Question

ПОМОГИТЕ НАЙТИ ПРОИЗВОДНЫЕ (КРОМЕ ПЕРВЫХ ДВУХ) все что сможете сами решить​

Answer 1

543

а)

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

б)

$y' = ln(7) \times {7}^{ \frac{x}{2} } \times \frac{1}{2} \times tg(3x) + \frac{1}{ {cos}^{2} (3x)} \times 3 \times {7}^{ \frac{x}{2} } = \\ = {7}^{ \frac{x}{2} } ( \frac{ ln(7) \times tg(3x)}{2} + \frac{3}{ {cos}^{2}(3x) } )$

550

а)

$y' = 2x log_{2}(x) + \frac{ {x}^{2} }{ ln(2) \times x} = \\ = 2x log_{2}(x) + \frac{x}{ ln(2) }$

б)

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

в)

$y' = ln(x) + \frac{1}{x} \times x = ln(x) + 1$

г)

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

552

а)

$y' = \frac{1}{x + 2}$

б)

$y' = \frac{1}{ ln(10) \times x}$

в)

$y' = \frac{2}{x}$

г)

$y' = \frac{1}{(x - 1)ln(2) }$

556

а)

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

б)

$y' = \frac{2lg(x) - \frac{1}{x ln(10) } \times x }{ {lg}^{2} (x)} = \\ = \frac{2lg(x) - \frac{1}{ ln(10) } }{ {lg}^{2}(x) }$

в)

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

г)

$y' = - {x}^{ - 2} + \frac{1}{x} = - \frac{1}{ {x}^{2} } + \frac{1}{x}$