Differentiating Inverse Functions - AP Calculus BC

If $f(x)=x^3+2x-1$ and $f(1)=2$ , what is $(f^{-1})'(2)$ ?
Answer: $\dfrac{1}{5}$ This problem requires finding the derivative of an inverse function at a specific point. The key formula is $(f^{-1})'(b) = \frac{1}{f'(a)}$ where $f(a) = b$ . Since we're given $f(1) = 2$ , we need $(f^{-1})'(2) = \frac{1}{f'(1)}$ . Computing $f'(x) = 3x^2 + 2$ , we get $f'(1) = 3(1)^2 + 2 = 5$ . A common error would be to evaluate $f'(2)$ instead of $f'(1)$ , which would give $f'(2) = 14$ and lead to the incorrect answer $\frac{1}{14}$ . The strategy is to always identify which input value maps to your target output before applying the reciprocal derivative formula.
Let $f(x)=x^3-x$ be one-to-one on $[1,\infty)$ with inverse $f^{-1}$ . If $f(2)=6$ , find $(f^{-1})'(6)$ .
Answer: $\dfrac{1}{11}$ 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) = 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, ( $f^{-1}$ )'(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.
Let $f(x)=e^x+x$ with $f(0)=1$ . What is $(f^{-1})'(1)$ ?
Answer: $\dfrac{1}{2}$ To find the derivative of an inverse function, we use the relationship $(f^{-1})'(b) = \frac{1}{f'(a)}$ where $f(a) = b$ . Given $f(0) = 1$ , we need $(f^{-1})'(1) = \frac{1}{f'(0)}$ . Taking the derivative, $f'(x) = e^x + 1$ , so $f'(0) = e^0 + 1 = 2$ . Therefore, $(f^{-1})'(1) = \frac{1}{2}$ . Students might mistakenly compute $f'(1) = e + 1$ by plugging in the wrong value, which would give approximately $\frac{1}{3.718}$ . Remember to use the input value that produces your target output, not the output value itself.
Let $f(x)=x^2+1$ restricted to $[0,\infty)$ with inverse $f^{-1}$ . Since $f(3)=10$ , what is $(f^{-1})'(10)$ ?
Answer: $\dfrac{1}{6}$ This problem tests the skill of differentiating inverse functions. To find the derivative of the inverse at a point, use the formula $(f^{-1})'(b) = \frac{1}{f'(a)}$ , where $f(a) = b$ . Here, a = 3 and b = 10, so compute $f'(x) = 2x$ , and $f'(3) = 6$ . Thus, $(f^{-1})'(10) = \frac{1}{6}$ . A tempting distractor is 6, which is $f'(3)$ 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.
For $f(x)=x^2+x-6$ on $[2,\infty)$ with $f(2)=0$ , what is $(f^{-1})'(0)$ ?
Answer: $\dfrac{1}{f'(2)}$ This problem tests the skill of differentiating inverse functions. To find the derivative of the inverse function at a point b, use the formula $(f^{-1})'(b) = \frac{1}{f'(a)}$ , where a is the value such that $f(a) = b$ . Here, $f(2) = 0$ , so a = 2 and b = 0. Thus, $(f^{-1})'(0) = \frac{1}{f'(2)}$ . A tempting distractor is choice A, $\frac{1}{f'(0)}$ , 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 $f'(a)$ .
Let $f(x)=x^3+1$ be invertible and $f(0)=1$ ; what is $(f^{-1})'(1)$ ?
Answer: $\dfrac{1}{f'(0)}$ This problem tests the skill of differentiating inverse functions. To find the derivative of the inverse function at a point b, use the formula $(f^{-1})'(b) = 1 / f'(a)$ , where a is the value such that $f(a) = b$ . In this case, $f(0) = 1$ , so a = 0 and b = 1. Therefore, $(f^{-1})'(1) = 1 / f'(0)$ . A tempting distractor is choice A, $1/f'(1)$ , which incorrectly uses the derivative at b instead of at a. A transferable strategy for inverse derivatives is to identify the point a where $f(a) = b$ and then compute the reciprocal of $f'(a)$ .
Let $f(x)=\sqrt{x+4}$ with inverse $f^{-1}$ . If $f(5)=3$ , what is $(f^{-1})'(3)$ ?
Answer: $6$ 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) = 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, ( $f^{-1}$ )'(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.
Given $f(x)=x^2+3$ for $x\ge0$ and $f(2)=7$ , what is $(f^{-1})'(7)$ ?
