Equal change of difference and of ratios of two linear functions Find all linear functions $f$, $g$, such that in their intersection point:
$\frac{d}{dx}\left(f(x)-g(x)\right) = \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)$
My ideas:
I started by writing $f(x)=ax+b$, $g(x)=cx+d$. Then $\frac{d}{dx}\left(f(x)-g(x)\right)=a-c$. Similarly one gets $\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=\left(b-\frac{ad}{c}\right)\left(\frac{c}{(cx+d)^2}\right)$, then substitute $x=\frac{b-d}{c-a}$. After that it just got messy.
 A: See my edit for a general solution.
We exclude the trivial case $a=c$; hence the lines will meet for $x=(b-d)/(a-c)$.
You've got the derivative wrong, using a CAS something like
factor(diff((a*x+b)/(c*x+d),x))

gives the quotient's derivative as
$$\frac{ad-bc}{(cx+d)^2}.$$
Now if $ad-bc=0$ (the right sides of $f$ and $g$ are multiples), there will be no solution to the problem  as $a-c\neq0$.
Substituting $x=(b-d)/(c-a)$ and subtracting $a-c$, that is
factor(subst(diff((a*x+b)/(c*x+d),x),x=(b-d)/(c-a))-(a-c))

yields
$$\frac{(a-c)(bc-c-ad+a)}{ad-bc}.$$
We want this fraction to be zero.  Excluding the trivial case $a=c$, we conclude 
$$ad-bc=a-c,$$
a fairly (unexpected) simple result.
And it gets even more simple.  Compute the second coordinate of the intersection to
$$y=a\frac{b-d}{c-a}+b=\frac{ad-bc}{a-c}.$$
Now the result is as simple as it could be: 

The derivatives of $f-g$ and $f/g$ are equal in the point of their intersection, if this point's second coordinate equals $1$.

BTW: Same result for $f+g$ and $f\cdot g$ (if  $-a-c\neq0$).
EDIT: Actually we can generalize the result.  Suppose for some $u$ we have $v=f(u)=g(u)$.  Then
$$\begin{align}
\bigl(f(u)+g(u)\bigr)'&=\bigl(f(u)\cdot g(u)\bigr)'\\
\iff f'(u)+g'(u)&=f'(u)\cdot g(u)+f(u)\cdot g'(u)\\
\iff f'(u)+g'(u)&=v\cdot\bigl(f'(u)+g'(u)\bigr)\\
\iff v&=1\quad\text{or}\quad f'(u)+g'(u)=0
\end{align}
$$
Similar result holds for the pair $f-g$ and $f/g$.
