Question

Найти производные: 1) y = sinx / 1+cosx 2) y = sin (1/x) 3) y = под корнем tg2x 4) y = ln^2 x 5) y = ln tgx

Answer 1

$1)\; \; y= \frac{sinx}{1+cosx} \\\\y'= \frac{cosx(1+cosx)-sinx\cdot (-sinx)}{(1+cosx)^2} = \frac{cosx+1}{(1+cosx)^2} = \frac{1}{1+cosx} \\\\2)\; \; y=sin(\frac{1}{x}) \\\\y'=cos( \frac{1}{x})\cdot \frac{-1\cdot x'}{x^2}=-\frac{1}{x^2}\cdot cos\frac{1}{x}\\\\3)\; \; y= \sqrt{tg2x} \\\\y'=\frac{1}{2\sqrt{tg2x}}\cdot \frac{1}{cos^22x} \cdot 2\\\\4)\; \; y=ln^2x\\\\y'=2\, lnx\cdot (lnx)'=2\, lnx\cdot \frac{1}{x}\\\\5)\; \; y=ln\, tgx\\\\y'= \frac{1}{tgx}\cdot \frac{1}{cos^2x} = \frac{1}{sinx\, cosx} =\frac{2}{sin2x}$