Question

Найти производную y = cos4x/ln^2(2x^2 +x+1)

Answer 1

Ответ:

$\boxed{ y' = \dfrac{ -4 \sin 4x \cdot \ln(2x^{2} + x + 1) - \dfrac{ \cos 4x \cdot 2(4x + 1) }{2x^{2} + x + 1}}{\ln^{3}(2x^{2} + x + 1)} }$

Пошаговое объяснение:

$y = \dfrac{\cos 4x}{\ln^{2}(2x^{2} + x +1)}$

Пусть $f(x) = \cos 4x$, $g(x) = 2x^{2} + x + 1$, $p(x) = \ln^{2}(2x^{2} + x + 1) = \ln^{2}(g(x))$

$y = \dfrac{f(x)}{p(x)}$, тогда $y' = \bigg (\dfrac{f(x)}{p(x)} \bigg)' = \dfrac{f'(x)p(x) - f(x)p'(x)}{p^{2}(x)};$

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

$g'(x) = (2x^{2} + x + 1)' = 4x + 1$

$p'(x) = (\ln^{2}(g(x))' = 2 \ln (g(x)) \cdot (\ln(g(x)))' = 2 \ln (g(x)) \cdot \dfrac{1}{g(x)} \cdot g'(x) =$

$= 2 \ln (2x^{2} + x + 1) \cdot \dfrac{1}{2x^{2} + x + 1} \cdot (4x + 1) = \dfrac{2(4x + 1)\ln (2x^{2} + x + 1)}{2x^{2} + x + 1}$

$y' = \dfrac{f'(x)p(x) - f(x)p'(x)}{p^{2}(x)} ;$

$y' = \dfrac{-4 \sin 4x \cdot \ln^{2}(2x^{2} + x + 1) - \dfrac{ \cos 4x \cdot 2(4x + 1)\ln (2x^{2} + x + 1)}{2x^{2} + x + 1}}{\ln^{4}(2x^{2} + x + 1)} =$

$= \dfrac{\ln(2x^{2} + x + 1) \bigg( -4 \sin 4x \cdot \ln(2x^{2} + x + 1) - \dfrac{ \cos 4x \cdot 2(4x + 1) }{2x^{2} + x + 1}\bigg)}{\ln^{3}(2x^{2} + x + 1) \cdot \ln(2x^{2} + x + 1) \cdot} =$

$= \dfrac{ -4 \sin4 x \cdot \ln(2x^{2} + x + 1) - \dfrac{ \cos 4x \cdot 2(4x + 1) }{2x^{2} + x + 1}}{\ln^{3}(2x^{2} + x + 1)} ;$