Question

Знайти похідні першого та другого порядку

1) y=x+ctg(1/x)
2) y= 3/x*arcsin*x/3

Answer 1

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

Answer 2

$1); ; y=x+ctgfrac{1}{x}; ; ,quad Big [; (ctgu)'=-frac{1}{sin^2u}cdot u'; ; ,; ; (frac{1}{u})'=-frac{u'}{u^2}; Big ]\\y'=1-frac{1}{sin^2frac{1}{x}}cdot (-frac{1}{x^2})=1+frac{1}{x^2cdot sin^2frac{1}{x}}; ;\\\y''=-frac{2xcdot sin^2frac{1}{x}+x^2cdot 2, sinfrac{1}{x}cdot cosfrac{1}{x}cdot (-frac{1}{x^2})}{x^4cdot sfrac{x}{y} in^4frac{1}{x}}=-frac{2cdot sinfrac{1}{x}cdot (xcdot sinfrac{1}{x}-cosfrac{1}{x})}{x^4cdot sin^4frac{1}{x}}=$

$=-frac{2cdot (xcdot sinfrac{1}{x}-cosfrac{1}{x})}{x^4cdot sin^3frac{1}{x}}\\iliqquad y''=-frac{2xcdot sin^2frac{1}{x}-sinfrac{2}{x}}{x^4cdot sin^3frac{1}{x}}; .$

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

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