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!

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.