Решите производные..

$f'(x) =( \sqrt[4]{3x+2})'*(3x-1) ^{4}+ \sqrt[4]{3x+2}*[(3x-1) ^{4})' = [(3x+2) ^{ \frac{1}{4} }]'*$ $(3x-1) ^{4} + \sqrt[4]{3x+2}*[(3x-1) ^{4}]' = \frac{1}{4}(3x+2) ^{- \frac{3}{4} }*(3x+2)'*(3x-1) ^{4}$ $+ \sqrt[4]{3x+2}*4(3x-1) ^{3}*(3x-1)' = \frac{3}{4} (3x+2) ^{- \frac{3}{4} } *(3x-1) ^{4}+$ $12 \sqrt[4]{3x+2}*(3x-1) ^{3} = \frac{3(3x-1) ^{4} }{4 \sqrt[4]{(3x+2) ^{3} } }+12 \sqrt[4]{3x+2}*(3x-1) ^{3}$
2) $f'(x) = ( \sqrt[3]{2x-1})'*(2x-3) ^{3} + \sqrt[3]{2x-1}*[(2x-3) ^{3}]'= \frac{1}{3} (2x-1) ^{- \frac{2}{3} } *$ $(2x-1)'*(2x-3) ^{3} + \sqrt[3]{2x-1} *3(2x-3) ^{2}*(2x-3)' = \frac{2}{3}(2x-1) ^{- \frac{2}{3} }*$ $(2x-3) ^{3} +6 \sqrt[3]{2x+1}*(2x-3) ^{2}=$ $\frac{2(2x-3) ^{ ^{3} } }{3 \sqrt[3]{(2x-1) ^{2} } }+6 \sqrt[3]{2x-1}*(2x-3) ^{2}$