Differentiating Inverse Functions - AP Calculus AB

Let $h(x)=x^2+2x$ (restricted to $x\ge -1$ ) and $h(3)=15$ . What is $(h^{-1})'(15)$ ?
Answer: $\dfrac{1}{8}$ 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.
For $f(x)=x^5-x$ , $f(1)=0$ . If $g=f^{-1}$ , what is $g'(0)$ ?
Answer: $\dfrac{1}{4}$ 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.
Let $f(x)=x^3+2x-5$ and $f(2)=7$ . What is $(f^{-1})'(7)$ ?
Answer: $\dfrac{1}{14}$ 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.
Let $g$ be differentiable and one-to-one with $g(-1)=4$ and $g'(-1)=-2$ . Find $(g^{-1})'(4)$ .
Answer: $\dfrac{1}{-2}$ This problem asks for the derivative of an inverse function using the relationship between a function and its inverse. The key formula is $(g^{-1})'(b) = \frac{1}{g'(a)}$ where $g(a) = b$ . We're told $g(-1) = 4$ , which means $g^{-1}(4) = -1$ , and we need $(g^{-1})'(4)$ . Substituting into the formula with $a = -1$ and $b = 4$ , we get $(g^{-1})'(4) = \frac{1}{g'(-1)} = \frac{1}{-2} = -\frac{1}{2}$ . Choice A ( $-2$ ) incorrectly uses $g'(-1)$ directly without taking the reciprocal. The strategy: identify the input-output pair, then take the reciprocal of the derivative at the input.
Let $q(x)=x^3+x^2$ and $q(1)=2$ . What is $(q^{-1})'(2)$ ?
Answer: $\dfrac{1}{5}$ 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 q(b) = a. Here, q(1) = 2, so b = 1 and a = 2. Compute q'(x) = 3x² + 2x, so q'(1) = 5, thus (q⁻¹)'(2) = 1/5. A tempting distractor is A, 5, 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.
Given $f(x)=e^x+x$ and $f(0)=1$ , let $g=f^{-1}$ . What is $g'(1)$ ?
Answer: $\dfrac{1}{2}$ 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.
Let $f$ be differentiable and one-to-one with $f(2)=5$ and $f'(2)=3$ ; what is $(f^{-1})'(5)$ ?
Answer: $\dfrac{1}{3}$ This problem requires finding the derivative of an inverse function at a specific point. The key relationship for inverse function derivatives is $(f^{-1})'(b) = \frac{1}{f'(a)}$ where $ f(a) = b $. Since we're given $ f(2) = 5 $ and $ f'(2) = 3 $, we need $ ( $f^{-1}$ )'(5) $. Because $ f(2) = 5 $, we know that $ $f^{-1}$ (5) = 2 $, so we evaluate at $ a = 2 $. Therefore, $(f^{-1})'(5) = \frac{1}{f'(2)} = \frac{1}{3}$ . Choice A ($ 3 $) is tempting because it's $ f'(2) $, but the inverse derivative requires the reciprocal. Remember: to find $ ( $f^{-1}$ )' $ at a point, take the reciprocal of $ f' $ at the corresponding input value.
A one-to-one differentiable function $s$ satisfies $s(-4)=9$ and $s'(-4)=\tfrac{2}{7}$ . Find $(s^{-1})'(9)$ .
Answer: $\dfrac{7}{2}$ This problem tests the inverse function derivative formula. Given $(s^{-1})'(b) = \frac{1}{s'(a)}$ where $s(a) = b$ , and knowing $s(-4) = 9$ , we have $s^{-1}(9) = -4$ . To find $(s^{-1})'(9)$ , we substitute $a = -4$ and $b = 9$ : $(s^{-1})'(9) = \frac{1}{s'(-4)} = \frac{1}{2/7} = \frac{7}{2}$ . Choice A ( $\frac{2}{7}$ ) incorrectly uses $s'(-4)$ without taking the reciprocal. The key principle: inverse function derivatives are reciprocals of the original function's derivatives at corresponding points.
If $p$ is one-to-one with $p(0)=7$ and $p'(0)=\sqrt{5}$ , what is $(p^{-1})'(7)$ ?
