Given:
[tex]y=\frac{1}{2-x}[/tex]
To Determine: Using the increament method the rate of change of y with respect to x
[tex]\begin{gathered} y+\Delta y=\frac{1}{2-(x+\Delta x)} \\ \Delta y=\frac{1}{2-(x+\Delta x)}-y \end{gathered}[/tex]
Substitute for y
[tex]\begin{gathered} \Delta y=\frac{1}{2-(x+\Delta x)}-\frac{1}{2-x} \\ \Delta y=\frac{2-x-(2-(x+\Delta x)}{(2-(x+\Delta x)(2-x)} \\ \Delta y=\frac{2-x-(2-x-\Delta x)}{(2-(x+\Delta x)(2-x)} \\ \Delta y=\frac{2-x-2+x+\Delta x}{(2-(x+\Delta x)(2-x)} \\ \Delta y=\frac{\Delta x}{(2-(x+\Delta x)(2-x)} \end{gathered}[/tex][tex]\begin{gathered} \text{Divide through by }\Delta x \\ \frac{\Delta y}{\Delta x}=\frac{\Delta x}{(2-(x+\Delta x)(2-x)}\times\frac{1}{\Delta x} \\ \frac{\Delta y}{\Delta x}=\frac{1}{(2-(x+\Delta x)(2-x)} \end{gathered}[/tex][tex]\frac{dy}{dx}=\frac{1}{(2-x)(2-x)}[/tex]
Hence, the rate of change of y with respect to x is
[tex]\frac{dy}{dx}=\frac{1}{(2-x)^2}[/tex]
