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(A) Given the functional equation $f(xy) = f(x) + f(y)$.Setting $x=1, y=1$,we get $f(1) = f(1) + f(1)$,which implies $f(1) = 0$.By the definition of t

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$\frac{f'(1)}{x}$

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(A) Given the functional equation $f(xy) = f(x) + f(y)$.Setting $x=1, y=1$,we get $f(1) = f(1) + f(1)$,which implies $f(1) = 0$.By the definition of the derivative:$f'(x) = \lim_{h 	o 0} \frac{f(x+h) - f(x)}{h}$Using the property $f(xy) = f(x) + f(y)$,we can write $f(x+h) = f(x(1 + \frac{h}{x})) = f(x) + f(1 + \frac{h}{x})$.Substituting this into the limit:$f'(x) = \lim_{h 	o 0} \frac{f(x) + f(1 + \frac{h}{x}) - f(x)}{h}$$f'(x) = \lim_{h 	o 0} \frac{f(1 + \frac{h}{x})}{h}$Since $f(1) = 0$,we can write $f(1 + \frac{h}{x}) = f(1 + \frac{h}{x}) - f(1)$.Let $t = \frac{h}{x}$. As $h 	o 0$,$t 	o 0$.$f'(x) = \lim_{t 	o 0} \frac{f(1+t) - f(1)}{xt} = \frac{1}{x} \lim_{t 	o 0} \frac{f(1+t) - f(1)}{t}$By the definition of the derivative at $x=1$,$\lim_{t 	o 0} \frac{f(1+t) - f(1)}{t} = f'(1)$.Therefore,$f'(x) = \frac{f'(1)}{x}$.

Let $f(x)$ be a differentiable function which satisfies the equation $f(xy) = f(x) + f(y)$ for all $x > 0, y > 0$. Then $f'(x)$ is equal to:

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