Let f : R rightarrow R be a differentiable fu | vedclass

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(B) Given $f(x+y)=f(x) f^{\prime}(y)+f^{\prime}(x) f(y)$ and $f(0)=1$.Setting $x=0$ and $y=0$ in the given equation:$f(0+0)=f(0)f^{\prime}(0)+f^{\prim

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$2$

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(B) Given $f(x+y)=f(x) f^{\prime}(y)+f^{\prime}(x) f(y)$ and $f(0)=1$.Setting $x=0$ and $y=0$ in the given equation:$f(0+0)=f(0)f^{\prime}(0)+f^{\prime}(0)f(0)$$f(0)=2f(0)f^{\prime}(0)$Since $f(0)=1$,we have $1=2(1)f^{\prime}(0)$,which implies $f^{\prime}(0)=\frac{1}{2}$.Now,setting $y=0$ in the original equation:$f(x+0)=f(x)f^{\prime}(0)+f^{\prime}(x)f(0)$$f(x)=f(x) \cdot \frac{1}{2} + f^{\prime}(x) \cdot 1$$f^{\prime}(x) = f(x) - \frac{1}{2}f(x) = \frac{1}{2}f(x)$.This is a first-order linear differential equation:$\frac{f^{\prime}(x)}{f(x)} = \frac{1}{2}$.Integrating both sides with respect to $x$:$\int \frac{f^{\prime}(x)}{f(x)} dx = \int \frac{1}{2} dx$$\ln(f(x)) = \frac{x}{2} + C$.Using $f(0)=1$,we get $\ln(1) = 0 + C$,so $C=0$.Thus,$f(x) = e^{x/2}$.Finally,we calculate $\log _e(f(4))$:$f(4) = e^{4/2} = e^2$.$\log _e(f(4)) = \log _e(e^2) = 2$.

Let $f : R \rightarrow R$ be a differentiable function with $f(0)=1$ and satisfying the equation $f(x+y)=f(x) f^{\prime}(y)+f^{\prime}(x) f(y)$ for all $x, y \in R$. Then,the value of $\log _e(f(4))$ is:

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