【微分積分】7-2-2 全微分の計算｜要点まとめ
このページでは、多変数関数における「全微分の計算」について整理します。全微分の定義に基づく基本的な計算手順や、偏微分を用いた計算の具体例をわかりやすく解説します。また、接平面の方程式の導出や計算のポイントも確認し、多変数微分の基礎力を確実に身につけることを目指します。
全微分の基本的な計算方法
【例題】次の関数が\((0,0)\)で全微分可能か調べなさい。
(1)\(f(x,y)=\left\{\begin{array}{l}\displaystyle xy\sin\frac{1}{\sqrt{x^2+y^2}}\ ((x,y)\neq(0,0)) \\ 0\ ((x,y)=(0,0))\end{array}\right.\)
偏微分係数を計算すると
\(\displaystyle f_x(0,0)=\lim_{h\to0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to0}\frac{0}{h}=0\)
\(\displaystyle f_y(0,0)=\lim_{k\to0}\frac{f(0,k)-f(0,0)}{k}=\lim_{k\to0}\frac{0}{k}=0\)
すなわち
\(\varepsilon(h,k)\)
\(=f(0+h,0+k)\)
\(\ \ \ -\{f(0,0)+f_x(0,0)h+f_y(0,0)k\}\)
\(=f(h,k)\)
\(\displaystyle =hk\sin\frac{1}{\sqrt{h^2+k^2}}\)
ここで、\(\displaystyle |\sin t|\leqq1,|hk|\leqq\frac{h^2+k^2}{2}\)より、
\(\displaystyle \lim_{(h,k)\to(0,0)}\frac{\varepsilon(h,k)}{\sqrt{h^2+k^2}}\)
\(\displaystyle =\lim_{(h,k)\to(0,0)}\frac{hk\sin\frac{1}{\sqrt{h^2+k^2}}}{\sqrt{h^2+k^2}}\)
\(\displaystyle \leqq\lim_{(h,k)\to(0,0)}\frac{h^2+k^2}{2\sqrt{h^2+k^2}}=0\)
よって、全微分可能。
\(\displaystyle f_x(0,0)=\lim_{h\to0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to0}\frac{0}{h}=0\)
\(\displaystyle f_y(0,0)=\lim_{k\to0}\frac{f(0,k)-f(0,0)}{k}=\lim_{k\to0}\frac{0}{k}=0\)
すなわち
\(\varepsilon(h,k)\)
\(=f(0+h,0+k)\)
\(\ \ \ -\{f(0,0)+f_x(0,0)h+f_y(0,0)k\}\)
\(=f(h,k)\)
\(\displaystyle =hk\sin\frac{1}{\sqrt{h^2+k^2}}\)
ここで、\(\displaystyle |\sin t|\leqq1,|hk|\leqq\frac{h^2+k^2}{2}\)より、
\(\displaystyle \lim_{(h,k)\to(0,0)}\frac{\varepsilon(h,k)}{\sqrt{h^2+k^2}}\)
\(\displaystyle =\lim_{(h,k)\to(0,0)}\frac{hk\sin\frac{1}{\sqrt{h^2+k^2}}}{\sqrt{h^2+k^2}}\)
\(\displaystyle \leqq\lim_{(h,k)\to(0,0)}\frac{h^2+k^2}{2\sqrt{h^2+k^2}}=0\)
よって、全微分可能。
(2)\(f(x,y)=\sqrt{|xy|}\)
偏微分係数を計算すると
\(\displaystyle f_x(0,0)=\lim_{h\to0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to0}\frac{0}{h}=0\)
\(\displaystyle f_y(0,0)=\lim_{k\to0}\frac{f(0,k)-f(0,0)}{k}=\lim_{k\to0}\frac{0}{k}=0\)
すなわち
\(\varepsilon(h,k)\)
\(=f(0+h,0+k)\)
\(\ \ \ -\{f(0,0)+f_x(0,0)h+f_y(0,0)k\}\)
\(=f(h,k)\)
