The derivative of the inverse of a function \( f \), denoted \( f^{-1}(x) \), is given by the following formula:

\[
\frac{d}{dx} \left( f^{-1}(x) \right) = \frac{1}{f'(f^{-1}(x))}
\]

This formula applies under the assumption that \( f \) is a one-to-one (bijective) function and that \( f'(x) \), the derivative of \( f(x) \), is non-zero at the relevant points. Here’s the step-by-step reasoning:

1. Suppose \( y = f^{-1}(x) \), which implies \( f(y) = x \).
2. Taking the derivative of both sides with respect to \( x \), we use implicit differentiation:
   \[
   \frac{d}{dx}(f(y)) = \frac{d}{dx}(x)
   \]
3. By the chain rule, this becomes:
   \[
   f'(y) \cdot \frac{dy}{dx} = 1
   \]
4. Solving for \( \frac{dy}{dx} \), we get:
   \[
   \frac{dy}{dx} = \frac{1}{f'(y)} = \frac{1}{f'(f^{-1}(x))}
   \]
   
This gives the derivative of the inverse function.

Would you like more details or have any questions?

Here are some related questions to further your understanding:
1. How do you find the derivative of an inverse trig function?
2. Can the formula for \( f^{-1}(x) \) derivative be used for all functions?
3. What happens if \( f'(x) = 0 \) at any point?
4. How do you compute the derivative of an inverse function for a given example?
5. How is implicit differentiation related to inverse functions?

**Tip:** Make sure that \( f(x) \) is one-to-one (bijective) before finding the derivative of its inverse!

Learn how to compute the derivative of the inverse of a function, f^−1(x), using implicit differentiation and the inverse function theorem. This detailed solution explains the formula d/dx(f^−1(x)) = 1 / f'(f^−1(x)) with step-by-step reasoning, ideal for Calculus I students.

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