Question

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

Answer 1

$P(x,y):f(x+y)=f\left(x\right)f^{\prime }\left(y\right)+f^{\prime }\left(x\right)f\left(y\right)$
$P(0,0):f\left(0\right)=f\left(0\right)f^{\prime }\left(0\right)+f^{\prime }\left(0\right)f\left(0\right)\Rightarrow f^{\prime }\left(0\right)=\frac{1}{2}(\because f(0)=1)$
$P(x,0):f\left(x\right)=f\left(x\right)f^{\prime }\left(0\right)+f^{\prime }\left(x\right)f\left(0\right)\Rightarrow f^{\prime }\left(x\right)=\frac{1}{2}f\left(x\right)$
$\Rightarrow \frac{f^{\prime }\left(x\right)}{f\left(x\right)}=\frac{1}{2}$
$\Rightarrow \int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx=\int \frac{1}{2}dx$
$\Rightarrow {\log }_{e}f\left(x\right)=\frac{x}{2}+C$
substituting $x=0,$ we get $C=0(\because f(0)=1)$
$\therefore {\log }_{e}f\left(x\right)=\frac{x}{2}$
$\Rightarrow {\log }_{e}f\left(4\right)=\frac{4}{2}=2$