\(=\sqrt{|hk|}\)
\(y=x\)軸上で原点に近づくとき、\(x(t)=t,y(t)=t\)として\(t\to0\)とすればよいので
\(\displaystyle \lim_{t\to0}\frac{\varepsilon(t,t)}{\sqrt{t^2+t^2}}\)
\(\displaystyle =\lim_{t\to0}\frac{|t|}{\sqrt{2}|t|}\)
\(\displaystyle =\frac{1}{\sqrt{2}}\)
よって、全微分可能ではない。
\(\displaystyle f_x(0,0)=\lim_{h\to0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to0}\frac{0}{h}=0\)
\(\displaystyle f_y(0,0)=\lim_{k\to0}\frac{f(0,k)-f(0,0)}{k}=\lim_{k\to0}\frac{0}{k}=0\)
すなわち
\(\varepsilon(h,k)\)
\(=f(0+h,0+k)\)
\(\ \ \ -\{f(0,0)+f_x(0,0)h+f_y(0,0)k\}\)
\(=f(h,k)\)
\(=\sqrt{|hk|}\)
\(y=x\)軸上で原点に近づくとき、\(x(t)=t,y(t)=t\)として\(t\to0\)とすればよいので
\(\displaystyle \lim_{t\to0}\frac{\varepsilon(t,t)}{\sqrt{t^2+t^2}}\)
\(\displaystyle =\lim_{t\to0}\frac{|t|}{\sqrt{2}|t|}\)
\(\displaystyle =\frac{1}{\sqrt{2}}\)
よって、全微分可能ではない。
(3)\(f(x,y)=\left\{\begin{array}{l}\displaystyle xy\sin^{-1}\frac{x^2-y^2}{x^2+y^2}\ ((x,y)\neq(0,0)) \\ 0\ ((x,y)=(0,0))\end{array}\right.\)
偏微分係数を計算すると
\(\displaystyle f_x(0,0)=\lim_{h\to0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to0}\frac{0}{h}=0\)
\(\displaystyle f_y(0,0)=\lim_{k\to0}\frac{f(0,k)-f(0,0)}{k}=\lim_{k\to0}\frac{0}{k}=0\)
すなわち
\(\varepsilon(h,k)\)
\(=f(0+h,0+k)\)
\(\ \ \ -\{f(0,0)+f_x(0,0)h+f_y(0,0)k\}\)
\(=f(h,k)\)
\(\displaystyle =hk\sin^{-1}\frac{h^2-k^2}{h^2+k^2}\)
ここで、\(\displaystyle |\sin^{-1}t|\leqq\frac{\pi}{2},|hk|\leqq\frac{h^2+k^2}{2}\)より、
\(\displaystyle \lim_{(h,k)\to(0,0)}\frac{\varepsilon(h,k)}{\sqrt{h^2+k^2}}\)
\(\displaystyle =\lim_{(h,k)\to(0,0)}\frac{hk\sin^{-1}\frac{h^2-k^2}{h^2+k^2}}{\sqrt{h^2+k^2}}\)
\(\displaystyle \leqq\lim_{(h,k)\to(0,0)}\frac{h^2+k^2}{2\sqrt{h^2+k^2}}=0\)
よって、全微分可能。
\(\displaystyle f_x(0,0)=\lim_{h\to0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to0}\frac{0}{h}=0\)
\(\displaystyle f_y(0,0)=\lim_{k\to0}\frac{f(0,k)-f(0,0)}{k}=\lim_{k\to0}\frac{0}{k}=0\)
すなわち
\(\varepsilon(h,k)\)
\(=f(0+h,0+k)\)
\(\ \ \ -\{f(0,0)+f_x(0,0)h+f_y(0,0)k\}\)
\(=f(h,k)\)
\(\displaystyle =hk\sin^{-1}\frac{h^2-k^2}{h^2+k^2}\)
