. 14\dfrac{1}{4}
  4. 44
  5. 13\dfrac{1}{3}

Explanation: This problem tests the skill of differentiating inverse functions. To find the derivative of the inverse at a point, use the formula (f−1f^{-1})'(b) = 1 / f'(a), where f(a) = b. Here, a = 1 and b = 4, so compute f'(x) = 3x^2 + 2, and f'(1) = 5. Thus, (f−1f^{-1})'(4) = 1/5. A tempting distractor is 5, which is f'(1) instead of its reciprocal. Always remember to take the reciprocal of the original function's derivative at the input point to find the inverse's derivative at the output point.

Question 19

Let f(x)=1xf(x)=\dfrac{1}{x} for x>0x>0 with inverse f−1(x)=1xf^{-1}(x)=\dfrac{1}{x}. Since f(2)=12f(2)=\tfrac{1}{2}, find (f−1)′(12)(f^{-1})'(\tfrac{1}{2}).

  1. −14-\dfrac{1}{4}
  2. −4-4 (correct answer)
  3. 44
  4. 14\dfrac{1}{4}
  5. 12\dfrac{1}{2}

Explanation: This problem tests the skill of differentiating inverse functions. To find the derivative of the inverse at a point, use the formula (f−1)′(b)=1f′(a)(f^{-1})'(b) = \frac{1}{f'(a)}, where f(a)=bf(a) = b. Here, a=2a = 2 and b=12b = \frac{1}{2}, so compute f′(x)=−1x2f'(x) = -\frac{1}{x^2}, and f′(2)=−14f'(2) = -\frac{1}{4}. Thus, (f−1)′(12)=−4(f^{-1})'(\frac{1}{2}) = -4. A tempting distractor is −14-\frac{1}{4}, which is f′(2)f'(2) instead of its reciprocal. Always remember to take the reciprocal of the original function's derivative at the input point to find the inverse's derivative at the output point.

Question 20

Let f(x)=ln⁡xf(x)=\ln x with inverse f−1(x)=exf^{-1}(x)=e^x. Since f(2)=ln⁡2f(2)=\ln 2, what is (f−1)′(ln⁡2)(f^{-1})'(\ln 2)?

  1. 12\dfrac{1}{2}
  2. ln⁡2\ln 2
  3. 22 (correct answer)
  4. 1ln⁡2\dfrac{1}{\ln 2}
  5. ln⁡22\dfrac{\ln 2}{2}

Explanation: This problem tests the skill of differentiating inverse functions. To find the derivative of the inverse at a point, use the formula (f−1f^{-1})'(b) = 1 / f'(a), where f(a) = b. Here, a = 2 and b = ln 2, so compute f'(x) = 1/x, and f'(2) = 1/2. Thus, (f−1f^{-1})'(ln 2) = 2. A tempting distractor is 1/2, which is f'(2) instead of its reciprocal. Always remember to take the reciprocal of the original function's derivative at the input point to find the inverse's derivative at the output point.