Найти y'(1), если y=((x^3)+x):((x^2)-2* корень из x

[latex]y`=( \frac{ x^{3}+ x}{ x^{2} -2 \sqrt{x} } )`=[ (\frac{u}{v})`= \frac{u`\cdot v-u\cdot v`}{v ^{2} }]= \\ =\frac{ (x^{3}+ x)`\cdot (x^{2} -2 \sqrt{x})-(x^{3}+ x)\cdot (x^{2} -2 \sqrt{x})`}{ (x^{2} -2 \sqrt{x}) ^{2} }= \\ = \frac{ (3x^{2}+ 1)\cdot (x^{2} -2 \sqrt{x})-(x^{3}+ x)\cdot (2x -2\cdot \frac{1}{2 \sqrt{x} } )}{ (x^{2} -2 \sqrt{x}) ^{2} } [/latex] [latex]y`(1)= \frac{ (3\cdot 1 ^{2} + 1)\cdot (1^{2} -2 \sqrt{1})-(1^{3}+ 1)\cdot (2\cdot 1 - \frac{1}{ \sqrt{1} } )}{ (1^{2} -2 \sqrt{1}) ^{2} } = \frac{-4-2}{1}=-6 [/latex]