Logarithmic Value of Differentiable Function Satisfying Functional Relation

To find the value of $\log_e(f(4))$, we analyze the given functional relation for the differentiable function $f: \mathbb{R} \to \mathbb{R}$: $f(x + y) = f(x)f'(y) + f'(x)f(y) \quad \text{for all } x, y \in \mathbb{R}$ with the initial condition $f(0) = 1$.

Step 1: Determine $f'(0)$ Substitute $x = 0$ and $y = 0$ into the given functional equation: $f(0 + 0) = f(0)f'(0) + f'(0)f(0)$ $f(0) = 2f(0)f'(0)$

Given $f(0) = 1$, we get: $1 = 2(1)f'(0) \implies f'(0) = \frac{1}{2}$

Step 2: Formulate the differential equation for $f(x)$ Substitute $y = 0$ into the original functional equation: $f(x + 0) = f(x)f'(0) + f'(x)f(0)$ $f(x) = f(x) \cdot \left(\frac{1}{2}\right) + f'(x) \cdot 1$

Rearranging terms: $f(x) - \frac{1}{2}f(x) = f'(x)$ $f'(x) = \frac{1}{2}f(x)$

Step 3: Solve the differential equation We can rewrite this first-order differential equation as: $\frac{f'(x)}{f(x)} = \frac{1}{2}$

Integrating both sides with respect to $x$: $\int \frac{f'(x)}{f(x)} \, dx = \int \frac{1}{2} \, dx$ $\log_e(f(x)) = \frac{x}{2} + C$ $f(x) = k \cdot e^{x/2}$

Using the condition $f(0) = 1$: $1 = k \cdot e^0 \implies k = 1$

Thus, the function is: $f(x) = e^{x/2}$

Step 4: Evaluate $\log_e(f(4))$ Substitute $x = 4$ into $f(x)$: $f(4) = e^{4/2} = e^2$

Taking the natural logarithm: $\log_e(f(4)) = \log_e(e^2) = 2$

Final Answer: The value of $\log_e(f(4))$ is 2.