Answer: $\dfrac{1}{\sqrt{5}}$ This question tests understanding of inverse function differentiation. The formula $(p^{-1})'(b) = \frac{1}{p'(a)}$ applies when $p(a) = b$ . Given $p(0) = 7$ , we know $p^{-1}(7) = 0$ , and we need $(p^{-1})'(7)$ . Using the formula with $a = 0$ and $b = 7$ : $(p^{-1})'(7) = \frac{1}{p'(0)} = \frac{1}{\sqrt{5}}$ . Choice A ( $\sqrt{5}$ ) represents $p'(0)$ without the reciprocal, which is a common error. The key insight: the derivative of an inverse function equals the reciprocal of the original function's derivative at the corresponding input.
Suppose $f(x)=\sin x+x$ and $f(0)=0$ . If $g=f^{-1}$ , what is $g'(0)$ ?
Answer: $\dfrac{1}{2}$ 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.
If $u$ is differentiable and one-to-one with $u(8)=2$ and $u'(8)=0.1$ , what is $(u^{-1})'(2)$ ?
Answer: $\dfrac{1}{0.1}$ To find the derivative of an inverse function, apply $(u^{-1})'(b) = \frac{1}{u'(a)}$ when $u(a) = b$ . Given $u(8) = 2$ and $u'(8) = 0.1$ , we seek $(u^{-1})'(2)$ . Since $u(8) = 2$ , we have $u^{-1}(2) = 8$ , so we evaluate at $a = 8$ . Therefore, $(u^{-1})'(2) = \frac{1}{u'(8)} = \frac{1}{0.1} = 10$ . Choice A ( $0.1$ ) represents $u'(8)$ itself rather than its reciprocal. The consistent pattern: inverse derivatives equal one divided by the original derivative at the pre-image point.
A one-to-one differentiable function $g$ satisfies $g(-1)=4$ and $g'(-1)=-2$ ; find $(g^{-1})'(4)$ .
Answer: $-\dfrac{1}{2}$ This question asks for the derivative of an inverse function using the inverse function derivative formula. The fundamental relationship states that $(g^{-1})'(b) = \frac{1}{g'(a)}$ when $ g(a) = b $. We're told $ g(-1) = 4 $ and $ g'(-1) = -2 $, and we need $ ( $g^{-1}$ )'(4) $. Since $ g(-1) = 4 $, we have $ $g^{-1}$ (4) = -1 $, so we use $ a = -1 $ in our formula. Thus $(g^{-1})'(4) = \frac{1}{g'(-1)} = \frac{1}{-2} = -\frac{1}{2}$ . Choice A ($ -2 $) incorrectly uses $ 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.
If $t(x)=x^3+3x$ and $t(1)=4$ , what is $(t^{-1})'(4)$ ?
Answer: $\dfrac{1}{6}$ This problem tests the skill of differentiating inverse functions. The derivative of the inverse function at a point $a$ is given by $\frac{1}{\text{derivative of the original function at } b}$ , where $ t(b) = a $. Here, $ t(1) = 4 $, so $ b = 1 $ and $ a = 4 $. Compute $ t'(x) = $3x^2$ + 3 $, so $ t'(1) = 6 $, thus $(t^{-1})'(4) = \frac{1}{6}$ . A tempting distractor is A, $ 6 $, which might be chosen if one forgets to take the reciprocal, but that gives the slope of the tangent to $ t $, not to $ $t^{-1}$ $. To solve similar problems, always remember that the derivatives of a function and its inverse are reciprocals at corresponding points.
If $g(x)=e^x+x$ and $g(0)=1$ , what is $(g^{-1})'(1)$ ?
Answer: $\dfrac{1}{2}$ 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(0) = 1, so b = 0 and a = 1. Compute g'(x) = eˣ + 1, so g'(0) = 2, thus (g⁻¹)'(1) = 1/2. A tempting distractor is A, 2, 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.
Suppose $q$ is differentiable and one-to-one with $q(6)=1$ and $q'(6)=-\tfrac{3}{5}$ . Find $(q^{-1})'(1)$ .
Answer: $-\dfrac{5}{3}$ This problem requires applying the inverse function derivative relationship. The formula states $(q^{-1})'(b) = \frac{1}{q'(a)}$ where $q(a) = b$ . Since $q(6) = 1$ , we have $q^{-1}(1) = 6$ , and we need $(q^{-1})'(1)$ . Applying the formula with $a = 6$ and $b = 1$ : $(q^{-1})'(1) = \frac{1}{q'(6)} = \frac{1}{-3/5} = -\frac{5}{3}$ . Choice A ( $-\frac{3}{5}$ ) incorrectly uses $q'(6)$ directly without reciprocating. Remember: to find the derivative of an inverse at a point, reciprocate the original derivative at the pre-image point.
