1.次の関数の2階偏導関数を全て求めなさい。
(1)\(f(x,y)=ax^2+2bxy+cy^2\)
1階偏導関数を求めると
\(f_x(x,y)=2ax+2by\)
\(f_y(x,y)=2bx+2cy\)
よって、2階偏導関数は
\(\displaystyle f_{xx}(x,y)=2a\)
\(\displaystyle f_{xy}(x,y)=2b\)
\(\displaystyle f_{yy}(x,y)=2c\)
(2)\(f(x,y,z)=(x+y^2+z^3)^2\)
1階偏導関数を求めると
\(f_x(x,y,z)=2(x+y^2+z^3)\)
\(f_y(x,y,z)=4y(x+y^2+z^3)\)
\(f_z(x,y,z)=6z^2(x+y^2+z^3)\)
よって、2階偏導関数は
\(\displaystyle f_{xx}(x,y,z)=2\)
\(\displaystyle f_{xy}(x,y,z)=4y\)
\(\displaystyle f_{xz}(x,y,z)=6z^2\)
\(\displaystyle f_{yy}(x,y,z)=4(x+y^2+z^3)+8y^2\)
\(\displaystyle f_{yz}(x,y,z)=12yz^2\)
\(\displaystyle f_{zz}(x,y,z)=12z(x+y^2+z^3)+18z^4\)
(3)\(f(x,y)=\sin(3x-2y)\)
1階偏導関数を求めると
\(f_x(x,y)=3\cos(3x-2y)\)
\(f_y(x,y)=-2\cos(3x-2y)\)
よって、2階偏導関数は
\(\displaystyle f_{xx}(x,y)=-9\sin(3x-2y)\)
\(\displaystyle f_{xy}(x,y)=6\sin(3x-2y)\)
\(\displaystyle f_{yy}(x,y)=-4\sin(3x-2y)\)
(4)\(f(x,y)=xe^{2y}\)
1階偏導関数を求めると
\(f_x(x,y)=e^{2y}\)
\(f_y(x,y)=2xe^{2y}\)
よって、2階偏導関数は
\(\displaystyle f_{xx}(x,y)=0\)
\(\displaystyle f_{xy}(x,y)=2e^{2y}\)
\(\displaystyle f_{yy}(x,y)=4xe^{2y}\)
(5)\(f(x,y)=x^3y+xy^2\)
1階偏導関数を求めると
\(f_x(x,y)=3x^2y+y^2\)
\(f_y(x,y)=x^3+2xy\)
よって、2階偏導関数は
\(\displaystyle f_{xx}(x,y)=6xy\)
\(\displaystyle f_{xy}(x,y)=3x^2+2y\)
\(\displaystyle f_{yy}(x,y)=2x\)
(6)\(f(x,y)=xy^2e^{\frac{x}{y}}\)
1階偏導関数を求めると
\(f_x(x,y)\)
\(\displaystyle =y^2e^{\frac{x}{y}}+xy^2・\frac{1}{y}・e^{\frac{x}{y}}\)
\(\displaystyle =(y^2+xy)e^{\frac{x}{y}}\)
\(f_y(x,y)\)
\(\displaystyle =2xye^{\frac{x}{y}}+xy^2e^{\frac{x}{y}}・\left(-\frac{x}{y^2}\right)\)
\(\displaystyle =(2xy-x^2)e^{\frac{x}{y}}\)
よって、2階偏導関数は
\(f_{xx}(x,y)\)
\(\displaystyle =ye^{\frac{x}{y}}+(y^2+xy)・\frac{1}{y}・e^{\frac{x}{y}}\)
\(\displaystyle =(x+2y)e^{\frac{x}{y}}\)
\(f_{xy}(x,y)\)
\(\displaystyle =(2y+x)e^{\frac{x}{y}}+(y^2+xy)・\left(-\frac{x}{y^2}\right)・e^{\frac{x}{y}}\)
\(\displaystyle =\left(2y-\frac{x^2}{y}\right)e^{\frac{x}{y}}\)
\(f_{yy}(x,y)\)
\(\displaystyle =2xe^{\frac{x}{y}}+(2xy-x^2)・\left(-\frac{x}{y^2}\right)・e^{\frac{x}{y}}\)
\(\displaystyle =\left(2x-\frac{2x^2}{y}+\frac{x^3}{y^2}\right)e^{\frac{x}{y}}\)
(7)\(f(x,y)=\tan^{-1}(x^2+y^2)\)
1階偏導関数を求めると
\(\displaystyle f_x(x,y)=\frac{2x}{1+(x^2+y^2)^2}\)
\(\displaystyle f_y(x,y)=\frac{2y}{1+(x^2+y^2)^2}\)
よって、2階偏導関数は
\(f_{xx}(x,y)\)
\(\displaystyle =\frac{2(1+(x^2+y^2)^2)-2x(2(x^2+y^2)・2x)}{(1+(x^2+y^2)^2)^2}\)
\(\displaystyle =\frac{2+2(x^2+y^2)^2-8x^2(x^2+y^2)}{(1+(x^2+y^2)^2)^2}\)
\(f_{xy}(x,y)\)
\(\displaystyle =-\frac{2x(2(x^2+y^2)・2y)}{(1+(x^2+y^2)^2)^2}\)
\(\displaystyle =\frac{-8xy(x^2+y^2)}{(1+(x^2+y^2)^2)^2}\)
\(f_{yy}(x,y)\)
\(\displaystyle =\frac{2(1+(x^2+y^2)^2)-2y(2(x^2+y^2)・2y)}{(1+(x^2+y^2)^2)^2}\)
\(\displaystyle =\frac{2+2(x^2+y^2)^2-8y^2(x^2+y^2)}{(1+(x^2+y^2)^2)^2}\)
