Let f(x) = x^3 + x and y = f(x). Express (f^{-1})'(y) in terms of x.

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Multiple Choice

Let f(x) = x^3 + x and y = f(x). Express (f^{-1})'(y) in terms of x.

Explanation:
The derivative of an inverse function tells us that (f^{-1})'(y) equals 1 divided by f'(f^{-1}(y)). Here y = f(x), so f^{-1}(y) = x. Therefore (f^{-1})'(y) = 1 / f'(x). For f(x) = x^3 + x, the derivative is f'(x) = 3x^2 + 1. Substituting gives (f^{-1})'(y) = 1 / (3x^2 + 1). This is valid for all real x because 3x^2 + 1 is always positive, so f is strictly increasing and invertible on all real numbers.

The derivative of an inverse function tells us that (f^{-1})'(y) equals 1 divided by f'(f^{-1}(y)). Here y = f(x), so f^{-1}(y) = x. Therefore (f^{-1})'(y) = 1 / f'(x). For f(x) = x^3 + x, the derivative is f'(x) = 3x^2 + 1. Substituting gives (f^{-1})'(y) = 1 / (3x^2 + 1). This is valid for all real x because 3x^2 + 1 is always positive, so f is strictly increasing and invertible on all real numbers.

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