Question

Найти производные: 1) y=x² · ∛x
2) y=$tg^{5}$·x
3) y=$\frac{1}{2}$x²- $\frac{1}{\sqrt{x} }$ + $\frac{1}{x^{2} }$ - 3
4) y= $2^{x^{2} }$ · arctg $2^{x}$
5) y= $\frac{sin3x}{arccos2x}$
6) y=㏑ tg 2x

Answer 1

Пошаговое объяснение:

1.

$y = {x}^{2} \sqrt[3]{x} = {x}^{ \frac{7}{3} } \\$

$y '= \frac{7}{3} {x}^{ \frac{4}{3} } = \frac{7}{3} x \sqrt[3]{ {x} } \\$

2.

$y = {tg}^{5} x$

$y' = 5 {tg}^{4} x \times (tgx)' = \\ = 5 {tg}^{2} x \times \frac{1}{ \cos {}^{2} (x) }$

3.

$y = \frac{1}{2} {x}^{2} - \frac{1}{ \sqrt{x} } + \frac{1}{ {x}^{2} } - 3 = \\ = \frac{1}{2} {x}^{2} - x {}^{ - \frac{1}{2} } + {x}^{ - 2} - 3$

$y '= \frac{1}{2} \times 2x - \frac{1}{2} {x}^{ - \frac{3}{2} } - 2 {x}^{ - 3} + 0 = \\ = x - \frac{1}{2x \sqrt{x} } - \frac{2}{ {x}^{3} }$

4.

$y = {2}^{ {x}^{2} } \times arctg( {2}^{x} )$

$y '= ( {2}^{ {x}^{2} } ) '\times \arctg( {2}^{x} ) + ( \arctg( {2}^{x} ))' \times {2}^{ {x}^{2} } = \\ = ln(2) \times {2}^{ {x}^{2} } \times 2x \times \arctg( {2}^{x} ) + \frac{1}{1 + {( {2}^{x}) }^{2} } \times ln(2) \times {2}^{x} \times {2}^{ {x}^{2} } = \\ = ln(2) \times {2}^{ {x}^{2} } (2x \arctg( {2}^{x} ) + \frac{ {2}^{x} }{1 + {4}^{x} } )$

5.

$y = \frac{ \sin(3x) }{ \arccos(2x)} \\$

$y '= \frac{( \sin(3x))' \times \arccos(2x) + ( \arccos(2x))' \sin(3x) }{ { \arccos}^{2} (2x)} = \\ = \frac{3 \cos(3x) \times \arccos(2x) - \frac{1}{ \sqrt{1 - 4 {x}^{2} } } \times 2 \times \sin(3x) }{ { \arccos}^{2} (2x)}$

6.

$y = ln(tg(2x))$

$y '= \frac{1}{tg(2x)} \times (tg2x)' \times (2x) '= \\ = \frac{ \cos(2x) }{ \sin(2x) } \times \frac{1}{ \cos {}^{2} (2x) } \times 2 = \frac{2}{ \sin(2x) \cos(2x) }$