How do you find the first and second derivatives of (3x-2)/(2x-5) using the quotient rule?

${\left[\frac{3 x - 2}{2 x - 5}\right]}^{'} = \frac{{\left[3 x - 2\right]}^{'} \left(2 x - 5\right) - \left(3 x - 2\right) {\left[2 x - 5\right]}^{'}}{2 x - 5} ^ 2 = - \frac{11}{2 x - 5} ^ 2$

Explanation:

First of all, let's recall the quotient rule.
Let f,g:]a,b[ to mathbb(R) be two functions and let $f$ and $g$ be differentiable in ]a,b[. Then forall x in ]a,b[ such that $g \left(x\right) \ne 0$ we get
${\left[f \frac{x}{g} \left(x\right)\right]}^{'} = \frac{{f}^{'} \left(x\right) g \left(x\right) - f \left(x\right) {g}^{'} \left(x\right)}{g \left(x\right)} ^ 2$

In this case we have
$f \left(x\right) = 3 x - 2$
$g \left(x\right) = 2 x - 5$

In order to get the derivative of their quotient, we need to compute their derivatives:
${f}^{'} \left(x\right) = 3$
${g}^{'} \left(x\right) = 2$

So, we substitute $f \left(x\right) , g \left(x\right) , f ' \left(x\right) , g ' \left(x\right)$ in the quotient rule's formula and get
${\left[\frac{3 x - 2}{2 x - 5}\right]}^{'} = \frac{3 \left(2 x - 5\right) - \left(3 x - 2\right) 2}{2 x - 5} ^ 2 = \frac{6 x - 15 - 6 x + 4}{2 x - 5} ^ 2 = - \frac{11}{2 x - 5} ^ 2$

Proof of the quotient rule:
Let $f , g$ be two functions that satisfy the hypotheses required buy the quotient rule - including that g(x) ne 0 forall x in ]a,b[ - and let $h$ be a real valued function of real variable, such that
$h \left(x\right) : = f \frac{x}{g} \left(x\right)$
Our aim is to compute ${h}^{'} \left(x\right)$.

First of all we can rewrite the quotient in the following manner
$f \left(x\right) = h \left(x\right) g \left(x\right)$
By differentiating both sides (product rule on the right side) we get that
${f}^{'} \left(x\right) = {h}^{'} \left(x\right) g \left(x\right) + h \left(x\right) {g}^{'} \left(x\right)$
We now isolate ${h}^{'} \left(x\right)$ on the left side
${h}^{'} \left(x\right) = \frac{{f}^{'} \left(x\right) - h \left(x\right) {g}^{'} \left(x\right)}{g} \left(x\right)$
and substitute $h \left(x\right)$ using its definition
${h}^{'} \left(x\right) = \frac{{f}^{'} \left(x\right) - \left[f \frac{x}{g} \left(x\right)\right] {g}^{'} \left(x\right)}{g} \left(x\right)$

We finally rewrite the last equation to get the rule's formula
${h}^{'} \left(x\right) = \frac{{f}^{'} \left(x\right) - \frac{f \left(x\right) {g}^{'} \left(x\right)}{g} \left(x\right)}{g} \left(x\right) = \frac{\frac{{f}^{'} \left(x\right) g \left(x\right) - f \left(x\right) {g}^{'} \left(x\right)}{g} \left(x\right)}{g} \left(x\right) = \frac{{f}^{'} \left(x\right) g \left(x\right) - f \left(x\right) {g}^{'} \left(x\right)}{g \left(x\right)} ^ 2$

NOTE:
Unlike the product rule, you can't invert the roles of $f$ and $g$ (even if $f$ is not null in ]a,b[) because the result would be different for every choice of $f$ and $g$ (real functions of real variable)!