Question

помогите пожалуйста

1. Найдите производную произведения функций
y = √xctgx;
y = (x^2+x)(2x+x^3);
y = √xcox.
2.Найдите производную сложной функции
f(x) = √2x^2-1;
f(x) = sin5x;
f(x) = (-2x^3-9)^6.
3.Найдите производную частного двух функций
y = (x^2/2x+1);
y = (1/x+2/√x);
y = (2x/sinx).

Answer 1

Ответ:

1.

$y = \sqrt{x} ctgx$

$y' = ( \sqrt{x} ) '\times ctgx + (ctgx)' \times \sqrt{x} = \\ = \frac{1}{2} {x}^ { - \frac{1}{2} } \times ctgx - \frac{1}{ { \sin }^{2}(x) } \sqrt{x} = \\ = \frac{ctgx}{2 \sqrt{x} } - \frac{ \sqrt{x} }{ { \sin }^{2}(x) }$

$y = ( {x}^{2} + x)(2x + {x}^{3} )$

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

$y = \sqrt{x} \cos(x)$

$y '= ( \sqrt{x} )' \cos(x) + ( \cos(x)) ' \times \sqrt{x} = \\ = \frac{ \cos(x) }{2 \sqrt{x} } - \sqrt{x} \sin(x)$

2.

$f(x) = \sqrt{2 {x}^{2} - 1 }$

$f'(x) = \frac{1}{2} {(2 {x}^{2} - 1) }^{ - \frac{1}{2} } \times (2 {x}^{2} - 1) '= \\ = \frac{4x}{2 \sqrt{2 {x}^{2} - 1} } = \frac{2x}{ \sqrt{2 {x}^{2} - 1} }$

$f(x) = \sin(5x ) \\ f'(x) = \cos(5x) \times (5x)' = 5 \cos(5x)$

$f(x) = {( - 2 {x}^{3} - 9) }^{ 6}$

$f'(x) = 6 {( - 2 {x}^{3} - 9) }^{5} \times ( - 2 {x}^{3} - 9) '= \\ = 6 \times ( - 6 {x}^{2} ) {( - 2 {x}^{3} - 9) }^{5} = \\ = - 36 {x}^{2} { (- 2 {x}^{3} - 9) }^{5}$

3.

$y = \frac{ {x}^{2} }{2x + 1}$

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

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

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

$y = \frac{2x}{ \sin(x) }$

$y' = \frac{2 \sin(x) - 2x \cos(x) }{ { \sin }^{2}(x) }$