【例題】次の関数の2階偏導関数を全て求めなさい。
(1)\(f(x,y)=x^3+3x^2y+y^2\)
1階偏導関数を求めると
\(f_x(x,y)=3x^2+6xy\)
\(f_y(x,y)=3x^2+2y\)
よって、2階偏導関数は
\(f_{xx}(x,y)=6x+6y\)
\(f_{xy}(x,y)=f_{yx}(x,y)=6x\)
\(f_{yy}(x,y)=2\)
(2)\(f(x,y)=e^x\cos y\)
1階偏導関数を求めると
\(f_x(x,y)=e^x\cos y\)
\(f_y(x,y)=-e^x\sin y\)
よって、2階偏導関数は
\(f_{xx}(x,y)=e^x\cos y\)
\(f_{xy}(x,y)=f_{yx}(x,y)=-e^x\sin y\)
\(f_{yy}(x,y)=-e^x\cos y\)
(3)\(f(x,y)=\log(x^2+y^2+1)\)
1階偏導関数を求めると
\(\displaystyle f_x(x,y)=\frac{2x}{x^2+y^2+1}\)
\(\displaystyle f_y(x,y)=\frac{2y}{x^2+y^2+1}\)
よって、2階偏導関数は
\(\displaystyle f_{xx}(x,y)=\frac{2(-x^2+y^2+1)}{(x^2+y^2+1)^2}\)
\(\displaystyle f_{xy}(x,y)=f_{yx}(x,y)=-\frac{4xy}{(x^2+y^2+1)^2}\)
\(\displaystyle f_{yy}(x,y)=\frac{2(x^2-y^2+1)}{(x^2+y^2+1)^2}\)
(4)\(\displaystyle f(x,y)=\tan^{-1}\frac{y}{x}\)
1階偏導関数を求めると
\(\displaystyle f_x(x,y)=\frac{-y}{x^2+y^2}\)
\(\displaystyle f_y(x,y)=\frac{x}{x^2+y^2}\)
よって、2階偏導関数は
\(\displaystyle f_{xx}(x,y)=\frac{2xy}{(x^2+y^2)^2}\)
\(\displaystyle f_{xy}(x,y)=f_{yx}(x,y)=\frac{y^2-x^2}{(x^2+y^2)^2}\)
\(\displaystyle f_{yy}(x,y)=\frac{-2xy}{(x^2+y^2)^2}\)
(5)\(\displaystyle f(x,y)=\frac{2x-y}{x^2+y^2}\)
1階偏導関数を求めると
\(f_x(x,y)\)
\(\displaystyle =\frac{2(x^2+y^2)-(2x-y)・2x}{(x^2+y^2)^2}\)
\(\displaystyle =\frac{-2x^2+2xy+2y^2}{(x^2+y^2)^2}\)
\(f_y(x,y)\)
\(\displaystyle =\frac{-(x^2+y^2)-(2x-y)・2y}{(x^2+y^2)^2}\)
\(\displaystyle =\frac{-x^2-4xy+y^2}{(x^2+y^2)^2}\)
よって、2階偏導関数は
\(f_{xx}(x,y)\)
\(\displaystyle =\frac{-4x+2y}{(x^2+y^2)^2}\)
\(\displaystyle \ \ \ +(-2x^2+2xy+2y^2)・\frac{-4x}{(x^2+y^2)^3}\)
\(\displaystyle =\frac{4x^3-6x^2y-12xy^2+2y^3}{(x^2+y^2)^3}\)
\(f_{xy}(x,y)=f_{yx}(x,y)\)
\(\displaystyle =\frac{2x+4y}{(x^2+y^2)^2}\)
\(\displaystyle \ \ \ +(-2x^2+2xy+2y^2)・\frac{-4y}{(x^2+y^2)^3}\)
\(\displaystyle =\frac{2x^3+12x^2y-6xy^2-4y^3}{(x^2+y^2)^3}\)
\(f_{yy}(x,y)\)
\(\displaystyle =\frac{-4x+2y}{(x^2+y^2)^2}\)
\(\displaystyle \ \ \ +(-x^2-4xy+y^2)・\frac{-4y}{(x^2+y^2)^3}\)
\(\displaystyle =\frac{-4x^3+6x^2y+12xy^2-2y^3}{(x^2+y^2)^3}\)
(6)\(\displaystyle f(x,y)=x^2\tan^{-1}\frac{y}{x}-y^2\tan^{-1}\frac{x}{y}\)
1階偏導関数を求めると
\(f_x(x,y)\)
\(\displaystyle =2x\tan^{-1}\frac{y}{x}+x^2・\frac{-y}{x^2+y^2}-y^2・\frac{y}{x^2+y^2}\)
\(\displaystyle =2x\tan^{-1}\frac{y}{x}-y\)
\(f_y(x,y)\)
\(\displaystyle =x^2・\frac{x}{x^2+y^2}-2y\tan^{-1}\frac{x}{y}-y^2・\frac{-x}{x^2+y^2}\)
\(\displaystyle =-2y\tan^{-1}\frac{x}{y}+x\)
よって、2階偏導関数は
\(f_{xx}(x,y)\)
\(\displaystyle =2\tan^{-1}\frac{y}{x}+2x・\frac{-y}{x^2+y^2}\)
\(\displaystyle =2\tan^{-1}\frac{y}{x}-\frac{2xy}{x^2+y^2}\)
\(f_{xy}(x,y)=f_{yx}(x,y)\)
\(\displaystyle =2x・\frac{x}{x^2+y^2}-1\)
\(\displaystyle =\frac{x^2-y^2}{x^2+y^2}\)
\(f_{yy}(x,y)\)
\(\displaystyle =-2\tan^{-1}\frac{x}{y}-2y・\frac{-x}{x^2+y^2}\)
\(\displaystyle =-2\tan^{-1}\frac{x}{y}+\frac{2xy}{x^2+y^2}\)