We are tasked with finding $(f^{-1})'(4)$, given the following:

• $f(-1) = 4$
• $f'(-1) = \frac{13}{5}$

We know the formula for the derivative of the inverse function is:

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

To find $(f^{-1})'(4)$, apply the steps:

1. First, identify that $f^{-1}(4) = -1$ because $f(-1) = 4$. Thus, $f^{-1}(4) = -1$.

2. Now substitute $x = 4$ into the derivative formula:

$(f^{-1})'(4) = \frac{1}{f'(f^{-1}(4))} = \frac{1}{f'(-1)}$

3. We are given $f'(-1) = \frac{13}{5}$, so:

$(f^{-1})'(4) = \frac{1}{\frac{13}{5}} = \frac{5}{13}$

Thus, $(f^{-1})'(4) = \frac{5}{13}$.

Would you like further details on this explanation or have any additional questions?

Here are five related questions to deepen your understanding:

1. How do you compute the derivative of the inverse function in general?
2. Why is the inverse function differentiable if the original function is differentiable?
3. How does the chain rule apply when differentiating inverse functions?
4. Can you explain a scenario where this method would fail, such as when $f'(x)$ is zero?
5. How would the calculation change if we were given different values for $f(x)$ and $f'(x)$?

Tip: When dealing with inverse functions, it's crucial to correctly match values between $f(x)$ and $f^{-1}(x)$. The two are related such that $f(f^{-1}(x)) = x$.