If `y=f(x)` and `x=g(y)` are inverse of each other. Then `g'(y)` and `g"(y)` in terms of derivative of f(y) is

A.`-(f"(x))/([f'(x)]^(3))`
B. `(f"(x))/([f'(x)]^(2))`
C. `-(f"(x))/([f'(x)]^(2))`
D. None of these

Options

A. `-(f"(x))/([f'(x)]^(3))`
B. `(f"(x))/([f'(x)]^(2))`
C. `-(f"(x))/([f'(x)]^(2))`
D. None of these

Answer

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

### Step 1: Differentiate \( y = f(x) \)
We start by differentiating \( y = f(x) \) with respect to \( x \):
\[ \frac{dy}{dx} = f'(x) \]

**Hint:** Remember that the derivative of a function gives the slope of the tangent line at any point on the curve.

### Step 2: Differentiate \( x = g(y) \)
Next, we differentiate \( x = g(y) \) with respect to \( y \):
\[ \frac{dx}{dy} = g'(y) \]

**Hint:** This derivative represents how \( x \) changes with respect to \( y \).

### Step 3: Relate \( g'(y) \) and \( f'(x) \)
Since \( y = f(x) \) and \( x = g(y) \) are inverses, we know that:
\[ \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} \implies g'(y) = \frac{1}{f'(x)} \]

**Hint:** The relationship between the derivatives of inverse functions is key to solving this problem.

### Step 4: Differentiate \( g'(y) \) to find \( g''(y) \)
Now, we differentiate \( g'(y) \) with respect to \( y \):
\[ g''(y) = \frac{d}{dy}\left(\frac{1}{f'(x)}\right) \]
Using the chain rule, we get:
\[ g''(y) = -\frac{f''(x)}{(f'(x))^2} \cdot \frac{dx}{dy} \]
Since \( \frac{dx}{dy} = g'(y) = \frac{1}{f'(x)} \), we substitute this into the equation:
\[ g''(y) = -\frac{f''(x)}{(f'(x))^2} \cdot \frac{1}{f'(x)} = -\frac{f''(x)}{(f'(x))^3} \]

**Hint:** When differentiating a quotient, remember to apply the quotient rule or the chain rule appropriately.

### Conclusion
Thus, we have:
\[ g''(y) = -\frac{f''(x)}{(f'(x))^3} \]

### Final Answer
The correct option is:
**A.** \(-\frac{f''(x)}{(f'(x))^3}\)