If a function is invertible the graph of its inverse is the mirror image of the function in `y=x.` If `f^(-1)(x)` is the inverse function of f(x), then `f(f^(-1)(x))=f^(-1)(f(x))=x`.
The value of `(d)/(dx)(f^(-1)(x))" at " x = 9 " for "f(x)=x^(3)-2x^(-3)+10" is "(x gt 0)`

Options

A. `-1//3`
B. `1//9`
C. `(1)/(27)`
D. none of these

Answer

To find the value of \(\frac{d}{dx}(f^{-1}(x))\) at \(x = 9\) for the function \(f(x) = x^3 - 2x^{-3} + 10\), we will follow these steps:

### Step-by-Step Solution

1. **Understand the relationship between \(f\) and \(f^{-1}\)**:
We know that if \(y = f(x)\), then \(x = f^{-1}(y)\). The derivatives of the inverse function can be expressed using the formula:
\[ \frac{d}{dx}(f^{-1}(x)) = \frac{1}{f'(f^{-1}(x))} \]

2. **Set \(f(x) = 9\)**:
We need to find \(x\) such that \(f(x) = 9\):
\[ x^3 - 2x^{-3} + 10 = 9 \]
Simplifying this gives:
\[ x^3 - 2x^{-3} + 1 = 0 \]

3. **Multiply through by \(x^3\)** to eliminate the negative exponent:
\[ x^6 + 1 - 2 = 0 \implies x^6 - 1 = 0 \]
This factors to:
\[ (x^3 - 1)(x^3 + 1) = 0 \]
Thus, \(x^3 = 1\) or \(x^3 = -1\). Since \(x > 0\), we take \(x^3 = 1\) which gives \(x = 1\).

4. **Find \(f'(x)\)**:
Now, we need to calculate the derivative \(f'(x)\):
\[ f'(x) = \frac{d}{dx}(x^3 - 2x^{-3} + 10) = 3x^2 + 6x^{-4} \]

5. **Evaluate \(f'(1)\)**:
We substitute \(x = 1\) into the derivative:
\[ f'(1) = 3(1^2) + 6(1^{-4}) = 3 + 6 = 9 \]

6. **Find \(\frac{d}{dx}(f^{-1}(x))\) at \(x = 9\)**:
Now we can substitute into the inverse derivative formula:
\[ \frac{d}{dx}(f^{-1}(9)) = \frac{1}{f'(f^{-1}(9))} = \frac{1}{f'(1)} = \frac{1}{9} \]

### Final Answer
Thus, the value of \(\frac{d}{dx}(f^{-1}(x))\) at \(x = 9\) is:
\[ \frac{1}{9} \]