Question

Найти производную функции

Answer 1

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