ここで、\(\displaystyle |\sin^{-1}t|\leqq\frac{\pi}{2},|hk|\leqq\frac{h^2+k^2}{2}\)より、
\(\displaystyle \lim_{(h,k)\to(0,0)}\frac{\varepsilon(h,k)}{\sqrt{h^2+k^2}}\)
\(\displaystyle =\lim_{(h,k)\to(0,0)}\frac{hk\sin^{-1}\frac{h^2-k^2}{h^2+k^2}}{\sqrt{h^2+k^2}}\)
\(\displaystyle \leqq\lim_{(h,k)\to(0,0)}\frac{h^2+k^2}{2\sqrt{h^2+k^2}}=0\)
よって、全微分可能。
(4)\(f(x,y)=\left\{\begin{array}{l}\displaystyle \frac{x^3+x^2-y^3+y^2}{x^2+y^2}\ ((x,y)\neq(0,0)) \\ 1\ ((x,y)=(0,0))\end{array}\right.\)
偏微分係数を計算すると
\(\displaystyle f_x(0,0)=\lim_{h\to0}\frac{f(h,0)-f(0,0)}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{(h+1)-1}{h}\)
\(=1\)
\(\displaystyle f_y(0,0)=\lim_{k\to0}\frac{f(0,k)-f(0,0)}{k}\)
\(\displaystyle =\lim_{k\to0}\frac{(-k+1)-1}{k}\)
\(=-1\)
すなわち
\(\varepsilon(h,k)\)
\(=f(0+h,0+k)\)
\(\ \ \ -\{f(0,0)+f_x(0,0)h+f_y(0,0)k\}\)
\(\displaystyle =\frac{h^3+h^2-k^3+k^2}{h^2+k^2}-(1+h-k)\)
\(\displaystyle =\frac{h^3-k^3}{h^2+k^2}\)
\(x\)軸上で原点に近づくとき、\(x(t)=t,y(t)=0\)として\(t\to0\)とすればよいので
\(\displaystyle \lim_{t\to+0}\frac{\varepsilon(t,0)}{\sqrt{t^2}}\)
\(\displaystyle =\lim_{t\to+0}\frac{t}{|t|}\)
\(\displaystyle =1\)
\(\displaystyle \lim_{t\to-0}\frac{\varepsilon(t,0)}{\sqrt{t^2}}\)
\(\displaystyle =\lim_{t\to-0}\frac{t}{|t|}\)
\(\displaystyle =-1\)
よって、全微分可能ではない。
\(\displaystyle f_x(0,0)=\lim_{h\to0}\frac{f(h,0)-f(0,0)}{h}\)
\(\displaystyle =\lim_{h\to0}\frac{(h+1)-1}{h}\)
\(=1\)
\(\displaystyle f_y(0,0)=\lim_{k\to0}\frac{f(0,k)-f(0,0)}{k}\)
\(\displaystyle =\lim_{k\to0}\frac{(-k+1)-1}{k}\)
\(=-1\)
すなわち
\(\varepsilon(h,k)\)
\(=f(0+h,0+k)\)
\(\ \ \ -\{f(0,0)+f_x(0,0)h+f_y(0,0)k\}\)
\(\displaystyle =\frac{h^3+h^2-k^3+k^2}{h^2+k^2}-(1+h-k)\)
\(\displaystyle =\frac{h^3-k^3}{h^2+k^2}\)
\(x\)軸上で原点に近づくとき、\(x(t)=t,y(t)=0\)として\(t\to0\)とすればよいので
\(\displaystyle \lim_{t\to+0}\frac{\varepsilon(t,0)}{\sqrt{t^2}}\)
\(\displaystyle =\lim_{t\to+0}\frac{t}{|t|}\)
\(\displaystyle =1\)
\(\displaystyle \lim_{t\to-0}\frac{\varepsilon(t,0)}{\sqrt{t^2}}\)
\(\displaystyle =\lim_{t\to-0}\frac{t}{|t|}\)
\(\displaystyle =-1\)
よって、全微分可能ではない。
【例題】次の関数の全微分を求めなさい。
