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）$z=xy^2+x^2y$、$x=u+v$、$y=u-v$
（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。$$\begin{cases} \dfrac{\partial z}{\partial u} =\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial u}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial u} \\ \dfrac{\partial z}{\partial v} =\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial v} \end{cases}$$
《解答例》
（１）$z=xy^2+x^2y$、$x=u+v$、$y=u-v$
$$\begin{align} \dfrac{\partial z}{\partial u} &=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial u}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial u} \\
&=(y^2+2xy) \cdot 1+(2xy+x^2) \cdot 1 \\
&=x^2+y^2+4xy \\
&=(x+y)^2+2xy \\
&=4u^2+2(u^2-v^2) \\
&=6u^2-2v^2 \ \ \cdots \cdots(\text{答}) \end{align}$$ $$\begin{align} \dfrac{\partial z}{\partial v} &=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial v} \\
&=(y^2+2xy) \cdot 1+(2xy+x^2) \cdot (-1) \\
&=y^2-x^2 \\
&=(y+x)(y-x) \\
&=2u(-2v) \\
&=-4uv \ \ \cdots \cdots(\text{答}) \end{align}$$
（２）$z=\sin (x-y)$、$x=u^2+v^2$、$y=2uv$
$$\begin{align} \dfrac{\partial z}{\partial u} &=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial u}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial u} \\ &=2u \cos (x-y)-2v \cos (x-y) \\ &=2(u-v) \cos (x-y) \\ &=2(u-v) \cos (u^2+v^2-2uv) \\ &=2(u-v) \cos (u-v)^2 \ \ \cdots \cdots(\text{答}) \end{align}$$ $$\begin{align} \dfrac{\partial z}{\partial v} &=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial v} \\ &=2v \cos (x-y)-2u \cos (x-y) \\ &=2(v-u) \cos (x-y) \\ &=2(v-u) \cos (u^2+v^2-2uv) \\ &=2(v-u) \cos (u-v)^2 \ \ \cdots \cdots(\text{答}) \end{align}$$
※$z_v$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）$z=f(x,y)$、$x=2u-3v$、$y=u-5v$
$$\begin{align} \dfrac{\partial z}{\partial u} &=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial u}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial u} \\
&=f_x(x,y) \cdot 2+f_y(x,y) \cdot 1 \\
&=2f_x(2u-3v,u-5v)+f_y(2u-3v,u-5v) \ \ \cdots \cdots(\text{答}) \end{align}$$ $$\begin{align} \dfrac{\partial z}{\partial v} &=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial v} \\
&=f_x(x,y) \cdot (-3)+f_y(x,y) \cdot (-5) \\
&=-3f_x(2u-3v,u-5v)-5f_y(2u-3v,u-5v) \ \ \cdots \cdots(\text{答}) \end{align}$$
（４）$z=f(x,y)$、$x=\cos (u+v)$、$y=\sin (u-v)$
$$\begin{align} \dfrac{\partial z}{\partial u} &=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial u}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial u} \\
&=f_x(x,y) \cdot (-\sin (u+v))+f_y(x,y) \cdot \cos (u-v) \\
&=-\sin (u+v)f_x(\cos (u+v),\sin (u-v)) \\ &\ \ \ \ \ +\cos (u-v)f_y(\cos (u+v),\sin (u-v)) \ \ \cdots \cdots(\text{答}) \end{align}$$ $$\begin{align} \dfrac{\partial z}{\partial v} &=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial v} \\
&=f_x(x,y) \cdot (-\sin (u+v))+f_y(x,y) \cdot (-\cos (u-v)) \\
&=-\sin (u+v)f_x(\cos (u+v),\sin (u-v)) \\ &\ \ \ \ \ -\cos (u-v)f_y(\cos (u+v),\sin (u-v)) \ \ \cdots \cdots(\text{答}) \end{align}$$
（５）$z=f(x+3y)$、$x=u-2v$、$y=3u-4v$
$$\begin{align} \dfrac{\partial z}{\partial u} &=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial u}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial u} \\
&=f^{\prime}(x+3y)\cdot 1 +3f^{\prime}(x+3y) \cdot 3 \\
&=10f^{\prime}(10u-14v) \ \ \cdots \cdots(\text{答}) \end{align}$$ $$\begin{align} \dfrac{\partial z}{\partial v} &=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial v} \\
&=f^{\prime}(x+3y)\cdot (-2) +3f^{\prime}(x+3y) \cdot (-4) \\
&=-14f^{\prime}(10u-14v) \ \ \cdots \cdots(\text{答}) \end{align}$$
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