Question

РЕБЯТА С МАТЕШОЙ а то не аттестация будет(

Ничо не понятно

Answer 1

$1)y =4 {tg}^{2} ( \sin(x))$

$y '= 8tg( \sin(x)) \times \frac{1}{ { \cos }^{2} (\sin(x)) } \times \cos(x) \\$

$2)y = \frac{ctg(3 {x}^{2} - 5)}{2 {x}^{3} } \\$

$y' = \frac{ - \frac{1}{ { \sin}^{2}(3 {x}^{2} - 5) } \times 2 {x}^{3} - 6 {x}^{2} ctg(3 {x}^{2} - 5) }{4 {x}^{6} } = \\ = - \frac{2 {x}^{3} }{ { \sin }^{2} (3 {x}^{2} - 5) } \times \frac{1}{4 {x}^{6} } - \frac{6 {x}^{2} ctg(3 {x}^{2} - 5)}{4 {x}^{6} } = \\ = - \frac{1}{2 {x}^{3} { \sin }^{2}( 3 {x}^{2} - 5)} - \frac{3ctg(3 {x}^{2} - 5)}{2 {x}^{4} }$

$3)y = arctg( ln(3x)) + 4 \times {2}^{ \sqrt[3]{x} }$

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

$4)y = {e}^{ {x}^{2} + 1} \sqrt{ {x}^{2} + 1 }$

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

$5)y = {(arcsin(2x))}^{x}$

по формуле:

$y '= ( ln(y)) ' \times y$

$( ln(y))' = ( ln( {arcsin2x)}^{x} )' = (x \times ln(arcsin2x)) ' = \\ = ln(arcsin(2x)) + \frac{x}{arcsin2x} \times \frac{1}{ \sqrt{1 - 4 {x}^{2} } } \times 2$

$y' = {(arcsin2x)}^{x} \times ( ln(arcsin(2x)) + \frac{2x}{ \sqrt{1 - 4{x}^{2} } arcsin2x} \\$

$6)xy = \sin(xy) \\ y + xy' = \cos(xy) \times (y + xy') \\ y + xy' = y \cos(xy) + xy' \cos(xy) \\ xy' - xy' \cos(xy) = y \cos(xy) - y \\ y'(x - x \cos(xy)) = y \cos(xy) - y \\ y '= \frac{y( \cos(xy) - 1)}{x(1 - \cos(xy)) } = \\ y' = - \frac{y}{x}$

$7)y = \sin(3t) \\ x = \cos(3t)$

формула:

$y'x = \frac{y't}{x't} \\$

$y't = 3 \cos(3t) \\ x't = - 3 \sin(3t)$

$y'x = \frac{3 \cos(3t) }{ - 3 \sin(3t) } = - ctg(3t)$