【微分積分】7-3-2 高階偏導関数の計算｜要点まとめ
このページでは、多変数関数に対する「高階偏導関数の計算方法」について整理します。2階偏導関数や混合偏導関数の具体的な計算手順、偏微分の順序に関する注意点、計算時に起こりやすいミスのポイントなどを中心に、大学数学で頻出の基本事項をわかりやすく解説します。今後の極値判定やテイラー展開、偏微分方程式の理解につながる基礎計算力を身につけていきましょう。
高階偏導関数の計算
【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
\(f_{xx}(x,y)=6x+6y\)
\(f_{xy}(x,y)=f_{yx}(x,y)=6x\)
\(f_{yy}(x,y)=2\)
\(f_x(x,y)=3x^2+6xy\)
\(f_y(x,y)=3x^2+2y\)
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
\(f_{xx}(x,y)=e^x\cos y\)
\(f_{xy}(x,y)=f_{yx}(x,y)=-e^x\sin y\)
\(f_{yy}(x,y)=-e^x\cos y\)
\(f_x(x,y)=e^x\cos y\)
\(f_y(x,y)=-e^x\sin y\)
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
\(\displaystyle f_{xx}(x,y)=\frac{2(-x^2+y^2+1)}{(x^2+y^2+1)^2}\)
\(\displaystyle f_{xy}(x,y)=f_{yx}(x,y)=-\frac{4xy}{(x^2+y^2+1)^2}\)
\(\displaystyle f_{yy}(x,y)=\frac{2(x^2-y^2+1)}{(x^2+y^2+1)^2}\)
\(\displaystyle f_x(x,y)=\frac{2x}{x^2+y^2+1}\)
\(\displaystyle f_y(x,y)=\frac{2y}{x^2+y^2+1}\)
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
\(\displaystyle f_{xx}(x,y)=\frac{2xy}{(x^2+y^2)^2}\)
\(\displaystyle f_{xy}(x,y)=f_{yx}(x,y)=\frac{y^2-x^2}{(x^2+y^2)^2}\)
\(\displaystyle f_{yy}(x,y)=\frac{-2xy}{(x^2+y^2)^2}\)
\(\displaystyle f_x(x,y)=\frac{-y}{x^2+y^2}\)
\(\displaystyle f_y(x,y)=\frac{x}{x^2+y^2}\)
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
\(f_{xx}(x,y)\)
\(\displaystyle =\frac{-4x+2y}{(x^2+y^2)^2}\)
\(\displaystyle \ \ \ +(-2x^2+2xy+2y^2)・\frac{-4x}{(x^2+y^2)^3}\)
\(\displaystyle =\frac{4x^3-6x^2y-12xy^2+2y^3}{(x^2+y^2)^3}\)
\(f_{xy}(x,y)=f_{yx}(x,y)\)
\(\displaystyle =\frac{2x+4y}{(x^2+y^2)^2}\)
\(\displaystyle \ \ \ +(-2x^2+2xy+2y^2)・\frac{-4y}{(x^2+y^2)^3}\)
\(\displaystyle =\frac{2x^3+12x^2y-6xy^2-4y^3}{(x^2+y^2)^3}\)
\(f_{yy}(x,y)\)
\(\displaystyle =\frac{-4x+2y}{(x^2+y^2)^2}\)
\(\displaystyle \ \ \ +(-x^2-4xy+y^2)・\frac{-4y}{(x^2+y^2)^3}\)
\(\displaystyle =\frac{-4x^3+6x^2y+12xy^2-2y^3}{(x^2+y^2)^3}\)
\(f_x(x,y)\)
\(\displaystyle =\frac{2(x^2+y^2)-(2x-y)・2x}{(x^2+y^2)^2}\)
\(\displaystyle =\frac{-2x^2+2xy+2y^2}{(x^2+y^2)^2}\)
\(f_y(x,y)\)
\(\displaystyle =\frac{-(x^2+y^2)-(2x-y)・2y}{(x^2+y^2)^2}\)
\(\displaystyle =\frac{-x^2-4xy+y^2}{(x^2+y^2)^2}\)
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
\(f_{xx}(x,y)\)
\(\displaystyle =2\tan^{-1}\frac{y}{x}+2x・\frac{-y}{x^2+y^2}\)
\(\displaystyle =2\tan^{-1}\frac{y}{x}-\frac{2xy}{x^2+y^2}\)
\(f_{xy}(x,y)=f_{yx}(x,y)\)
\(\displaystyle =2x・\frac{x}{x^2+y^2}-1\)
\(\displaystyle =\frac{x^2-y^2}{x^2+y^2}\)
\(f_{yy}(x,y)\)
\(\displaystyle =-2\tan^{-1}\frac{x}{y}-2y・\frac{-x}{x^2+y^2}\)
\(\displaystyle =-2\tan^{-1}\frac{x}{y}+\frac{2xy}{x^2+y^2}\)
\(f_x(x,y)\)
\(\displaystyle =2x\tan^{-1}\frac{y}{x}+x^2・\frac{-y}{x^2+y^2}-y^2・\frac{y}{x^2+y^2}\)
\(\displaystyle =2x\tan^{-1}\frac{y}{x}-y\)
\(f_y(x,y)\)
\(\displaystyle =x^2・\frac{x}{x^2+y^2}-2y\tan^{-1}\frac{x}{y}-y^2・\frac{-x}{x^2+y^2}\)
\(\displaystyle =-2y\tan^{-1}\frac{x}{y}+x\)
よって、2階偏導関数は
\(f_{xx}(x,y)\)
\(\displaystyle =2\tan^{-1}\frac{y}{x}+2x・\frac{-y}{x^2+y^2}\)
\(\displaystyle =2\tan^{-1}\frac{y}{x}-\frac{2xy}{x^2+y^2}\)
