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

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(C) Given the functional equation $f(x+y)=f(x) \cdot f^{\prime}(y)+f^{\prime}(x) \cdot f(y)$.Putting $x=0$ and $y=0$,we get $f(0)=f(0) \cdot f^{\prime

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$2$

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(C) Given the functional equation $f(x+y)=f(x) \cdot f^{\prime}(y)+f^{\prime}(x) \cdot f(y)$.Putting $x=0$ and $y=0$,we get $f(0)=f(0) \cdot f^{\prime}(0)+f^{\prime}(0) \cdot f(0) = 2 f(0) \cdot f^{\prime}(0)$.Since $f(0)=1$,we have $1 = 2(1) \cdot f^{\prime}(0)$,which implies $f^{\prime}(0)=\frac{1}{2}$.Now,keeping $x$ as a variable and setting $y=0$,we get $f(x+0)=f(x) \cdot f^{\prime}(0)+f^{\prime}(x) \cdot f(0)$.Substituting $f(0)=1$ and $f^{\prime}(0)=\frac{1}{2}$,we obtain $f(x) = f(x) \cdot \frac{1}{2} + f^{\prime}(x) \cdot 1$.This simplifies to $f^{\prime}(x) = \frac{1}{2} f(x)$,or $\frac{f^{\prime}(x)}{f(x)} = \frac{1}{2}$.Integrating both sides with respect to $x$,we get $\int \frac{f^{\prime}(x)}{f(x)} dx = \int \frac{1}{2} dx$,which results in $\log(f(x)) = \frac{1}{2}x + C$.Using $f(0)=1$,we have $\log(1) = 0 + C$,so $C=0$.Thus,$\log(f(x)) = \frac{1}{2}x$.For $x=4$,$\log(f(4)) = \frac{1}{2} 	imes 4 = 2$.

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

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