(1)\(\displaystyle f(x,y)=\tan^{-1}\frac{y}{x}\)
偏導関数を計算すると
\(\displaystyle f_x(x,y)=\frac{1}{1+(\frac{y}{x})^2}・\left(-\frac{y}{x^2}\right)=\frac{-y}{x^2+y^2}\)
\(\displaystyle f_y(x,y)=\frac{1}{1+(\frac{y}{x})^2}・\frac{1}{x}=\frac{x}{x^2+y^2}\)
よって、
\(df=f_x(x,y)dx+f_y(x,y)dy\)
\(\displaystyle =\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy\)
\(\displaystyle f_x(x,y)=\frac{1}{1+(\frac{y}{x})^2}・\left(-\frac{y}{x^2}\right)=\frac{-y}{x^2+y^2}\)
\(\displaystyle f_y(x,y)=\frac{1}{1+(\frac{y}{x})^2}・\frac{1}{x}=\frac{x}{x^2+y^2}\)
よって、
\(df=f_x(x,y)dx+f_y(x,y)dy\)
\(\displaystyle =\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy\)
(2)\(f(x,y)=\log(x^2+2y^2-xy)\)
偏導関数を計算すると
\(\displaystyle f_x(x,y)=\frac{2x-y}{x^2+2y^2-xy}\)
\(\displaystyle f_y(x,y)=\frac{4y-x}{x^2+2y^2-xy}\)
よって、
\(df=f_x(x,y)dx+f_y(x,y)dy\)
\(\displaystyle =\frac{2x-y}{x^2+2y^2-xy}dx+\frac{4y-x}{x^2+2y^2-xy}dy\)
\(\displaystyle f_x(x,y)=\frac{2x-y}{x^2+2y^2-xy}\)
\(\displaystyle f_y(x,y)=\frac{4y-x}{x^2+2y^2-xy}\)
よって、
\(df=f_x(x,y)dx+f_y(x,y)dy\)
\(\displaystyle =\frac{2x-y}{x^2+2y^2-xy}dx+\frac{4y-x}{x^2+2y^2-xy}dy\)
(3)\(f(x,y)=\cosh(x^2y)\)
偏導関数を計算すると
\(f_x(x,y)=2xy\sinh(x^2y)\)
\(f_y(x,y)=x^2\sinh(x^2y)\)
よって、
\(df=f_x(x,y)dx+f_y(x,y)dy\)
\(=2xy\sinh(x^2y)dx+x^2\sinh(x^2y)dy\)
\(f_x(x,y)=2xy\sinh(x^2y)\)
\(f_y(x,y)=x^2\sinh(x^2y)\)
よって、
\(df=f_x(x,y)dx+f_y(x,y)dy\)
\(=2xy\sinh(x^2y)dx+x^2\sinh(x^2y)dy\)
(4)\(f(x,y)=x^y\)
偏導関数を計算すると
\(f_x(x,y)=yx^{y-1}\)
\(f_y(x,y)=x^y\log x\)
よって、
\(df=f_x(x,y)dx+f_y(x,y)dy\)
\(=yx^{y-1}dx+x^y\log xdy\)
\(f_x(x,y)=yx^{y-1}\)
\(f_y(x,y)=x^y\log x\)
よって、
\(df=f_x(x,y)dx+f_y(x,y)dy\)
\(=yx^{y-1}dx+x^y\log xdy\)
【例題】次の曲面の与えられた点\(A\)における接平面の方程式を求めなさい。
(1)\(f(x,y)=x^2+y^2\)
\(A(1,2,f(1,2))\)
\(A(1,2,f(1,2))\)
偏導関数を計算すると
\(\displaystyle f_x(x,y)=2x\)
\(\displaystyle f_y(x,y)=2y\)
よって、求める接平面の方程式\(z\)は
\(z=f(1,2)+f_x(1,2)(x-1)+f_y(1,2)(y-2)\)
\(=5+2(x-1)+4(y-2)\)
\(=2x+4y-5\)
\(\displaystyle f_x(x,y)=2x\)
\(\displaystyle f_y(x,y)=2y\)
よって、求める接平面の方程式\(z\)は
