Question

вычислить производную функции

Answer 1

$y = (2x \sqrt[3]{ {x}^{2} } + \frac{4}{x \sqrt[4]{ {x}^{2} } } )( {x}^{2} + \frac{1}{ \sqrt[3]{x} } ) = \\ = (2 {x}^{ \frac{5}{3} } + 4 {x}^{ - \frac{3}{2} } )( {x}^{2} + {x}^{ - \frac{1}{3} } )$

$y '= (2 {x}^{ \frac{5}{3} } + 4 {x}^{ - \frac{3}{2} } )'( {x}^{2} + \frac{1}{ \sqrt[3]{x} } ) + ( {x}^{2} + {x}^{ - \frac{1}{3} })'(2x \sqrt[3]{ {x}^{2} } + \frac{4}{x \sqrt{x} } ) = \\ = (2 \times \frac{5}{3} {x}^{ \frac{2}{3} } - 4 \times \frac{3}{2} {x}^{ - \frac{5}{2} } )( {x}^{2} + \frac{1}{ \sqrt[3]{x} } ) + (2x - \frac{1}{3} {x}^{ - \frac{4}{3} } )(2x \sqrt[3]{ {x}^{2} } + \frac{4}{x \sqrt{x} } ) = \\ = ( \frac{10}{3} {x}^{ \frac{2}{3} } - 6 {x}^{ - \frac{5}{2} } )( {x}^{2} + {x}^{ - \frac{1}{3} } ) + (2x - \frac{1}{3} {x}^{ - \frac{4}{3} } )(2 {x}^{ \frac{5}{3} } + 4 {x}^{ - \frac{3}{2} } ) = \\ = \frac{10}{3} {x}^{ \frac{2}{3} + 2} + \frac{10}{3} {x}^{ \frac{2}{3} - \frac{1}{3} } - 6 {x}^{ - \frac{5}{2} + 2 } - 6 {x}^{ - \frac{5}{2} - \frac{1}{3} } + 4 {x}^{1 + \frac{5}{3} } + 8 {x}^{1 - \frac{3}{2} } - \frac{2}{3} {x}^{ - \frac{4}{3} + \frac{5}{3} } - \frac{4}{3} {x}^{ - \frac{4}{3} - \frac{3}{2} } = \\ = \frac{10}{3} {x}^{ \frac{8}{3} } + \frac{10}{3} {x}^{ \frac{1}{3} } - 6 {x}^{ - \frac{1}{2} } - 6 {x}^{ - \frac{17}{6} } + 4 {x}^{ \frac{8}{3} } + 8 {x}^{ - \frac{1}{2} } - \frac{2}{3} {x}^{ \frac{1}{3} } - \frac{4}{3} {x}^{ - \frac{17}{6} } = \\ = \frac{10}{3} {x}^{ \frac{8}{3} } + \frac{8}{3} {x}^{ \frac{1}{3} } + 2 {x}^{ - \frac{1}{2} } - \frac{22}{3} {x}^{ - \frac{17}{6} } = \\ = \frac{10}{3} {x}^{2} \sqrt[3]{ {x}^{2} } + \frac{8}{3} \sqrt[3]{x} + \frac{2}{ \sqrt{x} } - \frac{22}{3 {x}^{2} \sqrt[6]{ {x}^{5} } }$

Answer 2

$y=\Big(2x\cdot \sqrt[3]{x^2}+\dfrac{4}{x\sqrt[4]{x^2}}\Big)\cdot \Big(x^2+\dfrac{1}{\sqrt[3]{x}}\Big)\\\\\\y=\Big(2x^{\frac{5}{3}}+4x^{-\frac{3}{2}}\Big)\cdot \Big(x^2+x^{-\frac{1}{3}}\Big)\\\\\\y'=\Big(\dfrac{10}{3}\, x^{\frac{2}{3}}-6\, x^{-\frac{5}{3}}\Big)\cdot \Big(x^2+x^{-\frac{1}{3}}\Big)+\Big(2x^{\frac{5}{3}}+4x^{-\frac{3}{2}}\Big)\cdot \Big(2x-\dfrac{1}{3}\, x^{-\frac{4}{3}}\Big)=$

$=\Big(\dfrac{10}{3}\, \sqrt[3]{x^2}-\dfrac{6}{\sqrt[3]{x^5}}\Big)\cdot \Big(x^2-\dfrac{1}{\sqrt[3]{x}}\Big)+\Big(2\sqrt[3]{x^5}+\dfrac{4}{\sqrt{x^3}}\Big)\cdot \Big(2x-\dfrac{1}{3\sqrt[3]{x^4}}\Big)$