$y=\dfrac{ \sqrt{x^2-3x}}{x+1}$
$y'=\dfrac{ (\sqrt{x^2-3x})'(x+1)- \sqrt{x^2-3x}(x+1)' }{(x+1)^2}$
$=\dfrac{ \dfrac{ 2x-3}{2\sqrt{x^2-3x}}(x+1) -\sqrt{x^2-3x} }{(x+1)^2}$
$=\dfrac{ (2x-3)(x+1)-2(x^2-3x) }{(x+1)^2.2(\sqrt{x^2-3x})}$
$=\dfrac{ 2x^2-x-3-2x^2+6x}{2(x+1)^2\sqrt{x^2-3x}}$
$=\dfrac{5x-3}{2(x+1)^2\sqrt{x^2-3x}}$
$y'(4)=\dfrac{ 5.4-3}{2(4+1)^2\sqrt{16-3.4}}=\dfrac{17}{100}$
$y'(5)=\dfrac{5.5-3}{2(5+1)^2\sqrt{25-3.5}}=\dfrac{11}{36\sqrt{10}}$
