Помогите пожалуйста с решением

Приложения:

Ответы

Ответ дал: Miroslava227

Ответ:

$f'(x) = 5 {x}^{4} + 7 {x}^{6} + 12 {x}^{11}$

$f'( - 1) = 5 + 7 - 12 = 0$

$f'(x) = 3 \times 5 {(x + 4)}^{4} \times (x + 4) '= \\ = 15 {(x + 4)}^{4}$

$f'( - 3) = 15 {(4 - 3)}^{4} = 15$

$f'(x) = ( {(x + 3)}^{7} ) '\times {(x + 7)}^{3} + ( {(x + 7)}^{3} )' \times {(x + 3)}^{7} = \\ = 7 {(x + 3)}^{6} {(x + 7)}^{3} + 3 {(x + 7)}^{2} {(x + 3)}^{7} = \\ = {(x + 3)}^{6} {(x + 7)}^{2} (7(x + 7) + 3(x + 3)) = \\ = ( {x + 3)}^{6} {(x + 7)}^{2} (7x + 49 + 3x + 9) = \\ = {(x + 3)}^{6} {(x + 7)}^{2} (10x + 58)$

$f'( - 4) = {( - 1)}^{6} \times {3}^{2} \times ( - 40 + 58) = \\ = 9 \times 18 = 162$

$f'(x) = (7 {(x - 6)}^{ - 5} ) '= - 35 {(x - 6)}^{ - 6} = \\ = - \frac{35}{ {(x - 6)}^{6} }$

$f'(7) = - \frac{35}{ {(7 - 6)}^{6} } = - 35 \\$

$f'(x) = (4x + 3 + \frac{7}{x} )' = 4 - 7 {x}^{ - 2} = \\ = 4 - \frac{7}{ {x}^{2} }$

$f'( - 1) = 4 - 7 = - 3$

$f'(x) = 7 \times \frac{1}{2} ({6x + 19)}^{ - \frac{1}{2} } \times 6 = \\ = \frac{21}{ \sqrt{6x + 19} }$

$f'(5) = \frac{21}{ \sqrt{30 + 19} } = \frac{21}{7} = 3 \\$

$f'(x) = \frac{9}{ { \cos }^{2} (x)} + 8 \sin(x) \\$

$f'( \frac{5\pi}{6} ) = \frac{9}{ { \cos}^{2}( \frac{5\pi}{6} ) } + 8 \sin( \frac{5\pi}{6} ) = \\ = \frac{9}{ \frac{3}{4} } + 8 \times \frac{1}{2} = 12 + 4 = 16$

$f'(x) = \frac{( \sin(2x))' \times \cos(12x) - ( \cos(12x)) '\times \sin(2x) }{ \cos ^{2} (12x) } = \\ \frac{2 \cos(2x) \times \cos(12x) + 12 \sin(12x) \times \sin(2x) }{ \cos ^{2} (12x) }$

$f'( \frac{\pi}{36} ) = \frac{2 \cos( \frac{\pi}{18} ) \times \cos( \frac{\pi}{3} ) + 12 \sin( \frac{\pi}{3} ) \sin( \frac{\pi}{18} ) }{ { (\cos( \frac{\pi}{3} )) }^{2} } = \\ = \frac{2 \cos( \frac{\pi}{18} ) \times \frac{1}{2} +12 \times \frac{ \sqrt{3} }{2} \sin( \frac{\pi}{18} ) }{ \frac{1}{4} } = \\ = 4 \times ( \cos( \frac{\pi}{18} ) + 6 \sqrt{3} \sin( \frac{\pi}{18} ) ) = \\ = 4 \cos( \frac{\pi}{18} ) + 24 \sqrt{3} \sin( \frac{\pi}{18} )$

$f'(x) = \frac{(x + 3)' {e}^{x - 4} - ( {e}^{x - 4})'(x + 3) }{ {e}^{2(x - 4)} } = \\ = \frac{ {e}^{x - 4} (1 - (x + 3)}{ {e}^{2(x - 4)} } = \\ = \frac{1 - x - 3}{ {e}^{x - 4} } = \frac{ - x - 2}{ {e}^{x - 4} }$