Let f is differentiable and has the property f(xy) = f(x)+f(y). Show that f'(x) =a/x ,where a is a constant.

Question

Let f : R+ -> R such that f is differentiable and has the property f(xy) = f(x)+f(y).

Answer 1

Denote $g(x) = f(\exp(x))$ for $x \in \mathbb{R}$. Plugging $\exp(x)$ and $\exp(y)$ in the functional equation gives you $f(e^{x+y})=f(e^x)+f(e^y)$, so:

$$\forall x,y\in\mathbb{R},\ g(x+y)=g(x)+g(y)$$

You also know that $g$ is differentiable, so looking at standard results on the Cauchy functional equation, you find that $g(x) = a x$ for some constant $a$.

That gives you $f(x) = g(\ln(x)) = a \ln(x)$, so $f'(x) = \frac{a}{x}$.

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Alternative answer using that $f$ is differentiable (and not only continuous).

First note that $f(1\cdot 1)=f(1)+f(1)$, so $f(1)=0$. Then take $x>0$, and $h$ close to $0$. Then using the equation with $y = 1+\frac{h}{x}$ yields $f(x+h) = f(x) + f\big(1 + \frac{h}{x}\big)$, so subtracting $f(1)=0$ and dividing by $h$, we get: $$\frac{f(x+h)-f(x)}{h} = \frac{1}{x} \cdot \frac{f\big(1 + \frac{h}{x}\big) - f(1)}{\frac{h}{x}}$$

When $h$ goes to zero, the RHS goes to $\frac{1}{x}\cdot f'(1)$, and the LHS goes to $f'(x)$, so $f'(x)=\frac{f'(1)}{x}$. Your constant $a$ is $f'(1)$.

Answer 2

Define $g\colon \Bbb R\to\Bbb R$ as $g(x)=f(e^ x)$. Then we arrive at the Cauchy functional equation$$g(x+y)=f(e^{x+y})=f(e^xe^y)=f(e^x)+f(e^y)=g(x)+g(y).$$ Recall that the only continuous $g$ with ths property are of the form $g(x)=ax$. Then $$ a=g'(x)=e^xf'(e^x)$$ for all $x\in\Bbb R$.