Знайти похідні першого та другого порядку 1) y=x+ctg(1/x) 2) y= 3/x*arcsin*x/3

Решите задачу:

$1)\; \; y=x+ctg\frac{1}{x}\; \; ,\quad \Big [\; (ctgu)'=-\frac{1}{sin^2u}\cdot u'\; \; ,\; \; (\frac{1}{u})'=-\frac{u'}{u^2}\; \Big ]\\\\y'=1-\frac{1}{sin^2\frac{1}{x}}\cdot (-\frac{1}{x^2})=1+\frac{1}{x^2\cdot sin^2\frac{1}{x}}\; ;\\\\\\y''=-\frac{2x\cdot sin^2\frac{1}{x}+x^2\cdot 2\, sin\frac{1}{x}\cdot cos\frac{1}{x}\cdot (-\frac{1}{x^2})}{x^4\cdot s\frac{x}{y} in^4\frac{1}{x}}=-\frac{2\cdot sin\frac{1}{x}\cdot (x\cdot sin\frac{1}{x}-cos\frac{1}{x})}{x^4\cdot sin^4\frac{1}{x}}=$

$=-\frac{2\cdot (x\cdot sin\frac{1}{x}-cos\frac{1}{x})}{x^4\cdot sin^3\frac{1}{x}}\\\\ili\qquad y''=-\frac{2x\cdot sin^2\frac{1}{x}-sin\frac{2}{x}}{x^4\cdot sin^3\frac{1}{x}}\; .$

$2)\; \; y=\frac{3}{x}\cdot arcsin\frac{x}{3}\; \; ,\qquad \Big [\, (\frac{1}{u})'=-\frac{u'}{u^2}\; ,\; \; (arcsinu)'=\frac{1}{\sqrt{1-u^2}}\cdot u'\; ]\\\\y'=-\frac{3}{x^2}\cdot arcsin\frac{x}{3}+\frac{3}{x}\cdot \frac{1}{\sqrt{1-\frac{x^2}{9}}}\cdot \frac{1}{3}=\\\\=-\frac{3}{x^2}\cdot arcsin\frac{x}{3}+\frac{3}{x\cdot \sqrt{9-x^2}}\; ;$

$y''=\frac{3\cdot 2x}{x^4}\cdot arcsin\frac{x}{3}-\frac{3}{x^2}\cdot \frac{1}{\sqrt{1-\frac{x^2}{9}}}\cdot \frac{1}{3}-\frac{3\cdot (\sqrt{9-x^2}+x\cdot \frac{-2x}{2\sqrt{9-x^2}})}{x^2\cdot (9-x^2)}=\\\\=\frac{6}{x^3}\cdot arcsin\frac{x}{3}-\frac{3}{x^2\cdot \sqrt{9-x^2}}-\frac{3\cdot (2(9-x^2)-x^2)}{x^2\cdot \sqrt{(9-x^2)^3}}=\\\\=\frac{6}{x^3}\cdot arcsin\frac{x}{3}-\frac{3}{x^2\cdot \sqrt{9-x^2}}-\frac{54-9x^2}{x^2\cdot \sqrt{(9-x^2)^3}}\; .$