Решите пожалуйста производную сложной функции.

$y = \sqrt{(6-x)(5-2x)} = \sqrt{30-12x-5x+2 x^{2} } = \sqrt{2 x^{2} -17x+30}$
$y ' = ( \sqrt{2 x^{2} -17x+30} )'= \frac{1}{2 \sqrt{2 x^{2} -17x+30} }* (2 x^{2}-17x+30)' =$ $\frac{4x-17}{2 \sqrt{2 x^{2} -17x+30} } = \frac{2x-8,5}{ \sqrt{2 x^{2} -17x+30} }$
2) $y = \sqrt{ x^{3}-8 }$
$y' = ( \sqrt{ x^{3}-8 })' = \frac{1}{2 \sqrt{ x^{3}-8 } }*( x^{3} -8)' = \frac{3 x^{2} }{2 \sqrt{ x^{3}-8 } }$
3) $y = \sqrt[4]{ \frac{2}{2 x^{2} +1} }$
$y'= ( \sqrt[4]{ \frac{2}{2 x^{2} +1} })' =( \frac{2}{2 x^{2} +1} ) ^{ \frac{1}{4} }= \frac{1}{4} ( \frac{2}{2 x^{2} +1} ) ^{- \frac{3}{4} } *( \frac{2}{2 x^{2} +1} )'= \frac{1}{4} \sqrt[4]{ \frac{(2 x^{2} +1) ^{3} }{8} }*$ $2[(2 x^{2} +1) ^{-1}]'= \frac{1}{4} \sqrt[4]{ \frac{(2 x^{2} +1) ^{3} }{8} } * [-2(2 x^{2} +1) ^{-2}]= - \frac{ \sqrt[4]{ \frac{(2 x^{2} +1) ^{3} }{8} } }{(2 x^{2} +1) ^{2} }$