Алгебра: решить производные функции

а)б)в)г)___________________________а)б)в)г)___________________________а)б)в)г)__________________а)б)в)г)...

решить производные функции

Ответ:

schvarzengold100

14.01.2021 12:53

а)

f'(x) = 1

б)

f'(x) = 2(3x + 1) + 3(2x - 3) = 6x + 2 + 6x - 9 = 12x - 7

в)

$f'(x) = \frac{5(1 - 3x) + 3(5x + 3)}{ {(1 + 3x)}^{2} } = \frac{5 - 15x + 15x + 9}{ {(1 - 3x)}^{2} } = \frac{14}{ {(1 - 3x)}^{2} }$

г)

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

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а)

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

б)

$f'(x) = \cos(2 {x}^{2} - 5) \times 4x$

в)

$f'(x) = - \sin( {x}^{3} - 3 ) \times 3 {x}^{2}$

г)

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

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а)

$f'(x) = 100 {(3 - 5x + {x}^{2}) }^{99} \times (2x - 5)$

б)

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

в)

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

г)

$f'(x) = (3 {x}^{ \frac{7}{3} } )' = 7 {x}^{ \frac{4}{3} } = 7x \sqrt[3]{x}$

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а)

$f'(x) = 1 + \frac{1}{x}$

б)

$f'(x) = ln(3) \times { 3 }^{2 {x}^{2} } \times 4x$

в)

$f'(x) = 2 ln(5) \times {5}^{x} + 3 {e}^{x}$

г)

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