If `f(x-y)=f(x).g(y)-f(y).g(x)` and `g(x-y)=g(x).g(y)+f(x).f(y)` for all `x in R` . If right handed derivative at x=0 exists for f(x) find the derivative of g(x) at x =0

Options

A. `-1`
B. 0
C. 1
D. none of these

Explanation

To solve the problem step by step, we need to analyze the given functional equations and derive the necessary results to find the derivative of \( g(x) \) at \( x = 0 \).

### Step 1: Analyze the given equations
We have two functional equations:
1. \( f(x-y) = f(x)g(y) - f(y)g(x) \)
2. \( g(x-y) = g(x)g(y) + f(x)f(y) \)

### Step 2: Substitute specific values
Let's substitute \( y = 0 \) in both equations to simplify them.

For the first equation:
\[ f(x-0) = f(x)g(0) - f(0)g(x) \]
This simplifies to:
\[ f(x) = f(x)g(0) - f(0)g(x) \]
Rearranging gives us:
\[ f(x)(1 - g(0)) = -f(0)g(x) \]

For the second equation:
\[ g(x-0) = g(x)g(0) + f(x)f(0) \]
This simplifies to:
\[ g(x) = g(x)g(0) + f(x)f(0) \]
Rearranging gives us:
\[ g(x)(1 - g(0)) = f(x)f(0) \]

### Step 3: Analyze the implications
From the first equation, if \( g(0) \neq 1 \), we can express \( g(x) \) in terms of \( f(x) \):
\[ g(x) = -\frac{f(x)(1 - g(0))}{f(0)} \]

From the second equation, if \( g(0) \neq 1 \), we can express \( f(x) \) in terms of \( g(x) \):
\[ f(x) = \frac{g(x)(1 - g(0))}{f(0)} \]

### Step 4: Differentiate to find \( g'(0) \)
Since we are interested in \( g'(0) \), we can differentiate the expression for \( g(x) \) we derived:
\[ g'(x) = -\frac{f'(x)(1 - g(0))}{f(0)} \quad \text{(using the product rule)} \]

Now, evaluate at \( x = 0 \):
\[ g'(0) = -\frac{f'(0)(1 - g(0))}{f(0)} \]

### Step 5: Evaluate \( f'(0) \) and \( g(0) \)
Assuming \( f(x) = \sin(x) \) (as suggested in the video), we have:
\[ f'(x) = \cos(x) \quad \text{and thus} \quad f'(0) = \cos(0) = 1 \]
Assuming \( g(0) = 1 \) (as a reasonable assumption based on the functional equations), we have:
\[ g'(0) = -\frac{1(1 - 1)}{f(0)} = 0 \]

### Conclusion
Thus, the derivative of \( g(x) \) at \( x = 0 \) is:
\[ g'(0) = 0 \]