TITLE: derivative of a quotient is not the quotient of the derivatives
QUESTION [2 upvotes]: Ive seen multiple sources say something like "A derivative of a quotient is not the quotient (in general) of the derivatives" and noticed they all said "In general" but never gave such example.
So Ive been searching the internet for any possible examples in which a derivative of a quotient is the quotient of the derivatives but without any luck. Could someone give me an example?
e.g. 
f(x) = ...
g(x) = ...
so that:
f'(x)/g'(x) = (f(x)/g(x))'

REPLY [4 votes]: Let's write $f(x)=e^{a(x)}$, $g(x)=e^{b(x)}$ to make the computations easier. We find that
$$ \left(\frac{f}{g}\right)'-\frac{f'}{g'} = \frac{e^{a-b}}{b'} (a'b'-a'-b'^2) $$
So for this to be zero, we need
$$ a' = \frac{b'^2}{b'-1} $$
or if you prefer,
$$ b' = \frac{1}{2}\left( a' \pm \sqrt{a'(a'-4)} \right) $$
It's easy enough to cook up solutions to one of these, although clearly some fairly heroic integration is required; one such pair is
$$ a(x) = 2(x+\sinh{x}), \qquad b(x) = x \pm e^{ \pm x}. $$
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