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(x) ne 0# we get

#[f(x)/g(x)]^'= [f^'(x)g(x)-f(x)g^'(x)]/[g(x)]^2#

In this case we have

#f(x)=3x-2#

#g(x)=2x-5#

In order to get the derivative of their quotient, we need to compute their derivatives:

#f^'(x)=3#

#g^'(x)=2#

So, we substitute #f(x),g(x),f'(x),g'(x)# in the quotient rule's formula and get

#[(3x-2)/(2x-5)]^'=[3(2x-5)-(3x-2)2]/[2x-5]^2=(6x-15-6x+4)/(2x-5)^2=-11/(2x-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(x):=f(x)/g(x)#

Our aim is to compute #h^'(x)#.

First of all we can rewrite the quotient in the following manner

#f(x)=h(x)g(x)#

By differentiating both sides (product rule on the right side) we get that

#f^'(x)=h^'(x)g(x)+h(x)g^'(x)#

We now isolate #h^'(x)# on the left side

#h^'(x)=[f^'(x)-h(x)g^'(x)]/g(x)#

and substitute #h(x)# using its definition

#h^'(x)=[f^'(x)-[f(x)/g(x)]g^'(x)]/g(x)#

We finally rewrite the last equation to get the rule's formula

#h^'(x)=[f^'(x)-(f(x)g^'(x))/g(x)]/g(x)=[(f^'(x)g(x)-f(x)g^'(x))/g(x)]/g(x)=(f^'(x)g(x)-f(x)g^'(x))/[g(x)]^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)!