if ` phi` is inverse of f and f'(x) =`(1)/( 1 +x^n)` then` phi` '[x] equals

Options

A. `1+x^(n---)`
B. `1+ [f[x]^n)`
C. `1 + [phi [x] ]^n `
D. None of these

Answer

To find the derivative of the inverse function \( \phi \) of \( f \), we can use the relationship between the derivatives of inverse functions. If \( \phi \) is the inverse of \( f \), then the following relationship holds:

\[ \phi'(x) = \frac{1}{f'(\phi(x))} \]

Given that \( f'(x) = \frac{1}{1 + x^n} \), we can substitute this into our formula for \( \phi'(x) \).

### Step-by-Step Solution:

1. **Identify the relationship**: Since \( \phi \) is the inverse of \( f \), we have:
\[ f(\phi(x)) = x \]

2. **Differentiate both sides**: Differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}[f(\phi(x))] = \frac{d}{dx}[x] \]

3. **Apply the chain rule**: Using the chain rule on the left side:
\[ f'(\phi(x)) \cdot \phi'(x) = 1 \]

4. **Solve for \( \phi'(x) \)**: Rearranging gives:
\[ \phi'(x) = \frac{1}{f'(\phi(x))} \]

5. **Substitute \( f' \)**: Now substitute \( f'(\phi(x)) \) using the given derivative \( f'(x) = \frac{1}{1 + x^n} \):
\[ \phi'(x) = \frac{1}{\frac{1}{1 + \phi(x)^n}} = 1 + \phi(x)^n \]

### Final Result:
Thus, we have:
\[ \phi'(x) = 1 + \phi(x)^n \]