Answer: $\dfrac{1}{4}$ This problem involves finding the derivative of an inverse function using the formula $(f^{-1})'(b) = \frac{1}{f'(a)}$ where $f(a) = b$ . Since $f(2) = 7$ , we need $(f^{-1})'(7) = \frac{1}{f'(2)}$ . For $f(x) = x^2 + 3$ with $x geq 0$ , we have $f'(x) = 2x$ , giving $f'(2) = 4$ . Thus $(f^{-1})'(7) = \frac{1}{4}$ . A tempting error is to use $f'(7)$ instead, which would give $f'(7) = 14$ and the incorrect answer $\frac{1}{14}$ . Always identify the correct input-output pair before applying the inverse derivative formula.
Let $f(x)=x^5$ and $g=f^{-1}$ . What is $g'(32)$ ?
Answer: $\dfrac{1}{80}$ Finding the derivative of an inverse function requires careful application of the formula g'(y) = 1/f'(g(y)). Since f(x) = x⁵ and we need g'(32), we first find g(32): solving x⁵ = 32 gives x = 2, so g(32) = 2. The derivative f'(x) = 5x⁴, so f'(2) = 5(2)⁴ = 5(16) = 80. Therefore, g'(32) = 1/f'(g(32)) = 1/f'(2) = 1/80. A common mistake is computing f'(32) = 5(32)⁴, which is enormous, instead of f'(g(32)) = f'(2). The key insight is that inverse derivatives involve reciprocals evaluated at corresponding points, not at the given y-value.
If $f(x)=\ln(x)+x$ and $f(1)=1$ , what is $(f^{-1})'(1)$ ?
Answer: $\dfrac{1}{2}$ To find the derivative of an inverse function, we apply $(f^{-1})'(b) = \frac{1}{f'(a)}$ where $f(a) = b$ . Given $f(1) = 1$ , we need $(f^{-1})'(1) = \frac{1}{f'(1)}$ . Computing the derivative: $f'(x) = \frac{1}{x} + 1$ , so $f'(1) = 1 + 1 = 2$ . Therefore, $(f^{-1})'(1) = \frac{1}{2}$ . Students often confuse which value to substitute, potentially computing $f'(0)$ which is undefined due to $ln(0)$ . The key insight is that inverse derivatives flip the rate of change: if $f$ changes twice as fast at a point, then $f^{-1}$ changes half as fast at the corresponding point.
Let $f(x)=e^x+2$ be invertible and $f(0)=3$ ; what is $(f^{-1})'(3)$ ?
Answer: $\dfrac{1}{f'(0)}$ This problem tests the skill of differentiating inverse functions. To find the derivative of the inverse function at a point $b$ , use the formula $(f^{-1})'(b) = \frac{1}{f'(a)}$ , where $a$ is the value such that $f(a) = b$ . In this case, $f(0) = 3$ , so $a = 0$ and $b = 3$ . Therefore, $(f^{-1})'(3) = \frac{1}{f'(0)}$ . A tempting distractor is choice E, $\frac{1}{f'(3)}$ , 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 $f'(a)$ .
Let $f(x)=\sin x+x$ on $[-\pi/2,\pi/2]$ with $f(0)=0$ . Find $(f^{-1})'(0)$ .
Answer: $\dfrac{1}{2}$ This problem requires finding the derivative of an inverse function at a point. Using $(f^{-1})'(b) = \frac{1}{f'(a)}$ where $f(a) = b$ , and knowing $f(0) = 0$ , we need $(f^{-1})'(0) = \frac{1}{f'(0)}$ . Taking the derivative: $f'(x) = cos x + 1$ , we get $f'(0) = cos(0) + 1 = 1 + 1 = 2$ . Thus $(f^{-1})'(0) = \frac{1}{2}$ . A common mistake is computing $f'$ at the wrong point or forgetting that $cos(0) = 1$ , not 0. Remember that the inverse function derivative formula creates a reciprocal relationship between the rates of change.
Let $f(x)=x^3-2x+5$ be invertible and $f(1)=4$ ; what is $(f^{-1})'(4)$ ?
Answer: $\dfrac{1}{f'(1)}$ 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), where a is the value such that f(a) = b. In this case, f(1) = 4, so a = 1 and b = 4. Therefore, ( $f^{-1}$ )'(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).
If $f(x)=\ln(x-1)$ for $x>1$ and $f(e+1)=1$ , what is $(f^{-1})'(1)$ ?
Answer: $\dfrac{1}{f'(e+1)}$ This problem tests the skill of differentiating inverse functions. To find the derivative of the inverse function at a point b, use the formula $(f^{-1})'(b) = \frac{1}{f'(a)}$ , where a is the value such that $f(a) = b$ . Here, $f(e+1) = 1$ , so a = e+1 and b = 1. Thus, $(f^{-1})'(1) = \frac{1}{f'(e+1)}$ . A tempting distractor is choice A, $1/f'(1)$ , which mistakenly evaluates the derivative at b rather than at a. A transferable strategy for inverse derivatives is to identify the point a where $f(a) = b$ and then compute the reciprocal of $f'(a)$ .
