Question

Найти производную функции y= 4* arccos (sin x)

Answer 1

$y=4\cdot arccos (sinx)\\\\y'=4\cdot \frac{-1}{\sqrt{1-sin^2x}}\cdot (sinx)'=-\frac{4}{\sqrt{cos^2x}}\cdot cosx=-\frac{4\cdot cosx}{|cosx|}=\left \{ {{-4\; ,\; esli\; cosx>0} \atop {4\; ,\; esli\; cosx<0}} \right. \\\\\\\star cosx>0\; \; pri\; \; x\in (-\frac{\pi }{2}+2\pi n,\frac{\pi }{2}+2\pi n)\; ,\; n\in Z\\\\cosx<0\; \; pri\; \; x\in (\frac{\pi }{2}+2\pi n,\frac{3\pi }{2}+2\pi n)\; ,\; n\in Z\; \; \star$