Functional equation $f(xy) = f'(x)f(y)$, am I doing it right?

Question

So I want to find all functions $f : \mathbb R \to \mathbb R$ so that:

$$f(xy) = f'(x)f(y), \forall x,y \in \mathbb R$$

My try: $y=1$ gives: $f(x) = f'(x)f(1)$ or $f'(x) = \frac{f(x)}{f(1)}$

$$f(x) = e^{\frac{x}{f(1)}}$$

where $f(1)$ must fulfil $$f(1) = e^{\frac 1 {f(1)}}$$

We can conclude $f(1)>0$ and since $\cases{1/x & \text{is monotonic decaying}\\ e^x &\text{is monotonic increasing}}$

Therefore there must exist one and only such $f(1) \in \mathbb R$.

Does this make sense?

Answer 1

From $f(x)=f'(x)f(1)$, consider $f(1)=0$ before you start dividing by it. If $f(1)=0$, then $f(x)=0$. Assume it's not: then indeed, we have (for $c=\tfrac1{f(1)}$)

$$f'(x)=cf(x)$$

which is a widely known differential equation with the general solution

$$f(x)=Ke^{cx}$$

Now to find $c=\tfrac1{f(1)}$, simply substitute $f(x)=Ke^{cx}$ in the original equation to see

$$Ke^{cxy} = cKe^{cx}\cdot Ke^{cy}$$

assuming $K\neq 0$ (which would be $f(x)=0$), and setting $y=0$, we see $1 = cKe^{cx}$ for all $x$: this is impossible. Thus, $f(x)=0$ is the only option.

Answer 2

Choosing $x=0$ you get $f(0) = f'(0) f(y)$ for every $y\in\mathbb{R}$.

If $f'(0)\neq 0$ then $f$ is constant. As a consequence, $f'(x) = 0$ for every $x\in \mathbb{R}$, so that choosing $y=1$ in the relation you get $f(x) = 0$ for every $x\in\mathbb{R}$.

If $f'(0) = 0$, then by the above relation also $f(0) = 0$. Choosing $y=1$ you get $f(x) = f'(x) f(1)$. If $f(1) = 0$ then $f(x) = 0$ for every $x$. Otherwise, if $f(1) \neq 0$, then you get $f(x) = c e^{x/f(1)}$, but $c=0$ since $f(0) = 0$.