To solve $(f^{-1})'(8)$, we use the relationship between the derivatives of a function and its inverse:

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

Step 1: Identify the given information

• $f(1) = 8$, since the point $(1, 8)$ lies on the curve of $f$.
• The tangent line to $f$ at $(1, 8)$ is $y = 5x + 3$.

The slope of the tangent line represents the derivative of $f$ at $x = 1$: $f'(1) = 5$

Step 2: Evaluate $(f^{-1})'(8)$

Since $f(1) = 8$, it follows that $f^{-1}(8) = 1$. Using the inverse derivative formula:

$(f^{-1})'(8) = \frac{1}{f'(f^{-1}(8))} = \frac{1}{f'(1)}$

Substituting $f'(1) = 5$:

$(f^{-1})'(8) = \frac{1}{5}$

Final Answer:

$(f^{-1})'(8) = \frac{1}{5}$

Would you like further explanation on this problem or related topics?

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Here are 5 related questions for further exploration:

1. What is the geometric interpretation of the derivative of an inverse function?
2. How do you find the derivative of an inverse for more complex functions?
3. Can the formula $(f^{-1})'(a) = 1 / f'(f^{-1}(a))$ fail? Under what conditions?
4. How do tangent lines help in understanding local linearity of functions?
5. What are the applications of inverse function derivatives in real-world problems?

Tip: Always confirm that a function is one-to-one before applying inverse derivative formulas.