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Question:

Let f:R→R be a differentiable function with f(0)=1 and satisfying the equation f(x+y)=f(x)f'(y)+f'(x)f(y) for all x,y ∈ R. Then, value of loge(f(4)) is ________.

Solution:

P(x,y):f(x+y)=f(x)f'(y)+f'(x)f(y) ∀ x,y ∈ R
P(0,0):f(0)=f(0)f'(0)+f'(0)f(0) ⇒1=2f'(0)⇒f'(0)=1/2
P(x,0):f(x)=f(x).f'(0)+f'(x).f(0)⇒f(x)=1/2f(x)+f'(x)⇒f'(x)=1/2f(x)⇒f(x)=e^(x/2)⇒ln(f(4))=2.