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Dear experts, I have next initial expression: $$y=\dfrac{2x-3}{5-2x}$$ and I need find y'. In this case I use rule $$\left( \dfrac{u}{v} \right)' = \dfrac{u'v-uv'}{v^2}$$ and expect to obtain next result after simplification: $$y' = \dfrac{4}{(5-2x)^2}$$ But after diff((2x-3)/(5-2x),x) and simplification with factor() I get. $$y' = \dfrac{4}{(2x-5)^2}$$ How I can get $$(5-2x)^2$$ in denominator of derivative?