Question

пример 104, 133, 135 с подробным решением, пожалуйста

Answer 1

Ответ:

104

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

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

133

$y = {e}^{x} + x + 1$

$y' = {e}^{x} + 1 + 0 = {e}^{x} + 1$

$dy=({e}^{x} + 1)dx$

135

$y = \frac{2x - 4}{4x + 3}$

$y '= \frac{(2x - 4)'(4x + 3) - (4x + 3)'(2x - 4)}{ {(4x + 3)}^{2} } = \\ = \frac{2(4x + 3) - 4(2x - 4)}{ {(4x + 3)}^{2} } = \\ = \frac{8x + 6 - 8x + 16}{ {(4x + 3)}^{2} } = \frac{22}{ {(4x + 3)}^{2} }$

$dy=\frac{22}{ {(4x + 3)}^{2} } dx$