Question

Помогите пожалуйста!!!!!

Answer 1

Ответ:

1.

$y = 5 {x}^{4} - 3.5 {x}^{2} + x + 6$

$y' = 5 \times 4 {x}^{3} - 3.5 \times 2x + 1 = \\ = 20 {x}^{3} - 7x + 1$

2.

$y = ( \frac{8}{x} + {x}^{2} ) \sqrt{x} = \frac{8}{ \sqrt{x} } + {x}^{2} \sqrt{x} = \\ = 8 {x}^{ - \frac{1}{2} } + {x}^{ \frac{5}{2} }$

$y '= 8 \times ( - \frac{1}{2} ) {x}^{ - \frac{3}{2} } + \frac{5}{2} {x}^{ \frac{3}{2} } = \\ = - \frac{4}{x \sqrt{x} } + 2.5x \sqrt{x}$

3.

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

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

4.

$f(x) = \sqrt[4]{1 + {x}^{2} } = {(1 + {x}^{2}) }^{ \frac{1}{4} }$

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

5.

$f(x) = {5}^{2x}$

$f'(x) = ln(5) \times {5}^{2x} \times (2x)' = \\ = ln(5) \times {5}^{2x} \times 2$

6.

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

$f'(x) = \frac{( ln(x)) '( {e}^{x} + {e}^{ - x} ) -( {e}^{x} + {e}^{ - x} ) '( ln(x)) }{ {( {e}^{x} + {e}^{x}) }^{2} } = \\ = \frac{ \frac{1}{x} ( {e}^{x} + {e}^{ - x}) - ( {e}^{x} - {e}^{ - x}) ln(x) }{ {( {e}^{x} + {e}^{ - x} )}^{2} } = \\ = \frac{1}{x( {e}^{x} + {e}^{ - x} )} - \frac{ ( {e}^{x} - {e}^{ - x} ) ln(x) }{ {( {e}^{x} + {e}^{ - x} )}^{ 2} }$