Let $f(x)=x^3-2$ with inverse $g$ . What is $g'(6)$ ?
Answer: $\dfrac{1}{12}$ To find the derivative of the inverse function, we apply the formula g'(y) = 1/f'(g(y)). First, we determine g(6) by solving f(x) = 6: x³ - 2 = 6 gives x³ = 8, so x = 2 and g(6) = 2. The derivative of f is f'(x) = 3x², so f'(2) = 3(2)² = 12. Therefore, g'(6) = 1/f'(g(6)) = 1/f'(2) = 1/12. A tempting error is to compute f'(6) = 108 instead of f'(g(6)), which would give 1/108. The crucial step in inverse differentiation is always finding the input value that maps to your given output first.
Given $f(x)=\dfrac{x}{1+x}$ for $x>-1$ and $f(1)=\tfrac12$ , find $(f^{-1})'(\tfrac12)$ .
Answer: $4$ This problem involves finding the derivative of an inverse function using $(f^{-1})'(b) = \frac{1}{f'(a)}$ where $f(a) = b$ . Since $f(1) = \frac{1}{2}$ , we need $(f^{-1})'(\frac{1}{2}) = \frac{1}{f'(1)}$ . Using the quotient rule: $f'(x) = \frac{(1+x)(1) - x(1)}{(1+x)^2} = \frac{1}{(1+x)^2}$ . Thus $f'(1) = \frac{1}{(1+1)^2} = \frac{1}{4}$ , giving $(f^{-1})'(\frac{1}{2}) = 4$ . A common error is incorrectly applying the quotient rule or evaluating at $x = \frac{1}{2}$ instead of $x = 1$ . Always verify which input produces your target output before differentiating.
If $f(x)=x^3+2x$ and $g=f^{-1}$ , what is $g'(3)$ ?
Answer: $\dfrac{1}{5}$ This problem asks for the derivative of an inverse function at a specific point. Using the inverse function derivative formula g'(y) = 1/f'(g(y)), we first need to find g(3), meaning the x-value where f(x) = 3. Testing x = 1: f(1) = 1³ + 2(1) = 3, so g(3) = 1. Next, we compute f'(x) = 3x² + 2, giving f'(1) = 3(1)² + 2 = 5. Therefore, g'(3) = 1/f'(g(3)) = 1/f'(1) = 1/5. Students might mistakenly use f'(3) = 29 instead of f'(g(3)), leading to incorrect answers. Remember: always evaluate f' at the pre-image point, not at the given y-value.
Let $f(x)=\sin x$ on $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ with $f\left(\dfrac{\pi}{6}\right)=\dfrac12$ ; find $(f^{-1})'\left(\dfrac12\right)$ .
Answer: $\dfrac{1}{\cos\left(\dfrac{\pi}{6}\right)}$ 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), where a is the value such that f(a) = b. Here, f(π/6) = 1/2, so a = π/6 and b = 1/2. Thus, ( $f^{-1}$ )'(1/2) = 1 / f'(π/6) = 1 / cos(π/6). A tempting distractor is choice C, 1/cos(1/2), which confuses the evaluation point by using b directly in radians instead of 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).
Let $f(x)=\sqrt{x+4}$ (domain $x\ge -4$ ) and $g=f^{-1}$ . What is $g'(3)$ ?
Answer: $6$ This problem involves the inverse of a square root function. Using g'(y) = 1/f'(g(y)), we first find g(3) by solving √(x+4) = 3, which gives x + 4 = 9, so x = 5 and g(3) = 5. The derivative of f is f'(x) = 1/(2√(x+4)), so f'(5) = 1/(2√9) = 1/(2·3) = 1/6. Therefore, g'(3) = 1/f'(g(3)) = 1/f'(5) = 1/(1/6) = 6. A common error is forgetting the chain rule when differentiating the square root, or evaluating f' at 3 instead of g(3) = 5. For inverse derivatives of radical functions, carefully apply both the power rule and the reciprocal relationship.
Let $f(x)=x^5$ with inverse $f^{-1}(x)=x^{1/5}$ . Since $f(2)=32$ , what is $(f^{-1})'(32)$ ?
Answer: $\dfrac{1}{80}$ 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) = 1 / f'(a), where f(a) = b. Here, a = 2 and b = 32, so compute f'(x) = $5x^4$ , and f'(2) = 80. Thus, ( $f^{-1}$ )'(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.

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