\(z=f(1,2)+f_x(1,2)(x-1)+f_y(1,2)(y-2)\)
\(=5+2(x-1)+4(y-2)\)
\(=2x+4y-5\)
(2)\(f(x,y)=\log(x^2+y^2-1)\)
\(A(2,-1,f(2,-1))\)
\(A(2,-1,f(2,-1))\)
偏導関数を計算すると
\(\displaystyle f_x(x,y)=\frac{2x}{x^2+y^2-1}\)
\(\displaystyle f_y(x,y)=\frac{2y}{x^2+y^2-1}\)
よって、求める接平面の方程式\(z\)は
\(z=f(2,-1)\)
\(\ \ \ +f_x(2,-1)(x-2)+f_y(2,-1)(y+1)\)
\(\displaystyle =2\log2+1(x-2)-\frac{1}{2}(y+1)\)
\(\displaystyle =x-\frac{y}{2}-\frac{5}{2}+2\log2\)
\(\displaystyle f_x(x,y)=\frac{2x}{x^2+y^2-1}\)
\(\displaystyle f_y(x,y)=\frac{2y}{x^2+y^2-1}\)
よって、求める接平面の方程式\(z\)は
\(z=f(2,-1)\)
\(\ \ \ +f_x(2,-1)(x-2)+f_y(2,-1)(y+1)\)
\(\displaystyle =2\log2+1(x-2)-\frac{1}{2}(y+1)\)
\(\displaystyle =x-\frac{y}{2}-\frac{5}{2}+2\log2\)
(3)\(f(x,y)=e^{x+y}\cos(x-y)\)
\(A(0,\pi,f(0,\pi))\)
\(A(0,\pi,f(0,\pi))\)
偏導関数を計算すると
\(f_x(x,y)=e^{x+y}\{\cos(x-y)-\sin(x-y)\}\)
\(f_y(x,y)=e^{x+y}\{\cos(x-y)+\sin(x-y)\}\)
よって、求める接平面の方程式\(z\)は
\(z=f(0,\pi)+f_x(0,\pi)(x-0)+f_y(0,\pi)(y-\pi)\)
\(=-e^\pi-e^\pi(x-0)-e^\pi(y-\pi)\)
\(=-e^\pi x-e^\pi y+(\pi-1)e^\pi\)
\(f_x(x,y)=e^{x+y}\{\cos(x-y)-\sin(x-y)\}\)
\(f_y(x,y)=e^{x+y}\{\cos(x-y)+\sin(x-y)\}\)
よって、求める接平面の方程式\(z\)は
\(z=f(0,\pi)+f_x(0,\pi)(x-0)+f_y(0,\pi)(y-\pi)\)
\(=-e^\pi-e^\pi(x-0)-e^\pi(y-\pi)\)
\(=-e^\pi x-e^\pi y+(\pi-1)e^\pi\)
(4)\(\displaystyle f(x,y)=\sin^{-1}\frac{y}{x}\)
\(A(2,1,f(2,1))\)
\(A(2,1,f(2,1))\)
偏導関数を計算すると
\(\displaystyle f_x(x,y)=-\frac{y}{x\sqrt{x^2-y^2}}\)
\(\displaystyle f_y(x,y)=\frac{1}{\sqrt{x^2-y^2}}\)
よって、求める接平面の方程式\(z\)は
\(z=f(2,1)+f_x(2,1)(x-2)+f_y(2,1)(y-1)\)
\(\displaystyle =\frac{\pi}{6}-\frac{1}{2\sqrt{3}}(x-2)+\frac{1}{\sqrt{3}}(y-1)\)
\(\displaystyle =-\frac{x}{2\sqrt{3}}+\frac{y}{\sqrt{3}}+\frac{\pi}{6}\)
\(\displaystyle f_x(x,y)=-\frac{y}{x\sqrt{x^2-y^2}}\)
\(\displaystyle f_y(x,y)=\frac{1}{\sqrt{x^2-y^2}}\)
よって、求める接平面の方程式\(z\)は
\(z=f(2,1)+f_x(2,1)(x-2)+f_y(2,1)(y-1)\)
\(\displaystyle =\frac{\pi}{6}-\frac{1}{2\sqrt{3}}(x-2)+\frac{1}{\sqrt{3}}(y-1)\)
\(\displaystyle =-\frac{x}{2\sqrt{3}}+\frac{y}{\sqrt{3}}+\frac{\pi}{6}\)
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