Question

Знайти похідну функції y=(3х⁴+8х)⁷; у = 1/(х²+х)⁴; у=2(х-1)⁶+4(3-х)⁵; у=√3x-14;√2x³+4x

Answer 1

1

$y = {(3 {x}^{4} + 8x) }^{7}$

$y' = 7 {(3 {x}^{4} + 8x) }^{6} \times (3 {x}^{4} + 8x)' = \\ = 7 {(3 {x}^{4} + 8x)}^{6} \times (12 {x}^{3} + 8)$

2

$y = \frac{1}{ {( {x}^{2} + x)}^{4} } = {( {x}^{2} + x)}^{ - 4} \\$

$y' = - 4 {( {x}^{2} + x)}^{ - 5} \times ( {x}^{2} + x)' = \\ = - 4 {( {x}^{2} + x)}^{ - 5} \times (2x + 1) = \\ = - \frac{4(2x + 1)}{ {( {x}^{2} + x) }^{2} }$

3

$y = 2 {(x - 1)}^{6} + 4 {(3 - x)}^{5}$

$y' = 2 \times 6 {(x - 1)}^{5} \times (x - 1)' + 4 \times 5 {(3 - x)}^{4} \times (3 - x)' = \\ = 12 {(x - 1)}^{5} \times 1 + 20 {(3 - x)}^{4} \times ( - 1) = \\ = 12 {(x - 1)}^{5} - 20 {(3 - x)}^{4}$

4

$y = \sqrt{ \frac{3x - 14}{2 {x}^{3} + 4x } } \\$

$y' = \frac{1}{2} {( \frac{3x - 14}{2 {x}^{3} + 4x} )}^{ - \frac{1}{2} } \times ( \frac{3x - 14}{2 {x}^{3} + 4x} )' \\ = \\ = \frac{1}{2} \times \sqrt{ \frac{2 {x}^{3} + 4x }{3x - 14 } } \times \frac{(3x - 14)'(2 {x}^{3} + 4x) - (2 {x}^{3} + 4x)'(3x - 14) }{ {(2 {x}^{3} + 4x) }^{2} } = \\ = \frac{1}{2} \times \sqrt{ \frac{2 {x}^{3} + 4x}{3x - 14} } \times \frac{3(2 {x}^{3} + 4x) - (6 {x}^{2} + 4)(3x - 14) }{ {(2 {x}^{3} + 4x) }^{2} } = \\ = \frac{ \sqrt{2 {x}^{3} + 4x} }{2 \sqrt{3x - 14} } \times \frac{6 {x}^{3} + 12x - 18 {x}^{3} + 68 {x}^{2} - 12x + 56 }{ {(2 {x}^{3} + 4x) }^{2} } = \\ = \frac{ - 12 {x}^{3} + 68 {x}^{2} + 56 }{2 \sqrt{3x - 14 } \times \sqrt{ {(2 {x}^{3} + 4x) }^{3} } } = - \frac{6 {x}^{3} - 34 {x}^{2} - 28 }{ \sqrt{(3 x - 14) {(2 {x}^{3} + 4x)}^{3} } }$

Не уверена, что четвертая функция выглядит так