Question

помогите решить математику:



даю 35 баллов

Answer 1

Ответ:

а.

$y' = x'ctg ^{2} (7x) + ( {ctg}^{2} (7x))' \times (ctg(7x) )'\times (7x)' \times x \\ = {ctg}^{2} (7x) + 2ctg(7x) \times ( - \frac{1}{ { \sin }^{2} (7x)} ) \times 7 \times x = \\ = ctg(7x) \times (ctg(7x) - \frac{14x}{ { \sin}^{2}(7x) } )$

б.

$y' = ln(2) \times {2}^{ - { \cos }^{4}(5x) } \times ( - { \cos }^{4} (5x)) '\times ( \cos(5x))' \times (5x) '= \\ = ln(2) \times {2}^{ - { \cos }^{4} (5x)} \times ( - 4 { \cos}^{3} (5x)) \times ( - \sin(5x)) \times 5 = \\ = 20 ln(2) \times {2}^{ { - \cos }^{4} (5x)} { \cos }^{3} (5x) \sin(5x)$

в.

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

г.

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