Let's solve exercise 265, where we have $f(x) = 2x^3 - 3$ and $x = 1$. We need to find the derivative of the inverse function $(f^{-1})'(a)$.

Solution

For the derivative of the inverse function, we use the formula: $(f^{-1})'(a) = \frac{1}{f'(x)} \quad \text{where } a = f(x).$

1. Step 1: Find $f(1)$:

   Given $f(x) = 2x^3 - 3$, plug in $x = 1$: $f(1) = 2(1)^3 - 3 = 2 - 3 = -1.$

   So, $a = -1$.

2. Step 2: Find the derivative $f'(x)$:

   Differentiate $f(x)$: $f'(x) = \frac{d}{dx}(2x^3 - 3) = 6x^2.$

3. Step 3: Evaluate $f'(1)$:

   Plug in $x = 1$: $f'(1) = 6(1)^2 = 6.$

4. Step 4: Find $(f^{-1})'(-1)$:

   Using the formula: $(f^{-1})'(-1) = \frac{1}{f'(1)} = \frac{1}{6}.$

Final Answer:

$(f^{-1})'(-1) = \frac{1}{6}.$

Would you like a deeper explanation or have any questions about this problem? Here are some related problems:

1. How do you derive the formula $(f^{-1})'(a) = \frac{1}{f'(x)}$?
2. What is the meaning of the derivative of an inverse function in a practical context?
3. If $f(x) = 3x^2$, how would you find $(f^{-1})'(a)$?
4. What are the conditions under which a function has an inverse?
5. How do we determine the domain and range of an inverse function?

Tip: Remember, for a function to have an inverse, it must be one-to-one (bijective), meaning it should pass both the horizontal line test (injective) and cover the range (surjective).