A one-to-one differentiable function $s$ has $s(1)=9$ and $s'(1)=-\tfrac{3}{2}$ ; compute $(s^{-1})'(9)$ .
Answer: $-\dfrac{2}{3}$ This question requires applying the inverse function derivative formula $(s^{-1})'(b) = \frac{1}{s'(a)}$ when $s(a) = b$ . Given $s(1) = 9$ and $s'(1) = -\frac{3}{2}$ , we want $(s^{-1})'(9)$ . Since $s(1) = 9$ , we have $s^{-1}(9) = 1$ , so we evaluate at $a = 1$ . Therefore, $(s^{-1})'(9) = \frac{1}{s'(1)} = \frac{1}{-3/2} = -\frac{2}{3}$ . Choice B ( $-\frac{3}{2}$ ) incorrectly uses $s'(1)$ without taking the reciprocal. To find inverse derivatives: identify the input-output pair, then take the reciprocal of the derivative at that input.
If $f$ is differentiable with $f(2)=5$ and $f'(2)=3$ , what is $(f^{-1})'(5)$ ?
Answer: $\dfrac{1}{3}$ This problem requires finding the derivative of an inverse function at a specific point. The inverse function derivative formula states that $(f^{-1})'(b) = \frac{1}{f'(a)}$ where $f(a) = b$ . Since we're given $f(2) = 5$ , we know that $f^{-1}(5) = 2$ , so we need $(f^{-1})'(5)$ . Applying the formula with $a = 2$ and $b = 5$ , we get $(f^{-1})'(5) = \frac{1}{f'(2)} = \frac{1}{3}$ . Choice B ( $3$ ) is tempting because it's $f'(2)$ , but the inverse derivative is the reciprocal of the original derivative. Remember: to find $(f^{-1})'$ at a point, take the reciprocal of $f'$ at the corresponding input value.
A differentiable one-to-one function $v$ satisfies $v(5)=11$ and $v'(5)=12$ ; find $(v^{-1})'(11)$ .
Answer: $\dfrac{1}{12}$ This problem uses the inverse function derivative relationship $(v^{-1})'(b) = \frac{1}{v'(a)}$ where $v(a) = b$ . We're told $v(5) = 11$ and $v'(5) = 12$ , and must find $(v^{-1})'(11)$ . Because $v(5) = 11$ , we know $v^{-1}(11) = 5$ , so we use $a = 5$ in our formula. Thus $(v^{-1})'(11) = \frac{1}{v'(5)} = \frac{1}{12}$ . Choice A ( $12$ ) incorrectly gives $v'(5)$ without reciprocation. To master inverse derivatives: find where the original function produces your value, then take the reciprocal of its derivative there.
Let $f(x)=x^3+1$ and $f(2)=9$ . If $g=f^{-1}$ , what is $g'(9)$ ?
Answer: $\dfrac{1}{12}$ To find g'(9) where g is the inverse of f, we apply the inverse function derivative formula. Since f(2) = 9 and g = $f^{-1}$ , we have g(9) = 2. The relationship g'(9) = 1/f'(g(9)) = 1/f'(2) is key. Differentiating f(x) = x³ + 1 yields f'(x) = 3x², so f'(2) = 3(4) = 12. Therefore, g'(9) = 1/12. Students might mistakenly compute f'(9) = 3(81) = 243 and get 1/243, but we must evaluate f' at g(9) = 2, not at 9. The strategy: always track which point in the domain of f corresponds to your input in the domain of g.
Suppose $v$ is differentiable and one-to-one with $v(8)=-3$ and $v'(8)=\tfrac{5}{9}$ . Find $(v^{-1})'(-3)$ .
Answer: $\dfrac{9}{5}$ This question tests understanding of inverse function differentiation. The key formula is $(v^{-1})'(b) = \frac{1}{v'(a)}$ where $v(a) = b$ . Since $v(8) = -3$ , we know $v^{-1}(-3) = 8$ , and we need $(v^{-1})'(-3)$ . Applying the formula with $a = 8$ and $b = -3$ : $(v^{-1})'(-3) = \frac{1}{v'(8)} = \frac{1}{5/9} = \frac{9}{5}$ . Choice B ( $\frac{5}{9}$ ) is $v'(8)$ without reciprocation, a frequent error. Remember: inverse derivatives require reciprocating the original function's derivative at the pre-image point.

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