Question

1. Чему равно значение производной функции в точке x0, если
2. Найдите производную функции :
Помогите пожалуйста, друзья.

Answer 1

36.7

1

$f(x)= \frac{8}{x} + 5x - 2$

$f'(x) = - 8 {x}^{ - 2} + 5 - 0 = - \frac{8}{ {x}^{2} } + 5$

$f'(2) = - \frac{8}{4} + 5 = - 2 + 5 = 3$

2.

$f(x)= \frac{2 - 3x}{x + 2}$

$f'(x) = \frac{(2 - 3x)'(x + 2) - (x + 2)'(2 - 3x)}{ {(x + 2)}^{2} } = \\ = \frac{ - 3(x + 2) - (2 - 3x)}{ {(x + 2)}^{2} } = \frac{ - 3x - 6 - 2 + 3x}{ {(x + 2)}^{2} } = \\ = \frac{ - 8}{ {(x + 2)}^{2} }$

$f'( - 3) = \frac{ - 8}{1} = - 8$

3.

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

$f'(x)= \frac{( {x}^{2} + 2)'(x - 2) - (x - 2)'( {x}^{2} + 2)}{ {(x - 2)}^{2} } - 2 \cos(x) = \\ = \frac{2x(x - 2) - ( {x}^{2} + 2) }{ {(x - 2)}^{2} } - 2 \cos(x) = \\ = \frac{2 {x}^{2} - 4 {x}^{} - {x}^{2} - 2}{ {(x - 2)}^{2} } - 2 \cos(x) = \\ = \frac{ {x}^{2} - 4x - 2 }{ {(x - 2)}^{2} } - 2 \cos(x)$

$f'(0) = \frac{ - 2}{4} - 2 \cos(0) = - 0.5 - 2 = - 2.5$

4.

$f(x) = (1 + 3x) \sqrt{x} = \sqrt{x} + 3x \sqrt{x} = \\ = {x}^{ \frac{1}{2} } + 3 {x}^{ \frac{3}{2} }$

$f'(x) = \frac{1}{2} {x}^{ - \frac{1}{2} } + 3 \times \frac{3}{2} {x}^{ \frac{1}{2} } = \\ = \frac{1}{2 \sqrt{x} } + \frac{9}{2} \sqrt{x}$

$f'(9) = \frac{1}{2 \times 3} + \frac{9}{2} \times 3 = \\ = \frac{1}{6} + \frac{27}{2} = \frac{1 + 81}{6} = \frac{82}{6} = \frac{41}{3}$

5.

$f(x) = 3 \sqrt[3]{x} - 10 \sqrt[5]{x} = 3 {x}^{ \frac{1}{3} } - 10 {x}^{ \frac{1}{5} }$

$f'(x) = 3 \times \frac{1}{3} {x}^{ - \frac{2}{3} } - 10 \times \frac{1}{5} {x}^{ - \frac{4}{5} } = \\ = \frac{1}{ \sqrt[3]{ {x}^{2} } } - \frac{2}{ \sqrt[5]{ {x}^{4} } }$

$f'(1) = 1 - 2 = - 1$

6.

$f(x) = x \sin(x)$

$f'(x) = \sin(x) + x \cos(x)$

$f'(0) = 0$

36.12

1

$y' = 5 {(2x + 3)}^{4} \times (2x + 3) '= 10 {(2x + 3)}^{4}$

2

$y' = 18 {( \frac{x}{3} - 6) }^{17} \times \frac{1}{3} = 6 {( \frac{x}{3} - 6)}^{17}$

5

$y '= 3 \times ( - \frac{1}{ \sin {}^{2} ( \frac{x}{5} ) } ) \times \frac{1}{5} = - \frac{3}{5 \sin {}^{2} ( \frac{x}{5} ) }$

6

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

9

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

10

$y' = - 6( \frac{ {x}^{2} }{2} + 4x - 1) {}^{ - 7} \times (x + 4) = \\ = - \frac{6x + 24}{ {( \frac{ {x}^{2} }{2} + 4x - 1)}^{7} }$