Which formula correctly gives the derivative of the inverse function (f^{-1})'(y)?

Which formula correctly gives the derivative of the inverse function (f^{-1})'(y)?

• A. (f^{-1})'(y) = 1 / f'(f^{-1}(y))
• B. (f^{-1})'(y) = f'(f^{-1}(y))
• C. (f^{-1})'(y) = 1 / f(y)
• D. (f^{-1})'(y) = f(y)

Answer: (A)

The derivative of an inverse function tells us how the inverse changes as its input y changes. If y = f(x) and f is invertible with a nonzero slope at the corresponding x, then the rate at which the inverse changes with respect to y is the reciprocal of the slope of f at that x. In symbols, (f^{-1})'(y) = 1 / f'(f^{-1}(y)). This comes from differentiating y = f(x) with respect to y. Using the chain rule, dy/dx · dx/dy = 1, so dx/dy = 1 / (dy/dx). Since dy/dx is f'(x) and x = f^{-1}(y), we substitute to get (f^{-1})'(y) = 1 / f'(f^{-1}(y)). The condition f'(f^{-1}(y)) ≠ 0 ensures the inverse is differentiable there. Other forms that omit the reciprocal or replace the input with y in the slope fail to account for evaluating the slope at the inverse point, or misplace the reciprocal entirely.

The derivative of an inverse function tells us how the inverse changes as its input y changes. If y = f(x) and f is invertible with a nonzero slope at the corresponding x, then the rate at which the inverse changes with respect to y is the reciprocal of the slope of f at that x. In symbols, (f^{-1})'(y) = 1 / f'(f^{-1}(y)).

This comes from differentiating y = f(x) with respect to y. Using the chain rule, dy/dx · dx/dy = 1, so dx/dy = 1 / (dy/dx). Since dy/dx is f'(x) and x = f^{-1}(y), we substitute to get (f^{-1})'(y) = 1 / f'(f^{-1}(y)). The condition f'(f^{-1}(y)) ≠ 0 ensures the inverse is differentiable there.

Other forms that omit the reciprocal or replace the input with y in the slope fail to account for evaluating the slope at the inverse point, or misplace the reciprocal entirely.