Question

Помогите решить.Математика 1 курс

Answer 1

Ответ:

1.

$f(x) = \frac{ {x}^{2} - 2 }{ {x}^{2} + 2}$

$f'(x) = \frac{( {x}^{2} - 2)'( {x}^{2} + 2) - ( {x}^{2} + 2)'( {x}^{2} - 2)}{ {( {x}^{2} + 2)}^{2} } = \\ = \frac{2x( {x}^{2} + 2) - 2x( {x}^{2} - 2) }{ {( {x}^{2} + 2) }^{2} } = \frac{2x( {x}^{2} + 2 - {x}^{2} + 2) }{ {( {x}^{2} + 2)}^{2} } = \\ = \frac{8x}{ {( {x}^{2} + 2)}^{2} }$

$f'(1) = \frac{8}{ {3}^{2} } = \frac{8}{9}$

2.

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

$f'(x) = ( {x}^{2} - x) '\times \cos {}^{2} (x) + ( \cos {}^{2} (x)) ' \times ( {x}^{2} - x) = \\ =( 2x - 1) \cos {}^{2} (x) + 2 \cos(x) \times ( - \sin(x) ) \times ( {x}^{2} - x) = \\ = (2x - 1) \cos {}^{2} (x) - \sin(2x) \times ( {x}^{2} - x)$

$f'(0) = - 1 \times 1 - 0 = - 1$

3.

1.

$y = 2 {x}^{3} - {x}^{2} + \frac{1}{x {}^{4} }$

$y' = 6 {x}^{2} - 2x - 4 {x}^{ - 5} = 6 {x}^{2} - 2x - \frac{4}{ {x}^{5} }$

$y'(1) = 6 - 2 - 4 = 0$

$y'( - 2) = 6 \times 4 + 4 - \frac{4}{( - 32)} = 28 + \frac{1}{8} = 28 \frac{1}{8}$

2.

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

$y'( \frac{\pi}{6} ) = 2 \times \frac{ \sqrt{3} }{2} + \frac{1}{4} \times \frac{1}{2} = \sqrt{3} + \frac{1}{8}$

$y'( - \frac{\pi}{4} ) = 2 \times \frac{ \sqrt{2} }{2} + \frac{1}{4} \times ( - \frac{ \sqrt{2} }{2} ) = \\ = \sqrt{2} - \frac{ \sqrt{2} }{8} = \frac{7 \sqrt{2} }{8}$

3.

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

$y' = 9x - 3 {x}^{2} - 3 {x}^{ - 4} = 9x - 3 {x}^{2} - \frac{3}{ {x}^{4} }$

$y'(1) = 9 - 3 - 3 = 3$

$y'( - 2) = - 18 - 3 \times 4 - \frac{3}{16} = - 30 - \frac{3}{16} = - 30 \frac{3}{16}$

4.

$y = \frac{1}{2} \sin(x) + 3 \cos(x) \\ y '= \frac{1}{2} \cos(x) - 3 \sin(x)$

$y'( \frac{\pi}{6} ) = \frac{1}{2} \times \frac{ \sqrt{3} }{2} - 3 \times \frac{1}{2} = \frac{ \sqrt{3} }{4} - \frac{3}{2}$

$y'( - \frac{\pi}{4} ) = \frac{1}{2} \times \frac{ \sqrt{2} }{2} + \frac{3 \sqrt{2} }{2} = ( \frac{1}{4} + \frac{3}{2} ) \sqrt{2} = \frac{7 \sqrt{2} }{4}$

5.

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

$y' = 3 {x}^{2} - 6x + \frac{2}{ {x}^{3} }$

$y'(1) = 3 - 6 + 2 = - 1$

$y'( - 2) = 12 + 12 + \frac{2}{( - 8)} = 24 - 0.25 = 23.75$

6.

$y = 3 \sin(x) + \frac{1}{4} \cos(x) \\ y' = 3 \cos(x) - \frac{1}{4} \sin(x)$

$y'( \frac{\pi}{6} ) = \frac{3 \sqrt{3} }{2} - \frac{1}{8}$

$y'( - \frac{\pi}{4} ) = \frac{3 \sqrt{2} }{2} + \frac{ \sqrt{2} }{8} = \sqrt{2} ( \frac{3}{2} + \frac{1}{8} ) = \frac{13 \sqrt{2} }{8}$

4.

$f(x) = \sin(x) + {x}^{2} \\ f'(x) = \cos(x) + 2x$

5.

1.

$y' = {e}^{x} + x {e}^{x} = e {}^{x} (x + 1)$

2.

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

3.

$f'(x) = ln(7) \times {7}^{ - \cos(x) } \times \sin(x)$

4.

$f'(x) = \frac{1}{ \sin( \frac{x}{3} ) } \times \cos( \frac{x}{3} ) \times \frac{1}{3} = \frac{1}{3}ctg( \frac{x}{3} )$

5.

$y '= 12 {x}^{3} - 12 {x}^{2}$

6.

$y' = \frac{5}{x} - 2x$