1."f(x,y)=\\frac{x^3+y^3}{x+y}=\\frac{(x+y)(x^2-xy+y^2)}{x+y}=x^2-xy+y^2"
Degree is 2.
According to Euler's relation:
"x\\frac{\\partial f}{\\partial x}+y\\frac{\\partial f}{\\partial y}=nf(x,y)"
"x\\frac{\\partial f}{\\partial x}+y\\frac{\\partial f}{\\partial y}=x(2x-y)+y(2y-x)=2(x^2-xy+y^2)=nf(x,y)"
(Proved)
3."f(x,y)=10-x^2-y^2"
"0\t\\leqslant x^2+y^2\t\\leqslant9"
"-9\t\\leqslant -(x^2+y^2)\t\\leqslant0"
"10-9\t\\leqslant 10-(x^2+y^2)\t\\leqslant0+10"
"1\t\\leqslant f(x,y)\\leqslant10" is the range of the function.
Level Curve in the image.
2."f(x,y)=y^3+y Sin2x+e^{x+y}"
"f_x(x,y)=2ycos2x+e^{x+y}"
"f_{xx}(x,y)=-4y Sin2x+e^{x+y}"
"f_y(x,y)=3y^2+ Sin2x+e^{x+y}"
"f_{yy}(x,y)=6y+ e^{x+y}"
Partial derivatives are continuous and differentiable as sin (2x),cos(2x),ex+y,y2,y3,y are continuous and differentiable.
So,f(x,y) is differentiable at (1,-1).
(ii)"f(x,y)=|x-1| at (1,0)"
"f_y(x,y)=0"
"f_x(1,0)=lim_{h\\to0}\\frac{f((1+h),0)-f(1,0)}{h}"
"f_x(1,0)=lim_{h\\to0}\\frac{|h|-0}{h}=lim_{h\\to0}\\frac{|h|}{h}"
"lim_{h\\to0^+}\\frac{h}{h}=1" whereas "lim_{h\\to0^-}\\frac{-h}{h}=-1"
So,limit does not exist,partial derivative fx(x,y) does not exist at (1,0) .Thus, f(x,y) is not differentiable.
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