Question

Решите пожалуйста производную сложной функции.

Answer 1

$y = sqrt{(6-x)(5-2x)} = sqrt{30-12x-5x+2 x^{2} } = sqrt{2 x^{2} -17x+30}$
$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} }$