Question

Найти производную функции. Помогите решить пожалуйста

Answer 1

Ответ:

1.

$y = \frac{ {e}^{x} }{2x}$

$y '= \frac{( {e}^{x})' \times 2x - (2x)'e {}^{x} }{4 {x}^{2} } = \frac{ {e}^{x} \times 2x - 2 {e}^{x} }{4 {x}^{2} } = \\ = \frac{ {e}^{x} (2x - 1)}{4 {x}^{2} }$

2.

$y = \frac{ {x}^{3} + {x}^{6} }{ \sin(x) + 3}$

$y' = \frac{( {x}^{3} + {x}^{6})'( \sin(x) + 3) - (\sin(x) + 3)'( {x}^{3} + {x}^{6} ) }{ {( \sin(x) + 3) }^{2} } = \\ = \frac{(3 {x}^{2} + 6 {x}^{5})( \sin(x) + 3) - \cos(x) \times ( {x}^{3} + {x}^{6} ) }{ {( \sin(x) + 3)}^{2} }$

3.

$y = \frac{ \sin(x) - 3}{ \sin(x) + 3 }$

$y' = \frac{ \cos(x) \times ( \sin(x) + 3) - \cos(x)( \sin(x) - 3)}{ {( \sin(x) + 3) }^{2} } = \\ = \frac{ \cos(x) ( \sin(x) + 3 - \sin(x) + 3) }{( \sin(x) + 3) {}^{2} } = \\ = \frac{6 \cos(x) }{( \sin(x) + 3) {}^{2} }$

4.

$y' = ( \sin(x) - 3)'( \sin(x) + 3) + ( \sin(x) + 3)'( \sin(x) - 3) = \\ = \cos(x) ( \sin(x) + 3) + \cos(x) ( \sin(x) - 3) = \\ = \cos(x) ( \sin(x) + 3 + \sin(x) - 3) = \\ = 2 \sin(x) \cos(x) = \sin(2x)$