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To solve this problem, we need to find the derivative of the inverse function at a specific point, specifically \((f^{-1})'(-6)\), using the given values:

1. \( f(-1) = -6 \)
2. \( f'(-1) = \frac{2}{5} \)

### Step-by-Step Solution

1. **Recall the formula for the derivative of the inverse function:**

   \[
   (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}
   \]

2. **Identify \(f^{-1}(-6)\):**
   Since \( f(-1) = -6 \), it follows that \( f^{-1}(-6) = -1 \). This is because \(f^{-1}\) undoes what \(f\) does, and we are given that \(f(-1) = -6\).

3. **Apply the inverse derivative formula at \(x = -6\):**

   Substitute \(x = -6\) into the formula:
   
   \[
   (f^{-1})'(-6) = \frac{1}{f'(f^{-1}(-6))}
   \]

4. **Simplify using \(f^{-1}(-6) = -1\):**

   Substitute \(f^{-1}(-6) = -1\):
   
   \[
   (f^{-1})'(-6) = \frac{1}{f'(-1)}
   \]

5. **Substitute \(f'(-1) = \frac{2}{5}\):**

   Now, plug in the value \(f'(-1) = \frac{2}{5}\):
   
   \[
   (f^{-1})'(-6) = \frac{1}{\frac{2}{5}} = \frac{5}{2}
   \]

### Final Answer
\[
(f^{-1})'(-6) = \frac{5}{2}
\]

Let me know if you'd like further details or have any questions!

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### Related Questions

1. How do we derive the formula \((f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}\) using implicit differentiation?
2. What is the significance of the function \(f\) being one-to-one in the context of invertibility?
3. How would we approach this problem if \(f(-1)\) was not equal to \(-6\)?
4. Can we use this approach for functions that are not differentiable?
5. What changes if \(f'(-1)\) were negative instead of positive?

### Tip:
When working with inverses, remember that \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \). These relationships are key in understanding how inverse functions behave.

Find (f^{-1})'(-6) if f(-1) = -6 and f'(-1) = 2/5.

Learn how to find the derivative of the inverse of a function at a specific point. This calculus problem uses implicit differentiation and the inverse function theorem to solve for (f^{-1})'(-6) given f(-1) = -6 and f'(-1) = 2/5. Ideal for Grades 11-12.

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