Question

Знайти похідну функції:
а) y=3x^2+10x+7x^(-3/2) ;
б) y=(2x-3)∙√x ;
в) y=(x^2+7x)/(x-9) ;
г) y=(x^3-4x)^2 .

Answer 1

Ответ:

а

$y = 3 {x}^{2} + 10x + 7 {x}^{ - \frac{3}{2} }$

$y' = 3 \times 2x + 10 + 7 \times ( - \frac{3}{2} ) {x}^{ - \frac{5}{2} } = \\ = 6x + 10 - \frac{21}{2 {x}^{2} \sqrt{x} }$

б

$y = (2x - 3) \sqrt{x} = 2 x\sqrt{x} - 3 \sqrt{x} = \\ = 2 {x}^{ \frac{3}{2} } - 3 {x}^{ \frac{1}{2} }$

$y' = 2 \times \frac{3}{2} {x}^{ \frac{1}{2} } - 3 \times \frac{1}{2} {x}^{ - \frac{1}{2} } = \\ = 3 \sqrt{x} - \frac{3}{2 \sqrt{x} }$

в

$y = \frac{ {x}^{2} + 7x }{x - 9}$

$y '= \frac{( {x}^{2} + 7x)'(x - 9) - (x - 9)'( {x}^{2} + 7x) }{ {(x - 9)}^{2} } = \\ = \frac{(2x +7 )(x - 9) - ( {x}^{2} + 7x)}{(x - 9) {}^{2} } = \\ = \frac{2 {x}^{2} - 18x + 7x - 63 - {x}^{2} - 7x }{ {(x - 9)}^{2} } = \\ = \frac{ {x}^{2} - 18x - 63}{ {(x - 9)}^{2} }$

г

$y = {( {x}^{3} - 4x) }^{2}$

$y' = 2 {( {x}^{3} - 4x) }^{} \times ( {x}^{3} - 4x)' = \\ = 2( {x}^{3} - 4x) \times (3 {x}^{2} - 4) = \\ = (2 {x}^{3} - 4x)(3 {x}^{2} - 4) = \\ = 6 {x}^{5} - 8 {x}^{3} - 12 {x}^{3} + 16x = \\ = 6 {x}^{5} - 20 {x}^{3} + 16x$