Question

Найди производную функции:у(х)= $\frac{(x + 3)(2x - 2)}{ {x}^{2} } - 2 \cos2x$

Answer 1

$\boxed{y'(x) = \dfrac{4(3 - x)}{x^{3}} -4 \sin 2x}$

Объяснение:

$y(x) = \dfrac{(x + 3)(2x - 2)}{x^{2} } - 2 \cos 2x$

Пусть:

$f(x) = (x + 3)(2x - 2)$

$g(x) = {x^{2}$

а)

$f(x) = (x + 3)(2x - 2) = 2x^{2} - 2x + 6x - 6 = 2x^{2} + 4x - 6$

б)

$f'(x) = (2x^{2} + 4x - 6)' = 4x + 4$

в)

$g'(x) = (x^{2} )' = 2x$

$y'(x) = \bigg(\dfrac{(x + 3)(2x - 2)}{x^{2} } - 2 \cos 2x \bigg)' = \bigg ( \dfrac{(x + 3)(2x - 2)}{x^{2} } \bigg)' - \bigg(2 \cos 2x \bigg)'=$

$= \dfrac{4(3 - x)}{x^{3}} -4 \sin 2x$

1)

$\bigg ( \dfrac{(x + 3)(2x - 2)}{x^{2} } \bigg)' = \bigg ( \dfrac{f(x)}{g(x)} \bigg)' = \dfrac{f'(x)g(x) - f(x)g'(x)}{g^{2}(x)} =$

$= \dfrac{x^{2} (4x + 4) - 2x(2x^{2} + 4x - 6)}{(x^{2} )^{2}}= \dfrac{4x^{3} + 4x^{2} - (4x^{3} + 8x^{2} - 12x)}{x^{4}} =$

$= \dfrac{4x^{3} + 4x^{2} - 4x^{3} - 8x^{2} + 12x}{x^{4}} = \dfrac{12x - 4x^{2} }{x^{4}} = \dfrac{4x(3 - x)}{x \cdot x^{3}} = \dfrac{4(3 - x)}{x^{3}}$

2)

$\bigg(2 \cos 2x \bigg)'= 2 (\cos 2x)' = 2 \cdot (2x)' \cdot (-\sin 2x) = -4 \sin 2x$