Find $f'(-5)$ if $f(x-y)=\frac{f(x)}{f(y)}$ and $f'(5)=q.$ By the given condition $$f(x-y)=\frac{f(x)}{f(y)},\>\>\>\> \forall x,y \in \Bbb{R}.$$ Also given that  $f'(5)=q$  and $f'(0)=p$.
Then prove that $f'(-5)=q.$
 But if I take $f(x)=e^x$, then $f$ satisfies every condition and $p=1$ and $q=e^5$. Then $f'(-5) = e^{-5} \neq q.$
So this question is wrong I think. I think $f'(-5)=\frac{p^2}{q}.$ 
Please help me to solve this.
 A: Your answer/intuition is correct
If we put $z=x-y$ we get $$f(z)f(y)=f(y+z)$$ so by induction we get $f(x) = a^x$ for some $a$ and for all rational $x$. Since $f'(0)$ exists we have $f(x) = a^x$ for all real $x$. Now $p = f'(0) = a^0\ln a=\ln a$ and $q= a^{5}p$ so $\boxed{a^5 =q/p}$ and thus $$f'(-5) = a^{-5}\ln a = (p/q)\cdot p = p^2/q$$
A: HInt: take $f(x)=e^{kx}$
so $$f(x-y)=e^{k(x-y)}=e^{kx}e^{-ky}= \frac{e^{kx}}{e^{ky}}=\frac{f(x)}{f(y)}$$
A: Consider any such $f$ satisfies the condition $:$ $f^{\prime}(5)=q,f^{\prime}(0)=p,$ and for all $x,y\in\mathbb{R}$ that 
\begin{align}
f(x-y)=\frac{f(x)}{f(y)}.
\end{align}
Moreover, the last condition implies that $f(x+y)=f(x)f(y)$ for all $x,y\in\mathbb{R}.$
Observe that 
\begin{align}
q:=\lim_{h\rightarrow 0}\frac{f(5+h)-f(5)}{h}=\lim_{h\rightarrow 0}\frac{f(5)f(h)-f(5)}{h}=f(5)\lim_{h\rightarrow 0}\frac{f(h)-1}{h}=f(5)f^{\prime}(0)=f(5)p,
\end{align}
where we used $f(0)=1$ and this can be deduced by the last assumption. Further, we have $\displaystyle f(-5)=f(0-5)=\frac{f(0)}{f(5)}=\frac{1}{f(5)}$ and hence that
\begin{align}
f^{\prime}(-5)&=\lim_{h\rightarrow 0}\frac{f(-5+h)-f(-5)}{h}\\
&=\lim_{h\rightarrow 0}\frac{f(-5)(f(h)-1)}{h}\\
&=f(-5)f^{\prime}(0)\\
&=\frac{1}{f(5)}\cdot f^{\prime}(0)\\
&=\frac{p}{q}\cdot p
\end{align}
