Производная второго порядка: 1) sin(x)/(1+sin(x)) 2) xsin(x^3) 4) Частные производные...

Решите задачу:

$1)\quad y= \frac{sinx}{1+sinx} \\\\y'= \frac{cosx(1+sinx)-sinx\cdot cosx}{(1+sinx)^2} =\frac{cosx}{(1+sinx)^2} \\\\y''= \frac{-sinx(1+sinx)^2-cosx\cdot 2(1+sinx)\cdot cosx}{(1+sinx)^4} = \frac{-sinx(1+sinx)-2cos^2x}{(1+sinx)^3}\\\\2)\; \; y=x\cdot sin(x^3)\\\\y'=sin(x^3)+x\cdot cos(x^3)\cdot 3x^2=sin(x^3)+3x^3\cdot cos(x^3)\\\\y''=cos(x^3)\cdot 3x^2+9x^2\cdot cos(x^3)+3x^3\cdot (-sin(x^3))\cdot 3x^2=\\\\=3x^2\cdot cos(x^3)+9x^2\cdot cos(x^3)-9x^5\cdot sin(x^3)$

$3)\; \; u= \frac{x+y}{x^2+y^2} \\\\u'_{x}= \frac{x^2+y^2-(x+y)\cdot 2x}{(x^2+y^2)^2} = \frac{y^2-x^2-2xy}{(x^2+y^2)^2} \\\\u''_{xx}= \frac{(-2x-2y)(x^2+y^2)^2-(y^2-x^2-2xy)\cdot 2(x^2+y^2)\cdot 2x}{(x^2+y^2)^4} = \\\\=\frac{-2(x+y)(x^2+y^2)-4x(y^2-x^2-2xy)}{(x^2+y^2)^3} \\\\u''_{xy}= \frac{(2y-2x)(x^2+y^2)^2-(y^2-x^2-2xy)\cdot 2(x^2+y^2)\cdot 2y}{(x^2+y^2)^4}=\\\\=\frac{2(y-x)(x^2+y^2)-4y(y^2-x^2-2xy)}{(x^2+y^2)^3}$

$u'_{y}= \frac{x^2+y^2-(x+y)\cdot 2y}{(x^2+y^2)^2}= \frac{x^2-y^2-2xy}{(x^2+y^2)^2} \\\\u''_{yy}= \frac{(-2y-2x)(x^2+y^2)^2-(x^2-y^2-2xy)\cdot 2(x^2+y^2)\cdot 2y}{(x^2+y^2)^4}=\\\\= \frac{-2(x+y)(x^2+y^2)-4y(x^2-y^2-2xy)}{(x^2+y^2)^3}\\\\u''_{yx}=u''_{xy}$