\(\begin{array}{l}a)\,\,y = \dfrac{{x - 1}}{{5x - 2}}\\ \Rightarrow y' = \dfrac{{\left( {5x - 2} \right) - 5\left( {x - 1} \right)}}{{{{\left( {5x - 2} \right)}^2}}}\\\,\,\,\,\,\,y' = \dfrac{{5x - 2 - 5x + 5}}{{{{\left( {5x - 2} \right)}^2}}}\\\,\,\,\,\,\,y' = \dfrac{3}{{{{\left( {5x - 2} \right)}^2}}}\\b)\,\,y = \dfrac{{2x + 3}}{{7 - 3x}}\\ \Rightarrow y' = \dfrac{{2\left( {7 - 3x} \right) + 3\left( {2x + 3} \right)}}{{{{\left( {7 - 3x} \right)}^2}}}\\\,\,\,\,\,\,y' = \dfrac{{14 - 6x + 6x + 9}}{{{{\left( {7 - 3x} \right)}^2}}}\\\,\,\,\,\,\,y' = \dfrac{{23}}{{{{\left( {7 - 3x} \right)}^2}}}\\c)y = \dfrac{{{x^2} + 2x + 3}}{{3 - 4x}}\\y' = \dfrac{{\left( {2x + 2} \right)\left( {3 - 4x} \right) + 4\left( {{x^2} + 2x + 3} \right)}}{{{{\left( {3 - 4x} \right)}^2}}}\\= \dfrac{{6x - 8{x^2} + 6 - 8x + 4{x^2} + 8x + 12}}{{{{\left( {3 - 4x} \right)}^2}}}\\= \dfrac{{ - 4{x^2} + 6x + 18}}{{{{\left( {3 - 4x} \right)}^2}}}\end{array}\)
\(\begin{array}{l}d)\,\,y = \dfrac{{{x^2} + 7x + 3}}{{{x^2} - 3x}}\\\Rightarrow y' = \dfrac{{\left( {2x + 7} \right)\left( {{x^2} - 3x} \right) - \left( {{x^2} + 7x + 3} \right)\left( {2x - 3} \right)}}{{{{\left( {{x^2} - 3x} \right)}^2}}}\\= \dfrac{{2{x^3} - 6{x^2} + 7{x^2} - 21x - 2{x^3} - 14{x^2} - 6x + 3{x^2} + 21x + 9}}{{{{\left( {{x^2} - 3x} \right)}^2}}}\\= \dfrac{{ - 10{x^2} - 6x + 9}}{{{{\left( {{x^2} - 3x} \right)}^2}}}\end{array}\)