\(f_{xy}(x,y)=f_{yx}(x,y)\)
\(\displaystyle =2x・\frac{x}{x^2+y^2}-1\)
\(\displaystyle =\frac{x^2-y^2}{x^2+y^2}\)
\(f_{yy}(x,y)\)
\(\displaystyle =-2\tan^{-1}\frac{x}{y}-2y・\frac{-x}{x^2+y^2}\)
\(\displaystyle =-2\tan^{-1}\frac{x}{y}+\frac{2xy}{x^2+y^2}\)
【例題】\(f(x,y)=\left\{\begin{array}{l}\displaystyle \frac{xy(x^2-y^2)}{x^2+y^2}\ ((x,y)\neq(0,0)) \\ 0\ ((x,y)=(0,0))\end{array}\right.\)
のとき、\(f_{xy}(0,0)\)と\(f_{yx}(0,0)\)の値を求めなさい。
のとき、\(f_{xy}(0,0)\)と\(f_{yx}(0,0)\)の値を求めなさい。
1階偏導関数を求めると
\(\displaystyle f_x(x,y)=\frac{y(x^4-y^4+4x^2y^2)}{(x^2+y^2)^2}\)
\(\displaystyle f_y(x,y)=\frac{x(x^4-y^4-4x^2y^2)}{(x^2+y^2)^2}\)
微分係数を求めると
\(f_x(0,0)\)
\(\displaystyle =\lim_{h\to0}\frac{f(0+h,0)-f(0,0)}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{0-0}{h}\)
\(=0\)
\(f_y(0,0)\)
\(\displaystyle =\lim_{k\to0}\frac{f(0,0+k)-f(0,0)}{k}\)
\(\displaystyle =\lim_{k\to0}\frac{0-0}{k}\)
\(=0\)
よって、
\(f_{xy}(0,0)\)
\(\displaystyle =\lim_{k\to0}\frac{f_x(0,0+k)-f_x(0,0)}{k}\)
\(\displaystyle =\lim_{k\to0}\frac{-k-0}{k}\)
\(=-1\)
\(f_{yx}(0,0)\)
\(\displaystyle =\lim_{h\to0}\frac{f_y(0+h,0)-f_y(0,0)}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{h-0}{h}\)
\(=1\)
\(\displaystyle f_x(x,y)=\frac{y(x^4-y^4+4x^2y^2)}{(x^2+y^2)^2}\)
\(\displaystyle f_y(x,y)=\frac{x(x^4-y^4-4x^2y^2)}{(x^2+y^2)^2}\)
微分係数を求めると
\(f_x(0,0)\)
\(\displaystyle =\lim_{h\to0}\frac{f(0+h,0)-f(0,0)}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{0-0}{h}\)
\(=0\)
\(f_y(0,0)\)
\(\displaystyle =\lim_{k\to0}\frac{f(0,0+k)-f(0,0)}{k}\)
\(\displaystyle =\lim_{k\to0}\frac{0-0}{k}\)
\(=0\)
よって、
\(f_{xy}(0,0)\)
\(\displaystyle =\lim_{k\to0}\frac{f_x(0,0+k)-f_x(0,0)}{k}\)
\(\displaystyle =\lim_{k\to0}\frac{-k-0}{k}\)
\(=-1\)
\(f_{yx}(0,0)\)
\(\displaystyle =\lim_{h\to0}\frac{f_y(0+h,0)-f_y(0,0)}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{h-0}{h}\)
\(=1\)
【例題】次の関数は偏微分方程式\(\displaystyle \frac{\partial u}{\partial t}=\frac{\partial^2u}{\partial x^2}\)を証明しなさい。
\(\displaystyle u(x,t)=\frac{1}{\sqrt{t}}e^{-\frac{x^2}{4t}}\ \ \ (x\in\mathbb{R},t>0)\)
両辺の対数をとると
\(\displaystyle \log u(x,t)=-\frac{1}{2}\log t-\frac{x^2}{4t}\)
1階偏導関数を求めると
\(\displaystyle u_t(x,t)=\left(-\frac{1}{2t}+\frac{x^2}{4t^2}\right)u(x,t)\)
\(\displaystyle u_x(x,t)=-\frac{x}{2t}u(x,t)\)
よって、
\(u_{xx}(x,t)\)
\(\displaystyle =-\frac{1}{2t}u(x,t)-\frac{x}{2t}u_x(x,t)\)
\(\displaystyle =-\frac{1}{2t}u(x,t)-\frac{x}{2t}\left(-\frac{x}{2t}u(x,t)\right)\)
\(\displaystyle =\left(-\frac{1}{2t}+\frac{x^2}{4t^2}\right)u(x,t)\)
\(=u_t(x,t)\)
\(\displaystyle \log u(x,t)=-\frac{1}{2}\log t-\frac{x^2}{4t}\)
1階偏導関数を求めると
\(\displaystyle u_t(x,t)=\left(-\frac{1}{2t}+\frac{x^2}{4t^2}\right)u(x,t)\)
\(\displaystyle u_x(x,t)=-\frac{x}{2t}u(x,t)\)
よって、
\(u_{xx}(x,t)\)
\(\displaystyle =-\frac{1}{2t}u(x,t)-\frac{x}{2t}u_x(x,t)\)
\(\displaystyle =-\frac{1}{2t}u(x,t)-\frac{x}{2t}\left(-\frac{x}{2t}u(x,t)\right)\)
\(\displaystyle =\left(-\frac{1}{2t}+\frac{x^2}{4t^2}\right)u(x,t)\)
\(=u_t(x,t)\)
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