Question

похідна/производная найти)

Answer 1

$1)y' = 6 {x}^{5}$

$2)y' = - 2 {x}^{ - 3} = - \frac{2}{ {x}^{3} } \\$

$3)y' = 6 {x}^{2}$

$4)y' = - ( - 12) {x}^{ - 13} = \frac{12}{ {x}^{13} } \\$

$5)y' = \frac{1}{2} {x}^{ - \frac{1}{2} } = \frac{1}{2 \sqrt{x} } \\$

$6)y' = ( {x}^{ - 5} )' = - 5 {x}^{ - 6} = - \frac{5}{ {x}^{6} } \\$

$7)y' = ( {x}^{ - \frac{1}{3} } ) '= - \frac{1}{3} {x}^{ - \frac{4}{3} } = - \frac{1}{3x \sqrt[3]{x} } \\$

$8)y' = \frac{1}{3} {x}^{ - \frac{2}{3} } = \frac{1} { 3\sqrt[3]{ {x}^{2} } } \\$

$9)y' = ( {x}^{ \frac{5}{6} } )' = \frac{5}{6} {x}^{ - \frac{1}{6} } = \frac{5}{6 \sqrt[6]{x} } \\$

$10)y' = ( {x}^{ - 9} ) '= - 9 {x}^{ - 10} = - \frac{9}{ {x}^{10} } \\$

$11)y' = ( {x}^{ - \frac{3}{4} } ) '= - \frac{3}{4} {x}^ { - \frac{7}{4} } = - \frac{3}{4 x \sqrt[4]{ {x}^{3} } } \\$

$12)y' = - \frac{2}{7} {x}^{ - \frac{9}{7} } = - \frac{2}{7x \sqrt[7]{ {x}^{2} } } \\$

$13)y' = ( \frac{2 {x}^{3} }{ \sqrt[3]{x} } )' = (2 {x}^{3 - \frac{1}{3} } ) '= (2 {x}^{ \frac{8}{3} } ) '= \\ = 2 \times \frac{8}{3} {x}^{ \frac{5}{3} } = \frac{16}{3} x \sqrt[3]{ {x}^{2} }$

$14)y' = (4 {x}^{ - \frac{3}{2} } )' = - 4 \times \frac{3}{2} {x}^{ - \frac{5}{2} } = \\ = - \frac{6}{ {x}^{2} \sqrt{x} }$