Let f(x) be a real differentiable function such that f(0) = 1 and f(x+y)=f(x)f'(y)+f(y)f'(x) for all x, y ∈ R.

Question

Let f(x) be a real differentiable function such that f(0) = 1 and \(f(x + y) = f(x) f^{\prime}(y)+f(y) f^{\prime}(x)\text{ for all} \ x, y \in R\). Then \(\sum_{n=1}^{100} \log _e f(n)=\)

Answer 1

Answer is "2525" 

\(f(x+y)=f(x) f^{\prime}(y)+f(y) f^{\prime}(x)\)      ...(i)

Put \(y=0 \ In \quad(i)\)  

\(f(x)=f(x) f^{\prime}(0)+f^{\prime}(x)\)      ...(ii)  

put \(x=y=0 \) in (i)  

\(\mathrm{f}^{\prime}(0)=\frac{1}{2}\) put in (ii)