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

Given g(x)=x4+2g(x)=x^4+2 and g(1)=3g(1)=3, what is (g−1)′(3)(g^{-1})'(3)?

44
14\dfrac{1}{4}
1g′(3)\dfrac{1}{g'(3)}
g′(1)g'(1)
13\dfrac{1}{3}
← Back to quizzes

AP Calculus AB Quiz

AP Calculus AB Quiz: Differentiating Inverse Functions

Practice Differentiating Inverse Functions in AP Calculus AB 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 AB.

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

Given g(x)=x4+2g(x)=x^4+2 and g(1)=3g(1)=3, what is (g−1)′(3)(g^{-1})'(3)?

  1. 44
  2. 14\dfrac{1}{4} (correct answer)
  3. 1g′(3)\dfrac{1}{g'(3)}
  4. g′(1)g'(1)
  5. 13\dfrac{1}{3}
Explanation: This problem tests the skill of differentiating inverse functions. The derivative of the inverse function at a point a is given by 1 over the derivative of the original function at b, where g(b) = a. Here, g(1) = 3, so b = 1 and a = 3. Compute g'(x) = 4x³, so g'(1) = 4, thus (g⁻¹)'(3) = 1/4. A tempting distractor is A, 4, which might be chosen if one forgets to take the reciprocal, but that gives the slope of the tangent to g, not to g⁻¹. To solve similar problems, always remember that the derivatives of a function and its inverse are reciprocals at corresponding points.

Question 2

A one-to-one differentiable function ss has s(1)=9s(1)=9 and s′(1)=−32s'(1)=-\tfrac{3}{2}; compute (s−1)′(9)(s^{-1})'(9).

  1. −23-\dfrac{2}{3} (correct answer)
  2. −32-\dfrac{3}{2}
  3. 23\dfrac{2}{3}
  4. 1s′(9)\dfrac{1}{s'(9)}
  5. 19\dfrac{1}{9}
Explanation: This question requires applying the inverse function derivative formula (s−1)′(b)=1s′(a)(s^{-1})'(b) = \frac{1}{s'(a)} when s(a)=bs(a) = b. Given s(1)=9s(1) = 9 and s′(1)=−32s'(1) = -\frac{3}{2}, we want (s−1)′(9)(s^{-1})'(9). Since s(1)=9s(1) = 9, we have s−1(9)=1s^{-1}(9) = 1, so we evaluate at a=1a = 1. Therefore, (s−1)′(9)=1s′(1)=1−3/2=−23(s^{-1})'(9) = \frac{1}{s'(1)} = \frac{1}{-3/2} = -\frac{2}{3}. Choice B (−32-\frac{3}{2}) incorrectly uses s′(1)s'(1) without taking the reciprocal. To find inverse derivatives: identify the input-output pair, then take the reciprocal of the derivative at that input.

Question 3

For one-to-one differentiable rr, r(10)=0r(10)=0 and r′(10)=2r'(10)=\sqrt{2}; find (r−1)′(0)(r^{-1})'(0).

  1. 2\sqrt{2}
  2. 12\dfrac{1}{\sqrt{2}} (correct answer)
  3. 1r′(0)\dfrac{1}{r'(0)}
  4. 10\dfrac{1}{0}
  5. 1r′(10)\dfrac{1}{r'(10)}
Explanation: To solve this inverse derivative problem, we use (r−1)′(b)=1r′(a)(r^{-1})'(b) = \frac{1}{r'(a)} where r(a)=br(a) = b. We're given r(10)=0r(10) = 0 and r′(10)=sqrt2r'(10) = sqrt{2}, and must find (r−1)′(0)(r^{-1})'(0). Since r(10)=0r(10) = 0, we know r−1(0)=10r^{-1}(0) = 10, so we use a=10a = 10 in the formula. Thus (r−1)′(0)=1r′(10)=1sqrt2(r^{-1})'(0) = \frac{1}{r'(10)} = \frac{1}{sqrt{2}}. Choice A (sqrt2sqrt{2}) is the original derivative value, not its reciprocal. The key insight: inverse derivatives equal the reciprocal of the original derivative at the pre-image point.

Question 4

Let f(x)=x3+1f(x)=x^3+1 and f(2)=9f(2)=9. What is (f−1)′(9)(f^{-1})'(9)?

