Question

Помогите найти производные функций

Answer 1

Пошаговое объяснение:

а)

$f'(x) = - {(2 {x}^{2} - 4x + 7) }^{ - 2} \times (4x - 4) + \frac{5}{3} {(x - 1)}^{ \frac{2}{3} } = \\ = - \frac{4x - 4}{ {(2 {x}^{2} - 4 + 7)}^{2} } + \frac{5}{3} \sqrt[3]{ {(x - 1)}^{2} }$

б)

$f'(x) = ln(3) \times {3}^{ - {x}^{2} } \times ( - 2x)arccos(3 {x}^{4} ) + {3}^{ - {x}^{2} } \times ( - \frac{1}{ \sqrt{1 - 9 {x}^{8} } } ) \times 12 {x}^{3} = \\ = - {3}^{ - {x}^{2} } (2x ln(3) \times arccos(3 {x}^{4} ) + \frac{12 {x}^{3} }{ \sqrt{1 - 9 {x}^{8} } } )$

в)

$f'(x) = 6 {tg}^{5} (2x) \times \frac{1}{ { \cos }^{2} (2x)} \times 2 \times \cos(7 {x}^{3} ) - \sin(7 {x}^{3} ) \times 21 {x}^{2} \times {tg}^{6} (2x) + \\ + \frac{ {e}^{ \sin(5x) } \times 5\cos(5x) \times {(2 - 5x)}^{2} - 2(2 - 5x) \times ( - 5) {e}^{ \sin(5x) } }{ {(2 - 5x)}^{4} } = \\ = {tg}^{5} (2x)( \frac{12 \cos(7 {x}^{3} ) }{ { \cos}^{2} (2x) } - 21 {x}^{2} tg(2x) \sin(7 {x}^{3} ) ) + \frac{ 5{e}^{ \sin(5x) } (2 - 5x)((2 - 5x) \cos(5x) + 2) }{ {(2 - 5x)}^{2} } = \\ = {tg}^{5} (2x)( \frac{12 \cos(7 {x}^{3} ) }{ { \cos}^{2} (2x) } - 21 {x}^{2} tg(2x) \sin(7 {x}^{3} ) ) + \frac{ 5{e}^{ \sin(5x) } ((2 - 5x) \cos(5x) + 2) }{ {(2 - 5x)} }$

г)

$f'(x) = {( ln(x) + 1) }^{ \sin( \sqrt{x} ) }$

формула:

$f'(x) = ( ln(f(x)) )' \times f(x)$

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

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