Answer
The Derivative is:
$y'=-\frac{19}{(3x-2)^2}$
Work Step by Step
$y=\frac{2x+5}{3x-2}$
Using the Quocient Rule:
$y'=\frac{f'(x)⋅g(x)-f(x)⋅g'(x)}{g^2(x)}$
$y'=\frac{((1)2x^{1-1}+0)⋅(3x-2)-(2x+5)⋅((1)3x^{1-1}-0)}{(3x-2)^{2}}$
$y'=\frac{2(3x-2)-3(2x+5)}{(3x-2)^2}$
$y'=\frac{6x-4-6x-15}{(3x-2)^2}$
$y'=-\frac{19}{(3x-2)^2}$