  1. 1212
  2. 112\dfrac{1}{12} (correct answer)
  3. 1f′(9)\dfrac{1}{f'(9)}
  4. f′(2)f'(2)
  5. 19\dfrac{1}{9}
Explanation: This problem tests the skill of differentiating inverse functions. The derivative of the inverse function at a point a is given by 1 over the derivative of the original function at b, where f(b) = a. Here, f(2) = 9, so b = 2 and a = 9. Compute f′(x)=3x2f'(x) = 3x^2, so f'(2) = 12, thus (f−1)′(9)=112(f^{-1})'(9) = \frac{1}{12}. A tempting distractor is A, 12, which might be chosen if one forgets to take the reciprocal, but that gives the slope of the tangent to f, not to f−1f^{-1}. To solve similar problems, always remember that the derivatives of a function and its inverse are reciprocals at corresponding points.

Question 5

Let f(x)=x3+2x−5f(x)=x^3+2x-5 and f(2)=7f(2)=7. What is (f−1)′(7)(f^{-1})'(7)?

  1. 114\dfrac{1}{14} (correct answer)
  2. 1414
  3. 17\dfrac{1}{7}
  4. 1f′(7)\dfrac{1}{f'(7)}
  5. f′(2)f'(2)
Explanation: This problem tests the skill of differentiating inverse functions. The derivative of the inverse function at a point a is given by 1 over the derivative of the original function at b, where f(b) = a. Here, f(2) = 7, so b = 2 and a = 7. Compute f'(x) = 3x² + 2, so f'(2) = 14, thus (f⁻¹)'(7) = 1/14. A tempting distractor is B, 14, which might be chosen if one forgets to take the reciprocal, but that gives the slope of the tangent to f, not to f⁻¹. To solve similar problems, always remember that the derivatives of a function and its inverse are reciprocals at corresponding points.

Question 6

Let f(x)=x3+2x−5f(x)=x^3+2x-5 and f(2)=7f(2)=7. If g=f−1g=f^{-1}, what is g′(7)g'(7)?

  1. 114\dfrac{1}{14} (correct answer)
  2. 1414
  3. 172+2\dfrac{1}{7^2+2}
  4. 1f′(7)\dfrac{1}{f'(7)}
  5. f′(2)f'(2)
Explanation: This problem requires finding the derivative of an inverse function using the inverse function derivative theorem. Since g = f^(-1) and f(2) = 7, we know that g(7) = 2. The key relationship is that g'(7) = 1/f'(g(7)) = 1/f'(2). To find f'(2), we differentiate f(x) = x³ + 2x - 5 to get f'(x) = 3x² + 2, so f'(2) = 3(4) + 2 = 14. Therefore, g'(7) = 1/14. A common error is computing f'(7) instead of f'(2), which would give 1/149 rather than the correct 1/14. Remember: for inverse function derivatives, always evaluate f' at the point that maps to your input, not at the input itself.

Question 7

Suppose f(x)=sin⁡x+xf(x)=\sin x+x and f(0)=0f(0)=0. If g=f−1g=f^{-1}, what is g′(0)g'(0)?

  1. 1cos⁡(0)+1\dfrac{1}{\cos(0)+1}
  2. cos⁡(0)+1\cos(0)+1
  3. 1cos⁡(g(0))+1\dfrac{1}{\cos(g(0))+1}
  4. 22
  5. 12\dfrac{1}{2} (correct answer)
Explanation: This problem requires finding the derivative of an inverse function at a specific point. Given f(0) = 0 and g = f^(-1), we have g(0) = 0. The inverse derivative formula gives us g'(0) = 1/f'(g(0)) = 1/f'(0). To find f'(x) from f(x) = sin(x) + x, we differentiate to get f'(x) = cos(x) + 1, so f'(0) = cos(0) + 1 = 1 + 1 = 2. Thus g'(0) = 1/2. Students might mistakenly think the answer is 2 by computing cos(0) + 1 directly without taking the reciprocal. The strategy: always remember that inverse function derivatives involve taking the reciprocal of the original function's derivative.

Question 8

Let f(x)=ln⁡(x)+xf(x)=\ln(x)+x with f(1)=1f(1)=1. If g=f−1g=f^{-1}, what is g′(1)g'(1)?

  1. 12\dfrac{1}{2} (correct answer)
  2. 22
  3. 11+1g(1)\dfrac{1}{1+\frac{1}{g(1)}}
  4. 1+111+\dfrac{1}{1}
  5. 11+11\dfrac{1}{1+\frac{1}{1}}
Explanation: To find the derivative of the inverse function g at x = 1, we use the inverse derivative relationship. Given f(1) = 1 and g = f^(-1), we know g(1) = 1. The formula g'(1) = 1/f'(g(1)) = 1/f'(1) applies here. Differentiating f(x) = ln(x) + x yields f'(x) = 1/x + 1, so f'(1) = 1/1 + 1 = 2. Therefore, g'(1) = 1/2. A tempting error is to substitute into the expression 1 + 1/g(1), which would give 2 rather than 1/2. Always remember: the derivative of an inverse function equals the reciprocal of the original function's derivative, evaluated at the appropriate point.

