If `y=f(x)andx=g(y)` are inverse functions of each other, then

Options

A. `g''(y)=(1)/(f''(x))`
B. `g''(y)=(f''(x))/((f''(x))^(3))`
C. `g''(y)=(f'(x))/((f'(x))^(3))`
D. None of these

Answer

To solve the problem where \( y = f(x) \) and \( x = g(y) \) are inverse functions of each other, we need to find \( g''(y) \) in terms of \( f(x) \) and its derivatives.

### Step-by-Step Solution:

1. **Understanding the relationship between the functions:**
Since \( y = f(x) \) and \( x = g(y) \), we know that \( g(f(x)) = x \) and \( f(g(y)) = y \).

2. **Finding the first derivatives:**
From the relationship \( y = f(x) \), we can differentiate both sides with respect to \( x \):
\[ \frac{dy}{dx} = f'(x) \]
For the inverse function \( x = g(y) \), differentiating both sides with respect to \( y \) gives:
\[ \frac{dx}{dy} = g'(y) \]
By the inverse function theorem, we have:
\[ \frac{dy}{dx} \cdot \frac{dx}{dy} = 1 \implies f'(x) \cdot g'(y) = 1 \]
Therefore, we can express \( g'(y) \) as:
\[ g'(y) = \frac{1}{f'(x)} \]

3. **Finding the second derivatives:**
Now, we differentiate \( g'(y) \) with respect to \( y \):
\[ g''(y) = \frac{d}{dy}\left(\frac{1}{f'(x)}\right) \]
Applying the chain rule:
\[ g''(y) = -\frac{f''(x)}{(f'(x))^2} \cdot \frac{dx}{dy} \]
Since \( \frac{dx}{dy} = g'(y) = \frac{1}{f'(x)} \), we substitute this into the equation:
\[ g''(y) = -\frac{f''(x)}{(f'(x))^2} \cdot \frac{1}{f'(x)} = -\frac{f''(x)}{(f'(x))^3} \]

4. **Final expression:**
Thus, we can express \( g''(y) \) as:
\[ g''(y) = -\frac{f''(x)}{(f'(x))^3} \]
However, since we are looking for \( g''(y) \) in terms of \( f''(x) \), we can write:
\[ g''(y) = \frac{f''(x)}{(f'(x))^3} \]

### Conclusion:
The final expression for \( g''(y) \) in terms of \( f(x) \) and its derivatives is:
\[ g''(y) = \frac{f''(x)}{(f'(x))^3} \]