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 find \( g'''(0) \) for the function \( g(x) = f^{-1}(x) \), where

\[ f(x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \frac{x^5}{5}, \]

we will follow these steps:

### Step 1: Differentiate \( g(x) \)

Using the relationship between \( g(x) \) and \( f(x) \), we know that:

\[ g(f(x)) = x. \]

Differentiating both sides with respect to \( x \):

\[ g'(f(x)) \cdot f'(x) = 1. \]

Thus, we can express \( g'(f(x)) \) as:

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

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

Now, we differentiate \( g'(f(x)) \) again:

\[ g''(f(x)) \cdot f'(x) + g'(f(x)) \cdot f''(x) = 0. \]

Substituting \( g'(f(x)) = \frac{1}{f'(x)} \):

\[ g''(f(x)) \cdot f'(x) + \frac{f''(x)}{f'(x)} = 0. \]

This gives us:

\[ g''(f(x)) = -\frac{f''(x)}{(f'(x))^3}. \]

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

We differentiate \( g''(f(x)) \):

\[ g'''(f(x)) \cdot f'(x) + g''(f(x)) \cdot f''(x) = 0. \]

Substituting \( g''(f(x)) = -\frac{f''(x)}{(f'(x))^3} \):

\[ g'''(f(x)) \cdot f'(x) - \frac{f''(x) \cdot f''(x)}{(f'(x))^3} = 0. \]

Thus,

\[ g'''(f(x)) = \frac{f''(x) \cdot f''(x)}{(f'(x))^4}. \]

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

Now we need to find \( g'''(0) \). First, we calculate \( f(0) \), \( f'(0) \), \( f''(0) \), and \( f'''(0) \):

1. **Calculate \( f(0) \)**:

\[ f(0) = 0 + 0 + 0 + 0 + 0 = 0. \]

2. **Calculate \( f'(x) \)**:

\[ f'(x) = 1 + x + x^2 + x^3 + x^4. \]

So,

\[ f'(0) = 1. \]

3. **Calculate \( f''(x) \)**:

\[ f''(x) = 1 + 2x + 3x^2 + 4x^3. \]

Thus,

\[ f''(0) = 1. \]

4. **Calculate \( f'''(x) \)**:

\[ f'''(x) = 2 + 6x + 12x^2. \]

So,

\[ f'''(0) = 2. \]

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

Now substituting these values into our expression for \( g'''(f(0)) \):

\[ g'''(0) = \frac{f''(0) \cdot f''(0)}{(f'(0))^4} = \frac{1 \cdot 1}{1^4} = 1. \]

### Final Answer

Therefore, the value of \( g'''(0) \) is

\[ \boxed{1}. \]