Question

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

Answer 1

Пошаговое объяснение:

1.

$y = {x}^{3} ( \sqrt[4]{x} + 1) = ( {x}^{3} \sqrt[4]{x} + {x}^{3} ) = \\ = {x}^{ \frac{13}{4} } + {x}^{3}$

$y' = \frac{13}{4} {x}^{ \frac{9}{4} } + 3 {x}^{2} = \frac{13}{4} {x}^{2} \sqrt[4]{x} + 3 {x}^{2}$

2.

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

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