Question 9

Let ff be differentiable and one-to-one with f(2)=5f(2)=5 and f′(2)=3f'(2)=3; what is (f−1)′(5)(f^{-1})'(5)?

  1. 33
  2. 13\dfrac{1}{3} (correct answer)
  3. 1f′(5)\dfrac{1}{f'(5)}
  4. 1f′(2)\dfrac{1}{f'(2)}
  5. 15\dfrac{1}{5}
Explanation: This problem requires finding the derivative of an inverse function at a specific point. The key relationship for inverse function derivatives is (f−1)′(b)=1f′(a)(f^{-1})'(b) = \frac{1}{f'(a)} where f(a)=bf(a) = b. Since we're given f(2)=5f(2) = 5 and f′(2)=3f'(2) = 3, we need (f−1)′(5)(f^{-1})'(5). Because f(2)=5f(2) = 5, we know that f−1(5)=2f^{-1}(5) = 2, so we evaluate at a=2a = 2. Therefore, (f−1)′(5)=1f′(2)=13(f^{-1})'(5) = \frac{1}{f'(2)} = \frac{1}{3}. Choice A (33) is tempting because it's f′(2)f'(2), but the inverse derivative requires the reciprocal. Remember: to find (f−1)′(f^{-1})' at a point, take the reciprocal of f′f' at the corresponding input value.

Question 10

A differentiable one-to-one function pp satisfies p(0)=7p(0)=7 and p′(0)=5p'(0)=5; compute (p−1)′(7)(p^{-1})'(7).

  1. 15\dfrac{1}{5} (correct answer)
  2. 55
  3. 1p′(7)\dfrac{1}{p'(7)}
  4. 17\dfrac{1}{7}
  5. 1p′(0)\dfrac{1}{p'(0)}
Explanation: This problem involves finding the derivative of an inverse function at a specific value. The inverse function derivative formula tells us (p−1)′(b)=1p′(a)(p^{-1})'(b) = \frac{1}{p'(a)} when p(a)=bp(a) = b. We have p(0)=7p(0) = 7 and p′(0)=5p'(0) = 5, and need (p−1)′(7)(p^{-1})'(7). Since p(0)=7p(0) = 7, we get p−1(7)=0p^{-1}(7) = 0, so we use a=0a = 0. Thus (p−1)′(7)=1p′(0)=15(p^{-1})'(7) = \frac{1}{p'(0)} = \frac{1}{5}. Choice B (55) represents p′(0)p'(0) without the reciprocal, a common error. To find an inverse derivative: locate the input that produces your output, then take the reciprocal of the derivative there.

Question 11

If uu is differentiable and one-to-one with u(8)=2u(8)=2 and u′(8)=0.1u'(8)=0.1, what is (u−1)′(2)(u^{-1})'(2)?

  1. 0.10.1
  2. 10.1\dfrac{1}{0.1} (correct answer)
  3. 1u′(2)\dfrac{1}{u'(2)}
  4. 18\dfrac{1}{8}
  5. 1u′(8)\dfrac{1}{u'(8)}
Explanation: To find the derivative of an inverse function, apply (u−1)′(b)=1u′(a)(u^{-1})'(b) = \frac{1}{u'(a)} when u(a)=bu(a) = b. Given u(8)=2u(8) = 2 and u′(8)=0.1u'(8) = 0.1, we seek (u−1)′(2)(u^{-1})'(2). Since u(8)=2u(8) = 2, we have u−1(2)=8u^{-1}(2) = 8, so we evaluate at a=8a = 8. Therefore, (u−1)′(2)=1u′(8)=10.1=10(u^{-1})'(2) = \frac{1}{u'(8)} = \frac{1}{0.1} = 10. Choice A (0.10.1) represents u′(8)u'(8) itself rather than its reciprocal. The consistent pattern: inverse derivatives equal one divided by the original derivative at the pre-image point.

Question 12

A one-to-one differentiable function gg satisfies g(−1)=4g(-1)=4 and g′(−1)=−2g'(-1)=-2; find (g−1)′(4)(g^{-1})'(4).

