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.

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 $f'(x) = 3x^2$, so f'(2) = 12, thus $(f^{-1})'(9) = \frac{1}{12}$. 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 $f^{-1}$. To solve similar problems, always remember that the derivatives of a function and its inverse are reciprocals at corresponding points.

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.

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.

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.

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.

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.

This problem involves finding the derivative of an inverse function at a specific value. The inverse function derivative formula tells us $(p^{-1})'(b) = rac{1}{p'(a)}$ when $p(a) = b$. We have $p(0) = 7$ and $p'(0) = 5$, and need $(p^{-1})'(7)$. Since $p(0) = 7$, we get $p^{-1}(7) = 0$, so we use $a = 0$. Thus $(p^{-1})'(7) = rac{1}{p'(0)} = rac{1}{5}$. Choice B ($5$) represents $p'(0)$ 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.

To find the derivative of an inverse function, apply $(u^{-1})'(b) = rac{1}{u'(a)}$ 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, $(u^{-1})'(2) = rac{1}{u'(8)} = rac{1}{0.1} = 10$. 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.

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.