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

\(f(x+y)=f(x) \cdot f^{\prime}(y)+f^{\prime}(x) \cdot f(y) \forall x, y \in R\)
putting \(x=y=0\) we get
\(\begin{aligned} & f(0)=2 f(0) \cdot f^{\prime}(0) \\ & \Rightarrow f^{\prime}(0)=\frac{1}{2}[\because \text { given } f(0)=1]\end{aligned}\)
Now putting \(x=x\) and \(y=0\)
\(\begin{aligned} & f(x)=f(x) \cdot f^{\prime}(0)+f^{\prime}(x) \cdot f(0) \\ & \Rightarrow f(x)=\frac{1}{2} f(x)+f^{\prime}(x)\left[\because f(0)=1 \text { and } f^{\prime}(0)=\frac{1}{2}\right] \\ & \Rightarrow \frac{1}{2} f(x)=f^{\prime}(x) \\ & \Rightarrow \int \frac{f^{\prime}(x)}{f(x)} \mathrm{d} x=\int \frac{1}{2} \mathrm{~d} x \\ & \Rightarrow \log (f(x))=\frac{1}{2} x+c \\ & \because f(0)=1 \\ & \Rightarrow c=0\end{aligned}\)
i.e., \(\log (f(x))=\frac{1}{2} x\) putting \(x=4\)
\(\log (f(4))=\frac{1}{2} \times 4=2\)