Let $f:R\to R$ be a differentiable function$f\left( 0 \right)=1$. If $y=f\left( x \right)$ satisfies the differential equation $f\left( x+y \right)=f\left( x \right){{f}^{'}}\left( y \right)+{{f}^{'}}\left( x \right)f\left( y \right)$ for all real $x$ and y. Then find out the value of ${{\log }_{e}}\left( f\left( 4 \right) \right)$ \[\]

Hint: Use the given initial condition and the differential equation to find out the value of ${{f}^{'}}\left( 0 \right)$ and $f\left( 0 \right)$. Then try to find the expression for $\dfrac{{{f}^{'}}\left( x \right)}{f\left( x \right)}$ whose integration using a standard formula will result in a form of the required ${{\log }_{e}}\left( f\left( 4 \right) \right)$.

Complete step-by-step answer:
The given differential equation with $f:R\to R$ as a differentiable function is
\[f\left( x+y \right)=f\left( x \right){{f}^{'}}\left( y \right)+{{f}^{'}}\left( x \right)f\left( y \right)....(1)\]
with initial condition$f\left( 0 \right)=1$ . We begin by putting the initial condition that is when $y=1$ when $x=0$ in the differential equation \[\]
$\begin{align}
  & f\left( 0+0 \right)=f\left( 0 \right){{f}^{'}}\left( 0 \right)+{{f}^{'}}\left( 0 \right)f\left( 0 \right) \\
 & \Rightarrow f\left( 0 \right)={{f}^{'}}\left( 0 \right)+{{f}^{'}}\left( 0 \right) \\
 & \Rightarrow {{f}^{'}}\left( 0 \right)=\dfrac{1}{2} \\
\end{align}$ \[\]
For $y=0$ the differential equation transforms to
\[f\left( x \right)=f\left( x \right){{f}^{'}}\left( 0 \right)+{{f}^{'}}\left( x \right)f\left( 0 \right)....(2)\]
We put ${{f}^{'}}\left( 0 \right)=\dfrac{1}{2}$ and $f\left( 0 \right)=1$ in the given differential equation (2) to obtain,
\[\begin{align}
  & f\left( x \right)=f\left( x \right){{f}^{'}}\left( 0 \right)+{{f}^{'}}\left( x \right)f\left( 0 \right) \\
 & \Rightarrow f\left( x \right)=f\left( x \right)\cdot \dfrac{1}{2}+{{f}^{'}}\left( x \right)\cdot 1 \\
 & \Rightarrow {{f}^{'}}\left( x \right)=\dfrac{1}{2}f\left( x \right) \\
 & \Rightarrow \dfrac{{{f}^{'}}\left( x \right)}{f\left( x \right)}=\dfrac{1}{2} \\
\end{align}\] \[\]
We integrate both sides of the above result with respect to $x$ using the substitution of variable and the formula $\int{\dfrac{1}{t}}dt={{\log }_{e}}t$.
\[\begin{align}
  & \int{\dfrac{{{f}^{'}}\left( x \right)}{f\left( x \right)}}dx=\int{\dfrac{1}{2}}dx \\
 & \Rightarrow {{\log }_{e}}\left| f\left( x \right) \right|=\dfrac{x}{2}+c....(3) \\
\end{align}\]
\[\]We use again the given initial condition to find out the value of the constant occurred during integration,
\[\begin{align}
  & {{\log }_{e}}\left| f\left( x \right) \right|=\dfrac{x}{2}+c \\
 & \Rightarrow {{\log }_{e}}\left| y \right|=\dfrac{x}{2}+c\left( \because y=f\left( x \right) \right) \\
 & \Rightarrow {{\log }_{e}}\left| 1 \right|=\dfrac{0}{2}+c \\
 & \Rightarrow c=0 \\
\end{align}\]
Putting the value of $c$ in equation (3) we get
\[{{\log }_{e}}\left| f\left( x \right) \right|=\dfrac{x}{2}\]
As asked in the question we have to determine at $x=4$. So we put $x=4$ in the above result to get,
\[{{\log }_{e}}\left| f\left( 4 \right) \right|=\dfrac{4}{2}=2\]
So the answer is 2.

Note:Verification:
We can verify that the given function is an exponential function, say $ f\left( x \right)={{a}^{x}}$ . We can check the initial condition ${{a}^{0}}=1$ which satisfies. Also, ${{f}^{'}}\left( x \right)={{\log }_{e}}a\cdot {{a}^{x}}$ . At $x=0$
\[\begin{align}
  & {{f}^{'}}\left( 0 \right)={{\log }_{e}}a\cdot {{a}^{1}}=\dfrac{1}{2} \\
 & \Rightarrow a={{e}^{\dfrac{1}{2}}} \\
\end{align}\]
So the given function is obtained as $y=f\left( x \right)={{e}^{\dfrac{x}{2}}}$. We shall verify by putting it in the given differential equation . from the right hand side of equation (1)
\[\begin{align}
  & f\left( x \right){{f}^{'}}\left( y \right)+{{f}^{'}}\left( x \right)f\left( y \right) \\
 & ={{e}^{\dfrac{x}{2}}}\left( \dfrac{1}{2}{{e}^{\dfrac{y}{2}}} \right)+\left( \dfrac{1}{2}{{e}^{\dfrac{x}{2}}} \right){{e}^{\dfrac{y}{2}}} \\
 & =2\left( {{e}^{\dfrac{x+y}{2}}} \right)=f\left( x+y \right) \\
\end{align}\]
Which is equal to the right hand side of equation (1).
We need to be careful of wrong substitution and use of standard formula in this problem which may lead to incorrect results. We should not leave the modulus sign while integrating $\dfrac{1}{x}$