  1. −2-2
  2. 12\dfrac{1}{2}
  3. −12-\dfrac{1}{2} (correct answer)
  4. 1g′(4)\dfrac{1}{g'(4)}
  5. 14\dfrac{1}{4}
Explanation: This question asks for the derivative of an inverse function using the inverse function derivative formula. The fundamental relationship states that (g−1)′(b)=1g′(a)(g^{-1})'(b) = \frac{1}{g'(a)} when g(a)=bg(a) = b. We're told g(−1)=4g(-1) = 4 and g′(−1)=−2g'(-1) = -2, and we need (g−1)′(4)(g^{-1})'(4). Since g(−1)=4g(-1) = 4, we have g−1(4)=−1g^{-1}(4) = -1, so we use a=−1a = -1 in our formula. Thus (g−1)′(4)=1g′(−1)=1−2=−12(g^{-1})'(4) = \frac{1}{g'(-1)} = \frac{1}{-2} = -\frac{1}{2}. Choice A (−2-2) incorrectly uses g′(−1)g'(-1) directly without taking the reciprocal. The strategy: identify which original input gives your desired output, then take the reciprocal of the derivative at that input.

Question 13

Let q(x)=x+3q(x)=\sqrt{x+3} and q(6)=3q(6)=3. What is (q−1)′(3)(q^{-1})'(3)?

  1. 116\dfrac{1}{\tfrac{1}{6}}
  2. 16\dfrac{1}{6}
  3. 66 (correct answer)
  4. 1q′(3)\dfrac{1}{q'(3)}
  5. q′(6)q'(6)
Explanation: This problem tests the skill of differentiating inverse functions. The derivative of the inverse function at a point aa is given by 1q′(b)\frac{1}{q'(b)}, where q(b)=aq(b) = a. Here, q(6)=3q(6) = 3, so b=6b = 6 and a=3a = 3. Compute q′(x)=12x+3q'(x) = \frac{1}{2\sqrt{x+3}}, so q′(6)=16q'(6) = \frac{1}{6}, thus (q−1)′(3)=6(q^{-1})'(3) = 6. A tempting distractor is B, 16\frac{1}{6}, which might be chosen if one forgets to take the reciprocal, but that gives the slope of the tangent to q, not to q⁻¹. To solve similar problems, always remember that the derivatives of a function and its inverse are reciprocals at corresponding points.

Question 14

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

  1. 44
  2. 14\dfrac{1}{4} (correct answer)
  3. 1f′(0)\dfrac{1}{f'(0)}
  4. f′(1)f'(1)
  5. 10\dfrac{1}{0}
Explanation: This problem tests the skill of differentiating inverse functions. The derivative of the inverse function at a point a is given by 1 over the derivative of the original function at b, where f(b) = a. Here, f(1) = 0, so b = 1 and a = 0. Compute f′(x)=5x4−1f'(x) = 5x^4 - 1, so f′(1)=4f'(1) = 4, thus (f−1)′(0)=14(f^{-1})'(0) = \frac{1}{4}. A tempting distractor is A, 4, which might be chosen if one forgets to take the reciprocal, but that gives the slope of the tangent to f, not to f−1f^{-1}. To solve similar problems, always remember that the derivatives of a function and its inverse are reciprocals at corresponding points.

Question 15

Let h(x)=x2+2xh(x)=x^2+2x (restricted to x≥−1x\ge -1) and h(3)=15h(3)=15. What is (h−1)′(15)(h^{-1})'(15)?

  1. 88
  2. 18\dfrac{1}{8} (correct answer)
  3. 1h′(15)\dfrac{1}{h'(15)}
  4. h′(3)h'(3)
  5. 115\dfrac{1}{15}
Explanation: This problem tests the skill of differentiating inverse functions. The derivative of the inverse function at a point a is given by 1 over the derivative of the original function at b, where h(b) = a. Here, h(3) = 15, so b = 3 and a = 15. Compute h'(x) = 2x + 2, so h'(3) = 8, thus (h⁻¹)'(15) = 1/8. A tempting distractor is A, 8, which might be chosen if one forgets to take the reciprocal, but that gives the slope of the tangent to h, not to h⁻¹. To solve similar problems, always remember that the derivatives of a function and its inverse are reciprocals at corresponding points.

Question 16

Let tt be differentiable and one-to-one with t(−4)=−6t(-4)=-6 and t′(−4)=πt'(-4)=\pi; what is (t−1)′(−6)(t^{-1})'(-6)?

