Let `f: R->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 _______

Explanation

`f(x+y)=f(x)cdotf'(y)+f'(x)cdotf(y)" ...(i)"`
`"Putting "y=0`
`f(x)=f(x)f'(0)+f'(x)cdotf(0)=f(x)f'(0)+f'(x)" ...(2)"`
To find f'(0), in (1), put x=y=0.
`therefore" "f(0)=2f(0)cdotf'(0)`
`thereforef" "'(0)=(1)/(2)`
So, from (2). we get
`f'(x)=(f(x))/(2)`
`rArr" "int(f'(x))/(f(x))dx=int (1)/(2)dx`
`rArr" "log_(e)f(x)=(x//2)+c`
`rArr" "log_(e)f(x)=x//2" (as f(0) = 1)"`
`rArr" "log_(e)(f(4))=2`