Answer
$\frac{dy}{dx}=\frac{3}{(4-x)^2}$
Work Step by Step
Given that $y=\frac{2x-5}{4-x}$
$\frac{dy}{dx}=\frac{d\frac{2x-5}{4-x}}{dx}$
$\frac{dy}{dx}=\frac{(4-x)\frac{d(2x-5))}{dx}-(2x-5)\frac{d(4-x)}{dx}}{(4-x)^2}$
$\frac{dy}{dx}=\frac{8-2x+2x-5}{(4-x)^2}=\frac{3}{(4-x)^2}$
$\frac{dy}{dx}=\frac{3}{(4-x)^2}$