  1. π\pi
  2. 1t′(−6)\dfrac{1}{t'(-6)}
  3. −1π-\dfrac{1}{\pi}
  4. 1π\dfrac{1}{\pi} (correct answer)
  5. 1t′(−4)\dfrac{1}{t'(-4)}
Explanation: Finding this inverse derivative uses the formula (t−1)′(b)=1t′(a)(t^{-1})'(b) = \frac{1}{t'(a)} where t(a)=bt(a) = b. We have t(−4)=−6t(-4) = -6 and t′(−4)=pit'(-4) = pi, and need (t−1)′(−6)(t^{-1})'(-6). Because t(−4)=−6t(-4) = -6, we get t−1(−6)=−4t^{-1}(-6) = -4, so we use a=−4a = -4. Thus (t−1)′(−6)=1t′(−4)=1pi(t^{-1})'(-6) = \frac{1}{t'(-4)} = \frac{1}{pi}. Choice A (pipi) is the original derivative value without reciprocation, a frequent mistake. Remember: inverse function derivatives always involve taking the reciprocal of the original derivative at the corresponding point.

Question 17

Given f(x)=ex+xf(x)=e^x+x and f(0)=1f(0)=1, let g=f−1g=f^{-1}. What is g′(1)g'(1)?

  1. 1e1+1\dfrac{1}{e^1+1}
  2. e0+0e^0+0
  3. 12\dfrac{1}{2} (correct answer)
  4. 22
  5. f′(1)f'(1)
Explanation: This problem involves finding the derivative of an inverse function using the fundamental relationship between f and its inverse g. Since f(0) = 1 and g = f^(-1), we have g(1) = 0. The inverse derivative formula states g'(1) = 1/f'(g(1)) = 1/f'(0). Differentiating f(x) = e^x + x gives f'(x) = e^x + 1, so f'(0) = e^0 + 1 = 1 + 1 = 2. Thus g'(1) = 1/2. Students might incorrectly compute e^0 + 0 = 1, forgetting that the derivative of x is 1, not 0. The key insight: always differentiate the original function completely before evaluating at the required point.

Question 18

Let f(x)=x3−4xf(x)=x^3-4x and f(2)=0f(2)=0. If g=f−1g=f^{-1}, what is g′(0)g'(0)?

  1. 18\dfrac{1}{8} (correct answer)
  2. 88
  3. f′(2)f'(2)
  4. 1f′(0)\dfrac{1}{f'(0)}
  5. 13⋅22−4\dfrac{1}{3\cdot2^2-4}
Explanation: To find the derivative of the inverse function g at x = 0, we use the inverse derivative relationship. Since f(2) = 0 and g = f^(-1), we know g(0) = 2. The formula g'(0) = 1/f'(g(0)) = 1/f'(2) applies. Differentiating f(x) = x³ - 4x gives f'(x) = 3x² - 4, so f'(2) = 3(4) - 4 = 12 - 4 = 8. Therefore, g'(0) = 1/8. A common error is computing f'(0) = -4 and getting -1/4, but we must evaluate f' at the point g(0) = 2. Remember: the input to g' determines where we evaluate f', not the input value itself.

Question 19

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

  1. 1f′(0)\dfrac{1}{f'(0)}
  2. f′(1)f'(1)
  3. 14\dfrac{1}{4} (correct answer)
  4. 44
  5. 105−1\dfrac{1}{0^5-1}
Explanation: This problem asks for the derivative of an inverse function at a specific point. Since g = f^(-1) and f(1) = 0, we have g(0) = 1. The inverse function derivative formula tells us that g'(0) = 1/f'(g(0)) = 1/f'(1). Finding f'(x) from f(x) = x⁵ - x gives f'(x) = 5x⁴ - 1, so f'(1) = 5(1) - 1 = 4. Thus g'(0) = 1/4. Students often mistakenly compute f'(0) = -1 and get -1 as the answer, forgetting to evaluate f' at g(0) = 1, not at 0. The strategy is: find where the inverse function maps your input, then evaluate f' at that point.

Question 20

Given r(x)=ln⁡(x+1)r(x)=\ln(x+1) and r(e−1)=1r(e-1)=1, find (r−1)′(1)(r^{-1})'(1).

  1. 1e\dfrac{1}{e}
  2. ee (correct answer)
  3. 1r′(1)\dfrac{1}{r'(1)}
  4. r′(e−1)r'(e-1)
  5. 1r′(e)\dfrac{1}{r'(e)}
Explanation: This problem tests the skill of differentiating inverse functions. The derivative of the inverse function at a point a is given by 1 over the derivative of the original function at b, where r(b)=ar(b) = a. Here, r(e−1)=1r(e-1) = 1, so b = e-1 and a = 1. Compute r′(x)=1/(x+1)r'(x) = 1/(x+1), so r′(e−1)=1/er'(e-1) = 1/e, thus (r−1)′(1)=e(r^{-1})'(1) = e. A tempting distractor is A, 1/e1/e, which might be chosen if one forgets to take the reciprocal, but that gives the slope of the tangent to r, not to r⁻¹. To solve similar problems, always remember that the derivatives of a function and its inverse are reciprocals at corresponding points.