Question

Найти производную:
(Sin2x / x+3) над скобкой 1-х

Answer 1

$y=\Big (\frac{sin2x}{x+3}\Big )^{1-x}\\\\lny=ln\Big (\frac{sin2x}{x+3\frac{x}{y} }\Big )^{1-x}\\\\lny=(1-x)\cdot ln\frac{sin2x}{x+3}\; \; ,\; \; \; \; \; \; \; (lnu)'=\frac{u'}{u}\\\\\frac{y'}{y}=(1-x)'\cdot ln\frac{sin2x}{x+3}+(1-x)\cdot \Big (ln\frac{sin2x}{x+3}\Big )'\\\\\frac{y'}{y}=-ln\frac{sin2x}{x+3}+(1-x)\cdot \frac{x+3}{sin2x}\cdot \frac{2\, cos2x\cdot (x+3)-sin2x}{(x+3)^2}\\\\y'=y\cdot \Big (-ln\frac{sin2x}{x+3}+(1-x)\cdot \frac{2\, cos2x\cdot (x+3)-sin2x}{(x+3)\cdot sin2x}\Big )$

$y'=\Big (\frac{sin2x}{x+3}\Big )^{1-x}\cdot \Big (-ln\frac{sin2x}{x+3}+(1-x)\cdot \frac{2\, cos2x\cdot (x+3)-sin2x}{(x+3)\cdot sin2x}\Big )\\\\y'=\Big (\frac{sin2x}{x+3}\Big )^{1-x}\cdot \Big (-ln\frac{sin2x}{x+3}+2\, (1-x)\cdot ctg2x-\frac{(1-x)}{(x+3)}\Big )$