Derivative of a fraction with respect to another I've found this derivative on a textbook
$\dfrac{d(c_{t+1}/c_t)}{d(\dfrac{\gamma}{c_t}/\dfrac{1-\gamma}{c_{t+1}})}=\dfrac{1-\gamma}{\gamma} \dfrac{d(c_{t+1}/c_t)}{d(c_{t+1}/c_t)}=\dfrac{1-\gamma}{\gamma}$
I would like to understand the first passage. Was $\dfrac{1-\gamma}{\gamma}$ just brought out of the $d()$ at the denominator?
 A: A more rigorous way to do this is as follows. Define a new variable: $$S\equiv\frac{\dfrac{\gamma}{c_t}}{\dfrac{1-\gamma}{c_{t+1}}}.$$ By simple rearrangement, you can compute that $$\frac{c_{t+1}}{c_t}=\frac{1-\gamma}{\gamma}\times S.$$ Hence, if you imagine $c_{t+1}/c_t$ as a function of $S$, you have that $$\frac{\mathrm{d}\left(\dfrac{c_{t+1}}{c_t}\right)}{\mathrm d S}=\frac{1-\gamma}{\gamma},$$ since $c_{t+1}/c_t$ is just a linear function of $S$ with coefficient $(1-\gamma)/\gamma$.
In your original question, “bringing out $(1-\gamma)/\gamma$ out of the denominator” is an informal operation, since what appears there is, in fact, a differential operator and, rigorously speaking, there is no “denominator;” it's just a conventional notation. However, the informal trick of “bringing out $(1-\gamma)/\gamma$ out of the denominator” does work in most cases, which serves as a justification for denoting the differential operation as though it were a fraction.
In general, you may want to use the implicit function theorem to take the derivative of one quantity with respect to another if the functional relationship between these two quantities is complicated.
A: You must refer to the definition of derivative (coefficient brings out from derivation) and notice that in the fraction you mentioned above (in fact a notation just!) the denominator is an independent object and nominator -usually- is a function of it.
finally Do derivative to obtain final result!
A: Thank you to the answerer. I've been thinking on it, I could have used also the inverse function theorem.
$\dfrac{d(c_{t+1}/c_t)}{d(\dfrac{\gamma}{c_t}/\dfrac{1-\gamma}{c_{t+1}})}\quad (1)$
is the inverse of
$\dfrac{d(\dfrac{\gamma}{c_t}/\dfrac{1-\gamma}{c_{t+1}})}{d(c_{t+1}/c_t)}= \dfrac{d (\dfrac{c_{t+1}}{c_t} \cdot \dfrac{\gamma}{1-\gamma})}{d(c_{t+1}/c_t)}=\dfrac{\gamma}{1-\gamma} \quad (2) $
So since the result of (1) is equal to the reciprocal of the result of (2), the derivative will be
$\dfrac{1-\gamma}{\gamma}$
