Let `g: RvecR` be a differentiable function satisfying `g(x)=g(y)g(x-y)AAx , y in R` and `g^(prime)(0)=aa n dg^(prime)(3)=bdot` Then find the value of `g^(prime)(-3)dot`

Explanation

To solve the problem, we need to find \( g'(-3) \) given the differentiable function \( g \) that satisfies the equation \( g(x) = g(y) g(x-y) \) for all \( x, y \in \mathbb{R} \), along with the conditions \( g'(0) = a \) and \( g'(3) = b \).

### Step-by-Step Solution:

1. **Start with the functional equation**:
\[ g(x) = g(y) g(x-y) \]

2. **Differentiate both sides with respect to \( x \)** (treating \( y \) as a constant):
\[ g'(x) = g(y) g'(x-y) \]

3. **Set \( y = x \)**:
\[ g'(x) = g(x) g'(0) \]
Since \( g'(0) = a \), we can rewrite this as:
\[ g'(x) = g(x) a \]

4. **Rearranging gives us**:
\[ \frac{g'(x)}{g(x)} = a \]

5. **Integrate both sides**:
\[ \int \frac{g'(x)}{g(x)} \, dx = \int a \, dx \]
This leads to:
\[ \ln |g(x)| = ax + C \]
where \( C \) is a constant.

6. **Exponentiating both sides** gives us:
\[ g(x) = e^{ax + C} = e^C e^{ax} \]
Let \( e^C = k \), then:
\[ g(x) = k e^{ax} \]

7. **Finding \( g(0) \)**:
\[ g(0) = k e^{0} = k \]

8. **Using the functional equation again with \( x = 0 \)**:
\[ g(0) = g(0) g(0) \Rightarrow k = k^2 \]
This implies \( k = 0 \) or \( k = 1 \). Since \( g'(0) = a \) and cannot be zero, we have \( k = 1 \):
\[ g(0) = 1 \]

9. **Thus, we have**:
\[ g(x) = e^{ax} \]

10. **Now differentiate \( g(x) \)**:
\[ g'(x) = a e^{ax} \]

11. **Find \( g'(-3) \)**:
\[ g'(-3) = a e^{-3a} \]

12. **Using the given condition \( g'(3) = b \)**:
\[ g'(3) = a e^{3a} = b \]
From this, we can express \( e^{3a} \):
\[ e^{3a} = \frac{b}{a} \]

13. **Substituting back into \( g'(-3) \)**:
\[ g'(-3) = a e^{-3a} = a \cdot \frac{1}{e^{3a}} = a \cdot \frac{a}{b} = \frac{a^2}{b} \]

### Final Result:
\[ g'(-3) = \frac{a^2}{b} \]