Solving a functional equation ( $ f(x-y) = f(x)/f(y)$ ) 
Consider the functional equation $$f(x-y)=f(x)/f(y)$$
  If $f'(0)= p$ and $f'(5)=q$, then what is the value of $f'(-5)$ ?

My attempt. Using the equation written above I was able to determine the value $f(0)$, which is $1$. Now to find out the value of $f'(-5)$ I need to find the value of $f(5)$: then I believe I should use the chain rule of differentiation in order to get my final answer. Any hints on how to go about that?
 A: When $x=y=0$ we have $f(0)=f(0)/f(0)$, which implies $f(0)=1$. 
Let $x-y=a$: then $f(a)f(y)=f(a+y)$.
For fixed $a$ we have $f(a)f'(y)=f'(a+y)$.
When $y=0$, we get   $f(a)f'(0)=f'(a+0)$, or $pf(a)=f'(a)$, or $$f(a)=\frac{f'(a)}{p} \tag{1}$$
When $y=5$, we get  $f(a)f'(5)=f'(a+5)$, or $f(a) q=f'(a+5)$, or $$f(a)=\frac{f'(a+5)}{q} \tag{2}$$
From (1) and (2) we have $\frac{f'(a)}{p} = \frac{f'(a+5)}{q}$. 
Let $a=-5$. Then $$\frac{f'(-5)}{p} = \frac{f'(-5+5)}{q}, \text{ i.e. }\quad {f'(-5)}=\frac{p}{q}\cdot f'(0)=\frac{p^2}{q}.$$
A: You already know $f(0)=1.$ Also, $f(-y)=f(0-y)=1/f(y)$ is the inverse of $f(y).$ Hence, $f(x+y)=f(x-(-y))=f(x)f(y).$ This shows that $f$ is a homomorphism from $(\mathbb R,+,0)$ to $(\mathbb R^*,\times,1).$ Then we shall show that $f(q)=f(1)^q,\forall q\in\mathbb Q.$  

First, for $q=0,$ we already know that $f(0)=1=f(1)^0.$
Then, for any $\alpha\in\mathbb N_{\gt0},$ we have $f(\alpha x)=f(x+\cdots+x)=f(x)^\alpha.$
Hence, $\forall\alpha,\beta\in\mathbb N_{\gt0},$ substituing $1/\alpha$ for $x$ and raising the equation to the power $\beta/\alpha,$ we find $$f(1)^{\beta/\alpha}=f(1/\alpha)^\beta=f(\beta/\alpha).$$
Moreover, we have shown that $f(-y)=1/f(y),$ thus $f(-\beta/\alpha)=1/f(\beta/\alpha)=f(1)^{-\beta/\alpha}.$ This shows that $f(q)=f(1)^q,\forall q\in\mathbb Q.$
Since it is implicitly assumed that $f$ is differentiable, we may calculate the derivative $f'$ using sequences in $\mathbb Q.$ So $\frac{df(x)}{dx}\mid_{y}=\lim\limits_{\substack{q\rightarrow y\\q\in\mathbb Q}}(f(1)^q-f(1)^y)/(q-y)=f(1)^y\cdot\ln(f(1)),$ for $y\in\mathbb Q,$ by elementary calculus. Thus the hypothesis shows that $$\begin{align}p&=f'(0)=f(1)^0\ln(f(1))=\ln(f(1))\\q&=f'(5)=f(1)^5\ln(f(1))\end{align}$$
Hence $f(1)^5=q/p,$ and
$$f'(-5)=f(1)^{-5}\ln(f(1))=(p/q)\cdot p=p^2/q.$$

Hope this helps.
A: For sure, $f(0) = 1$. Indeed $$f(x-x) = \frac{f(x)}{f(x)} = 1.$$
Moreover:
$$f(-x) = f(0-x) = \frac{f(0)}{f(x)} = \frac{1}{f(x)} \Rightarrow f(x)f(-x) = 1.$$
Using logarithms, we get:
$$\log(f(x)f(-x)) = \log(f(x)) + \log(f(-x)) = \log(1) = 0.$$
This suggests that $f(x) = g(x)^{x}$, where $g(x)$ is a even function (i.e. $g(-x) = g(x)$).
Indeed:
$$\log(f(x)) + \log(f(-x)) = \log(g(x)^x) + \log(g(-x)^{-x}) = \\x\log(g(x)) - x\log(g(-x)) = x\log(g(x)) - x\log(g(x)) = 0.$$
Let's try if this work!
$$f(x-y) = g(x-y)^{x-y} = \frac{g(x-y)^x}{g(x-y)^y}.$$
We would like that this is equal to $\frac{f(x)}{f(y)}$, and then $g(x)$ must be a constant $g(x) = a$. Finally:
$$f(x-y) = a^{x-y} = \frac{a^x}{a^y} = \frac{f(x)}{f(y)}.$$
Let's find $f'(0)$:
$$f'(0) = \left. \log(a) a^x \right|_{x = 0} = \log(a) = p \Rightarrow a = e^{p} \Rightarrow f(x) = e^{px}.$$
The derivative of $f(x)$ is $$f'(x) = pe^{px}$$
and then:
$$f'(5) = pe^{5p} = q,$$
while $$f'(-5) = pe^{-5p} = pe^{5p-10p} = pe^{5p}e^{-10p} = qe^{-10p}$$
Alternatively:
$$f'(-5) = pe^{-5p} = \frac{p}{e^{5p}} = \frac{p^2}{pe^{5p}} = \frac{p^2}{q}.$$
A: Implicit in your identity $f(x-y)=f(x)/f(y)$ is $f(y)\ne 0$.
Letting $u=x-y$, $v=y$ gives the standard exponential property
$$
                 f(u)f(v)=f(u+v). \tag{*}
$$
And $f(0)f(0)=f(0)$ implies $f(0)=1$.
Differentiating with respect to $u$ and then setting $u=0$:
$$
                         f'(u)f(v)=f'(u+v) \\
                         f'(0)f(v)=f'(v) \\
                         p=f'(0)=\frac{f'(v)}{f(v)}
$$
Because this holds for all $v$,
$$
      f'(-5)=\frac{f'(-5)}{f(-5)}\frac{1}{f(5)}=\frac{f'(-5)}{f(-5)}\frac{f'(5)}{f(5)}\frac{1}{f'(5)}=p^2\frac{1}{f'(5)}=\frac{p^2}{q}.
$$
