Question

Найти производную функции

Answer 1

$y=(sin2x+x)^{lnx}\\\\lny=ln(sin2x+x)^{lnx}\; \qquad \; \; \star \; ln\, x^{k}=k\cdot lnx\; \star \\\\lny=lnx\cdot ln(sin2x+x)\\\\\Big (lny\Big )'=\Big (lnx\cdot ln(sin2x+x)\Big )'\\\\\frac{y'}{y}=\frac{1}{x}\cdot ln(sin2x+x)+lnx\cdot \frac{1}{sin2x+x}\cdot (2\, cos2x+1)\\\\y'=y\cdot \Big (\frac{1}{x}\cdot ln(sin2x+x)+\frac{lnx\cdot (2\, cos2x+1)}{sin2x+x}\Big )\\\\y'=(sin2x+x)^{lnx}\cdot \Big (\frac{1}{x}\cdot ln(sin2x+x)+\frac{lnx\cdot (2\, cos2x+1)}{sin2x+x}\Big )$