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(B) Given the functional equation $f(xy) = f(x) + f(y)$.Setting $x = 1$ and $y = 1$,we get $f(1) = f(1) + f(1)$,which implies $f(1) = 0$.Now,different

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(B) Given the functional equation $f(xy) = f(x) + f(y)$.Setting $x = 1$ and $y = 1$,we get $f(1) = f(1) + f(1)$,which implies $f(1) = 0$.Now,differentiate the given equation with respect to $y$ keeping $x$ constant:$\frac{d}{dy} f(xy) = \frac{d}{dy} (f(x) + f(y))$$f'(xy) \cdot x = f'(y)$.Setting $y = 1$,we get $x \cdot f'(x) = f'(1)$.Let $f'(1) = k$,then $f'(x) = \frac{k}{x}$.Integrating both sides with respect to $x$,we get $f(x) = k \ln(x) + C$.Since $f(1) = 0$,we have $k \ln(1) + C = 0$,which implies $C = 0$.Thus,$f(x) = k \ln(x)$.Now,calculate $f(e) + f(1/e)$:$f(e) = k \ln(e) = k(1) = k$.$f(1/e) = k \ln(1/e) = k \ln(e^{-1}) = -k \ln(e) = -k$.Therefore,$f(e) + f(1/e) = k + (-k) = 0$.

If $f(x)$ is a differentiable function such that $f(xy) = f(x) + f(y)$ for all $x, y > 0$,then $f(e) + f(1/e) = ?$

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