![1)\; \; y=x+ctg\frac{1}{x}\; \; ,\quad \Big [\; (ctgu)'=-\frac{1}{sin^2u}\cdot u'\; \; ,\; \; (\frac{1}{u})'=-\frac{u'}{u^2}\; \Big ]\\\\y'=1-\frac{1}{sin^2\frac{1}{x}}\cdot (-\frac{1}{x^2})=1+\frac{1}{x^2\cdot sin^2\frac{1}{x}}\; ;\\\\\\y''=-\frac{2x\cdot sin^2\frac{1}{x}+x^2\cdot 2\, sin\frac{1}{x}\cdot cos\frac{1}{x}\cdot (-\frac{1}{x^2})}{x^4\cdot s\frac{x}{y} in^4\frac{1}{x}}=-\frac{2\cdot sin\frac{1}{x}\cdot (x\cdot sin\frac{1}{x}-cos\frac{1}{x})}{x^4\cdot sin^4\frac{1}{x}}=](/tpl/images/0947/8417/477fb.png)
![2)\; \; y=\frac{3}{x}\cdot arcsin\frac{x}{3}\; \; ,\qquad \Big [\, (\frac{1}{u})'=-\frac{u'}{u^2}\; ,\; \; (arcsinu)'=\frac{1}{\sqrt{1-u^2}}\cdot u'\; ]\\\\y'=-\frac{3}{x^2}\cdot arcsin\frac{x}{3}+\frac{3}{x}\cdot \frac{1}{\sqrt{1-\frac{x^2}{9}}}\cdot \frac{1}{3}=\\\\=-\frac{3}{x^2}\cdot arcsin\frac{x}{3}+\frac{3}{x\cdot \sqrt{9-x^2}}\; ;](/tpl/images/0947/8417/02efb.png)
1) y=x+ctg(1/x)
у' = 1 + 1/Sin²(1/x) * 1/x² = 1 + 1/(x²*sin(1/x) )
y'= 1 + x⁻²*Sin⁻¹(1/x)
y'' = (x⁻²)' *Sin⁻¹(1/x) + x⁻²*(Sin⁻¹(1/x))' =
= -2x⁻³*Sin(1/x) + x⁻²* (-Sin⁻²(1/x)*Cos(1/x)*(-1/x²) ) =
=-2Sin(1/x) /x³ +Cos(1/x)/ (x²*Sin²(1/x))
2) y= 3/x*arcSinx/3
y' =(3/x)' *arcSinx/3 + 3/x*(arcSinx/3)' =
= -3/х² * arcSinx/3 + 3/x*1/√(1 - x²/9) * (1/3)=
=-3arcSinx/3 /х² + 1/√(1 - х²/9)
y'' = (-x²/√(1 - x²/9) -6x*arcSinx/3 )/x⁴
