Question

y= ln x/tg x помогите решить?

Answer 1

$y=\frac{lnx}{tgx}\\\\y'=\frac{\frac{1}{x}\cdot tgx-lnx\cdot \frac{1}{cos^2x}}{tg^2x}=\frac{cos^2x\cdot tgx-x\cdot lnx}{x\cdot cos^2x\cdot tg^2x}=\frac{cosx\cdot sinx-x\cdot inx}{x\cdot sin^2x}=\frac{0,5\, sin2x-x\cdot lnx}{x\cdot sin^2x}$