If $f(\frac{x}{y})=\frac{f(x)}{f(y)} \, , f(y),y \neq 0$ and $f'(1)=2$ then $f(x)=$? If $f(\frac{x}{y})=\frac{f(x)}{f(y)} \, , f(y),y \neq 0$ and $f'(1)=2$ then $f(x)=$?
I am not sure where to begin, any hints on starting and steps is apreciated.
Thank you
 A: $f(x/y)=f(x)/f(y),\forall x,y$


*

*if you take $x=y$ you obtain $f(1)=1$. As a consequence $f(1/y)=1/f(y)$.

*$x=0$ and $f$ not constant implies $f(0)=0$.

*if $a,b \neq 0$ then $f(ab)=f(a/(1/b))=f(a)/(f(1/b))=f(a)f(b)$. If for an element $x \neq 0$ we have $f(x)=0$ then $f$ would be constant, which contradicts $f'(1)=2$. 

*In particular you have $f(x^2)=(f(x))^2>0$.

*Define for $x>0$ $g(x)=\log(f(e^x))$. Then $$g(x+y)=\log(f(e^xe^y))=\log(f(e^x))+\log(f(e^y))=g(x)+g(y)$$

*This means that $g$ satisfies a Cauchy functional equation and it is continuous (since you assume $f$ to be differentiable at $1$). This means that there exists $a$ such that $g(x)=ax$.

*This leads to: $\log(f(e^x))=ax \Rightarrow f(e^x)=(e^{x})^a \Rightarrow f(x)=x^a$ for $x>0$.

*Now the condition $f'(1)=2$ implies that $a=2$ and $f(x)=x^2$ for $x>0$.

*Because $f(x^2)=f(x)^2$ even for negative $x$ you have $f(-x)=\pm f(x)$.
A: Let $f(x)=x^2$. Then it is one.  
For $f(x/y)=(x/y)^2=x^2/y^2=f(x)/f(y)$ and $f'(1)=2$. 

Edit:
  First we find that $f(1)=f(y/y)=f(y)/f(y)=1$. So, $$\begin{align}f(y^{-1}) &=f(1/y)=f(1)/f(y)=f(y)^{-1}\end{align}$$ and $$\begin{align} f(xy) &=f(x/y^{-1})=f(x)f(y)  \,\,,\text{when}\,\,\,y\neq 0.\end{align}$$  That is, when restricted to $\mathbb R^*$, it is an endomomorphism of $\mathbb R^*$. Of course linear functions are all such endomorphisms. Notice that the word "linear" here means linear in the multiplicative case: $x\cdot x\cdot \cdot\cdot x=x^k$. Together with the derivative condition, ths gives us the answer.  

Thanks.
A: Hint For $x=y$ you get $f(1)=1$. What can you figure out from that and $f'(1)=2$?
A: Linear functions satisfied this function-equation trivially.
Let $f(x)=2x$ then $f'(x)=2$ for all $x$ and $f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$.
That's all :-)
R
