If `g` is the inverse of `f` and `f'(x)=1/(1+x^n)` , prove that `g^(prime)(x)=1+(g(x))^n`

Explanation

To prove that \( g'(x) = 1 + (g(x))^n \) given that \( g \) is the inverse of \( f \) and \( f'(x) = \frac{1}{1 + x^n} \), we can follow these steps:

### Step 1: Understand the relationship between \( g \) and \( f \)
Since \( g \) is the inverse of \( f \), we have:
\[ f(g(x)) = x \]

### Step 2: Differentiate both sides
Differentiating both sides with respect to \( x \) using the chain rule gives:
\[ \frac{d}{dx}[f(g(x))] = \frac{d}{dx}[x] \]
This simplifies to:
\[ f'(g(x)) \cdot g'(x) = 1 \]

### Step 3: Solve for \( g'(x) \)
From the equation \( f'(g(x)) \cdot g'(x) = 1 \), we can isolate \( g'(x) \):
\[ g'(x) = \frac{1}{f'(g(x))} \]

### Step 4: Substitute \( f'(x) \)
We know that \( f'(x) = \frac{1}{1 + x^n} \). Therefore, substituting \( g(x) \) into \( f' \):
\[ f'(g(x)) = \frac{1}{1 + (g(x))^n} \]

### Step 5: Substitute back into the equation for \( g'(x) \)
Now substituting this back into our expression for \( g'(x) \):
\[ g'(x) = \frac{1}{\frac{1}{1 + (g(x))^n}} = 1 + (g(x))^n \]

### Conclusion
Thus, we have shown that:
\[ g'(x) = 1 + (g(x))^n \]
This completes the proof.

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