Question

ради христа Используя основные правила дифференцирования и таблицу производных
основных элементарных функций, найти производную для функции
y=y(x)=f(x);

Answer 1

1

$y' = 2 \cos(x) - 7 \sin(x)$

2

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

3

$y = \sqrt{x} + \frac{1}{ \sqrt[3]{x} } - 2 \cos(2 {x}^{2} + 1 ) = \\ {x}^{ \frac{1}{2} } + {x}^{ - \frac{1}{3} } - 2 \cos(2 {x}^{2} + 1)$

$y '= \frac{1}{2} {x}^{ - \frac{1}{2} } - \frac{1}{3} {x}^{ - \frac{4}{3} } + 2 \sin(2 {x}^{2} + 1) \times (2 {x}^{2} + 1)' = \\ = \frac{1}{2 \sqrt{x} } - \frac{1}{3x \sqrt[3]{x} } + 2 \sin(2 {x}^{2} + 1 ) \times 4x = \\ = \frac{1}{2 \sqrt{x} } - \frac{1}{3x \sqrt[3]{x} } + 8x \sin(2 {x}^{2} + 1 )$

4

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

5

$y' = 4 \cos {}^{3} ( ln(x) ) \times ( \cos( ln(x)) ) ' \times (ln(x))' = \\ = 4 \cos {}^{3} ( ln(x) ) \times ( - \sin( ln(x)) ) \times \frac{1}{x} = = - \frac{4}{x} \cos {}^{3} ( ln(x) ) \sin( ln(x) )$