Question

Похідна першого порядку

Answer 1

Ответ:

а

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

б

$y '= 2 ln(arcsin \sqrt{x} ) \times ( ln(arcsin \sqrt{x} )) '\times ( arcsin( \sqrt{x} )) '\times ( \sqrt{x} ) '= \\ = 2 ln(arcsin \sqrt{x} ) \times \frac{1}{arcsin \sqrt{x} } \times \frac{1}{ \sqrt{1 - {x}^{2} } } \times \frac{1}{2 \sqrt{x} } = \\ = \frac{ ln(arcsin \sqrt{x} ) }{ \sqrt{x - {x}^{3} } arcsin \sqrt{x} }$

в

$y '= ln(2) \times {2}^{ \sin(x) + \cos {}^{2} (x) } \times (\sin(x) + \cos {}^{2} (x)) ' = \\ = ln(2) \times {2}^{ \sin(x) + \cos {}^{2} (x) } \times ( \cos(x) - 2 \cos(x) \sin(x) ) = \\ = ln(2) \times {2}^{ \sin(x) + \cos {}^{2} (x) } \times ( \cos(x) - \sin(2x))$

г

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