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問題4.2.5

合成関数の微分を用いて$z_u$、$z_v$を求めよ。

（１）$z=x{y}^2+x^{2}y$、$x=u+v$、$y=u-v$

（２）$z=\sin (x-y)$、$x=u^2+v^2$、$y=2uv$

（３）$z=f(x,y)$、$x=2u-3v$、$y=u-5v$

（４）$z=f(x,y)$、$x=\cos (u+v)$、$y=\sin (u-v)$

（５）$z=f(x+3y)$、$x=u-2v$、$y=3u-4v$

 

《ポイント》

合成関数の偏微分は以下の公式で与えられます。これは「連鎖律」（chain rule）と呼ばれる合成関数の微分に関する性質で、高校数学の内容ではいわゆる「缶詰めの微分」として登場するものです。$$\{\begin{array}{l}{\displaystyle \frac{\partial z}{\partial u}}={\displaystyle \frac{\partial z}{\partial x}}{\displaystyle \frac{\partial x}{\partial u}}+{\displaystyle \frac{\partial z}{\partial y}}{\displaystyle \frac{\partial y}{\partial u}}\\ {\displaystyle \frac{\partial z}{\partial v}}={\displaystyle \frac{\partial z}{\partial x}}{\displaystyle \frac{\partial x}{\partial v}}+{\displaystyle \frac{\partial z}{\partial y}}{\displaystyle \frac{\partial y}{\partial v}}\end{array}$$

 

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《解答例》

（１）$z=x{y}^2+x^{2}y$、$x=u+v$、$y=u-v$

$$\begin{array}{rl}{\displaystyle \frac{\partial z}{\partial u}}& ={\displaystyle \frac{\partial z}{\partial x}}{\displaystyle \frac{\partial x}{\partial u}}+{\displaystyle \frac{\partial z}{\partial y}}{\displaystyle \frac{\partial y}{\partial u}}\\ & =(y^2+2xy)\cdot 1+(2xy+x^2)\cdot 1\\ & =x^2+y^2+4xy\\ & =(x+y{)}^2+2xy\\ & =4u^2+2(u^2-v^2)\\ & =6u^2-2v^2\cdots \cdots (\text{答})\end{array}$$ $$\begin{array}{rl}{\displaystyle \frac{\partial z}{\partial v}}& ={\displaystyle \frac{\partial z}{\partial x}}{\displaystyle \frac{\partial x}{\partial v}}+{\displaystyle \frac{\partial z}{\partial y}}{\displaystyle \frac{\partial y}{\partial v}}\\ & =(y^2+2xy)\cdot 1+(2xy+x^2)\cdot (-1)\\ & =y^2-x^2\\ & =(y+x)(y-x)\\ & =2u(-2v)\\ & =-4uv\cdots \cdots (\text{答})\end{array}$$

 

（２）$z=\sin (x-y)$、$x=u^2+v^2$、$y=2uv$

$$\begin{array}{rl}{\displaystyle \frac{\partial z}{\partial u}}& ={\displaystyle \frac{\partial z}{\partial x}}{\displaystyle \frac{\partial x}{\partial u}}+{\displaystyle \frac{\partial z}{\partial y}}{\displaystyle \frac{\partial y}{\partial u}}\\ & =2u\cos (x-y)-2v\cos (x-y)\\ & =2(u-v)\cos (x-y)\\ & =2(u-v)\cos (u^2+v^2-2uv)\\ & =2(u-v)\cos (u-v{)}^2\cdots \cdots (\text{答})\end{array}$$ $$\begin{array}{rl}{\displaystyle \frac{\partial z}{\partial v}}& ={\displaystyle \frac{\partial z}{\partial x}}{\displaystyle \frac{\partial x}{\partial v}}+{\displaystyle \frac{\partial z}{\partial y}}{\displaystyle \frac{\partial y}{\partial v}}\\ & =2v\cos (x-y)-2u\cos (x-y)\\ & =2(v-u)\cos (x-y)\\ & =2(v-u)\cos (u^2+v^2-2uv)\\ & =2(v-u)\cos (u-v{)}^2\cdots \cdots (\text{答})\end{array}$$
※$z_v$ですが、$2(v-u)\cos (v-u{)}^2$と文字の並びを揃えた形で解答しても良いでしょう。

