関数 $f(x, y) = 2x^3 + x^2y - 3xy^2 - 4y^3$ の2階偏微分 $f_{xx}$, $f_{yy}$, $f_{xy}$, $f_{yx}$ をそれぞれ求める。解析学偏微分多変数関数2階偏微分2025/5/141. 問題の内容関数 f(x,y)=2x3+x2y−3xy2−4y3f(x, y) = 2x^3 + x^2y - 3xy^2 - 4y^3f(x,y)=2x3+x2y−3xy2−4y3 の2階偏微分 fxxf_{xx}fxx, fyyf_{yy}fyy, fxyf_{xy}fxy, fyxf_{yx}fyx をそれぞれ求める。2. 解き方の手順まず、1階偏微分 fxf_xfx と fyf_yfy を計算します。fx=∂f∂x=6x2+2xy−3y2f_x = \frac{\partial f}{\partial x} = 6x^2 + 2xy - 3y^2fx=∂x∂f=6x2+2xy−3y2fy=∂f∂y=x2−6xy−12y2f_y = \frac{\partial f}{\partial y} = x^2 - 6xy - 12y^2fy=∂y∂f=x2−6xy−12y2次に、2階偏微分を計算します。fxx=∂2f∂x2=∂fx∂x=12x+2yf_{xx} = \frac{\partial^2 f}{\partial x^2} = \frac{\partial f_x}{\partial x} = 12x + 2yfxx=∂x2∂2f=∂x∂fx=12x+2yfyy=∂2f∂y2=∂fy∂y=−6x−24yf_{yy} = \frac{\partial^2 f}{\partial y^2} = \frac{\partial f_y}{\partial y} = -6x - 24yfyy=∂y2∂2f=∂y∂fy=−6x−24yfxy=∂2f∂x∂y=∂fx∂y=2x−6yf_{xy} = \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial f_x}{\partial y} = 2x - 6yfxy=∂x∂y∂2f=∂y∂fx=2x−6yfyx=∂2f∂y∂x=∂fy∂x=2x−6yf_{yx} = \frac{\partial^2 f}{\partial y \partial x} = \frac{\partial f_y}{\partial x} = 2x - 6yfyx=∂y∂x∂2f=∂x∂fy=2x−6y3. 最終的な答えfxx=12x+2yf_{xx} = 12x + 2yfxx=12x+2yfyy=−6x−24yf_{yy} = -6x - 24yfyy=−6x−24yfxy=2x−6yf_{xy} = 2x - 6yfxy=2x−6yfyx=2x−6yf_{yx} = 2x - 6yfyx=2x−6y