Let `f(x)=(x^2-x)/(x^2+2x)` then `d(f^(-1)x)/(dx)` is equal to

Options

A. `(-3)/((1-x)^(2))`
B. `(3)/((1-x)^(2))`
C. `(1)/((1-x)^(2))`
D. none of these

Explanation

To solve the problem, we need to find the derivative of the inverse function \( f^{-1}(x) \) for the given function \( f(x) = \frac{x^2 - x}{x^2 + 2x} \).

### Step-by-Step Solution:

1. **Identify the function**:
\[ f(x) = \frac{x^2 - x}{x^2 + 2x} \]

2. **Determine where \( f(x) \) is undefined**:
The function \( f(x) \) is undefined when the denominator is zero:
\[ x^2 + 2x = 0 \]
Factoring gives:
\[ x(x + 2) = 0 \implies x = 0 \text{ or } x = -2 \]
Thus, the domain of \( f(x) \) is \( \mathbb{R} \setminus \{-2, 0\} \).

3. **Simplify the function**:
We can simplify \( f(x) \):
\[ f(x) = \frac{x(x - 1)}{x(x + 2)} = \frac{x - 1}{x + 2} \quad \text{(for } x \neq 0\text{)} \]

4. **Express \( f(x) \) in terms of \( y \)**:
Set \( y = f(x) \):
\[ y = \frac{x - 1}{x + 2} \]
Rearranging gives:
\[ y(x + 2) = x - 1 \implies yx + 2y = x - 1 \]
Rearranging to isolate \( x \):
\[ yx - x = -1 - 2y \implies x(y - 1) = -1 - 2y \implies x = \frac{-1 - 2y}{y - 1} \]
Thus, we have:
\[ f^{-1}(y) = \frac{-1 - 2y}{y - 1} \]

5. **Differentiate \( f^{-1}(x) \)**:
Now we need to differentiate \( f^{-1}(x) \):
\[ f^{-1}(x) = \frac{-1 - 2x}{x - 1} \]
We apply the quotient rule:
\[ \frac{d}{dx} f^{-1}(x) = \frac{(x - 1)(-2) - (-1 - 2x)(1)}{(x - 1)^2} \]

6. **Simplify the derivative**:
Simplifying the numerator:
\[ = \frac{-2(x - 1) + (1 + 2x)}{(x - 1)^2} = \frac{-2x + 2 + 1 + 2x}{(x - 1)^2} = \frac{3}{(x - 1)^2} \]

7. **Final result**:
Therefore, the derivative of the inverse function is:
\[ \frac{d}{dx} f^{-1}(x) = \frac{3}{(1 - x)^2} \]

### Conclusion:
The answer is:
\[ \frac{d(f^{-1}(x))}{dx} = \frac{3}{(1 - x)^2} \]