 

（３）$z=f(x,y)$、$x=2u-3v$、$y=u-5v$

$$\begin{array}{rl}{\displaystyle \frac{\partial z}{\partial u}}& ={\displaystyle \frac{\partial z}{\partial x}}{\displaystyle \frac{\partial x}{\partial u}}+{\displaystyle \frac{\partial z}{\partial y}}{\displaystyle \frac{\partial y}{\partial u}}\\ & =f_x(x,y)\cdot 2+f_y(x,y)\cdot 1\\ & =2f_x(2u-3v,u-5v)+f_y(2u-3v,u-5v)\cdots \cdots (\text{答})\end{array}$$ $$\begin{array}{rl}{\displaystyle \frac{\partial z}{\partial v}}& ={\displaystyle \frac{\partial z}{\partial x}}{\displaystyle \frac{\partial x}{\partial v}}+{\displaystyle \frac{\partial z}{\partial y}}{\displaystyle \frac{\partial y}{\partial v}}\\ & =f_x(x,y)\cdot (-3)+f_y(x,y)\cdot (-5)\\ & =-3f_x(2u-3v,u-5v)-5f_y(2u-3v,u-5v)\cdots \cdots (\text{答})\end{array}$$

 

（４）$z=f(x,y)$、$x=\cos (u+v)$、$y=\sin (u-v)$

$$\begin{array}{rl}{\displaystyle \frac{\partial z}{\partial u}}& ={\displaystyle \frac{\partial z}{\partial x}}{\displaystyle \frac{\partial x}{\partial u}}+{\displaystyle \frac{\partial z}{\partial y}}{\displaystyle \frac{\partial y}{\partial u}}\\ & =f_x(x,y)\cdot (-\sin (u+v))+f_y(x,y)\cdot \cos (u-v)\\ & =-\sin (u+v)f_x(\cos (u+v),\sin (u-v))\\ & +\cos (u-v)f_y(\cos (u+v),\sin (u-v))\cdots \cdots (\text{答})\end{array}$$ $$\begin{array}{rl}{\displaystyle \frac{\partial z}{\partial v}}& ={\displaystyle \frac{\partial z}{\partial x}}{\displaystyle \frac{\partial x}{\partial v}}+{\displaystyle \frac{\partial z}{\partial y}}{\displaystyle \frac{\partial y}{\partial v}}\\ & =f_x(x,y)\cdot (-\sin (u+v))+f_y(x,y)\cdot (-\cos (u-v))\\ & =-\sin (u+v)f_x(\cos (u+v),\sin (u-v))\\ & -\cos (u-v)f_y(\cos (u+v),\sin (u-v))\cdots \cdots (\text{答})\end{array}$$

 

（５）$z=f(x+3y)$、$x=u-2v$、$y=3u-4v$

$$\begin{array}{rl}{\displaystyle \frac{\partial z}{\partial u}}& ={\displaystyle \frac{\partial z}{\partial x}}{\displaystyle \frac{\partial x}{\partial u}}+{\displaystyle \frac{\partial z}{\partial y}}{\displaystyle \frac{\partial y}{\partial u}}\\ & =f^{\prime }(x+3y)\cdot 1+3f^{\prime }(x+3y)\cdot 3\\ & =10f^{\prime }(10u-14v)\cdots \cdots (\text{答})\end{array}$$ $$\begin{array}{rl}{\displaystyle \frac{\partial z}{\partial v}}& ={\displaystyle \frac{\partial z}{\partial x}}{\displaystyle \frac{\partial x}{\partial v}}+{\displaystyle \frac{\partial z}{\partial y}}{\displaystyle \frac{\partial y}{\partial v}}\\ & =f^{\prime }(x+3y)\cdot (-2)+3f^{\prime }(x+3y)\cdot (-4)\\ & =-14f^{\prime }(10u-14v)\cdots \cdots (\text{答})\end{array}$$

 

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復習例題は設定していません。

 

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