Let `f (x)=x+ (x ^(2))/(2 )+ (x ^(3))/(3 )+ (x ^(4))/(4 ) +(x ^(5))/(5)` and let `g (x) = f ^(-1) (x).` Find `g'''(o).`

Answer

To solve the problem, we need to find \( g'''(0) \) where \( g(x) = f^{-1}(x) \) and \( f(x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \frac{x^5}{5} \).

### Step 1: Understand the relationship between \( f(x) \) and \( g(x) \)

Since \( g(x) = f^{-1}(x) \), we have:
\[ f(g(x)) = x \]
Differentiating both sides with respect to \( x \):
\[ f'(g(x)) \cdot g'(x) = 1 \]
Thus,
\[ g'(x) = \frac{1}{f'(g(x))} \]

### Step 2: Differentiate to find \( g''(x) \)

Now differentiate \( g'(x) \):
\[ g''(x) = \frac{d}{dx}\left(\frac{1}{f'(g(x))}\right) \]
Using the chain rule:
\[ g''(x) = -\frac{f''(g(x)) \cdot g'(x)}{(f'(g(x)))^2} \]
Substituting \( g'(x) \):
\[ g''(x) = -\frac{f''(g(x)) \cdot \frac{1}{f'(g(x))}}{(f'(g(x)))^2} = -\frac{f''(g(x))}{(f'(g(x)))^3} \]

### Step 3: Differentiate to find \( g'''(x) \)

Now differentiate \( g''(x) \):
\[ g'''(x) = -\frac{d}{dx}\left(\frac{f''(g(x))}{(f'(g(x)))^3}\right) \]
Using the quotient rule:
\[ g'''(x) = -\left(\frac{(f'''(g(x)) \cdot g'(x))(f'(g(x)))^3 - f''(g(x)) \cdot 3(f'(g(x)))^2 f''(g(x)) \cdot g'(x)}{(f'(g(x)))^6}\right) \]
Substituting \( g'(x) \):
\[ g'''(x) = -\frac{(f'''(g(x)) \cdot \frac{1}{f'(g(x))})(f'(g(x)))^3 - f''(g(x)) \cdot 3(f'(g(x)))^2 f''(g(x)) \cdot \frac{1}{f'(g(x)))}}{(f'(g(x)))^6} \]

### Step 4: Evaluate at \( x = 0 \)

To find \( g'''(0) \), we need \( g(0) \), \( f'(0) \), \( f''(0) \), and \( f'''(0) \).

1. **Calculate \( f(0) \)**:
\[ f(0) = 0 + 0 + 0 + 0 + 0 = 0 \quad \Rightarrow \quad g(0) = 0 \]

2. **Calculate \( f'(x) \)**:
\[ f'(x) = 1 + x + x^2 + x^3 + x^4 \]
Thus,
\[ f'(0) = 1 \]

3. **Calculate \( f''(x) \)**:
\[ f''(x) = 1 + 2x + 3x^2 + 4x^3 \]
Thus,
\[ f''(0) = 1 \]

4. **Calculate \( f'''(x) \)**:
\[ f'''(x) = 2 + 6x + 12x^2 \]
Thus,
\[ f'''(0) = 2 \]

### Step 5: Substitute back into \( g'''(0) \)

Now substituting into the expression for \( g'''(0) \):
\[ g'''(0) = -\frac{(f'''(0) \cdot 1)(1)^3 - (1) \cdot 3(1)^2(1) \cdot 1}{(1)^6} \]
\[ g'''(0) = -\frac{(2)(1) - 3(1)(1)}{1} = -\frac{2 - 3}{1} = -(-1) = 1 \]

### Final Result
Thus, the final result is:
\[ \boxed{1} \]