Triple product rule

$ \frac{\partial x}{\partial y}\frac{\partial y}{\partial z}\frac{\partial z}{\partial x}=-1

が成り立つ

$ x=g_x(y,z)

$ y=g_y(z,x)

$ z=g_z(x,y)

$ \mathrm d g_x=\frac{\partial g_x}{\partial y}\mathrm dy+\frac{\partial g_x}{\partial z}\mathrm dz

$ \mathrm d g_y=\frac{\partial g_y}{\partial z}\mathrm dz+\frac{\partial g_y}{\partial x}\mathrm dx

$ \mathrm d g_z=\frac{\partial g_z}{\partial x}\mathrm dx+\frac{\partial g_z}{\partial y}\mathrm dy

$ \mathrm d g_z(g_x(y,z),y)=\left.\left(\frac{\partial g_z}{\partial x}\mathrm dx+\frac{\partial g_z}{\partial y}\mathrm dy\right)\right|_{x=g_x}

$ \iff \mathrm d g_x=\left.\left(\frac{\partial g_z}{\partial x}\right)^{-1}\mathrm d g_z\right|_{x=g_x}-\left.\left({\frac{\partial g_z}{\partial x}}^{-1}\frac{\partial g_z}{\partial y}\right)\right|_{x=g_x}\mathrm dy

$ \implies\frac{\partial g_x}{\partial y}=0-\left(\left.\frac{\partial g_z}{\partial x}\right|_{x=g_x}\right)^{-1}\frac{\partial g_z}{\partial y}

$ \because z=g_z(g_x(y,z),y)\implies\frac{\partial}{\partial y}g_z(g_x(y,z),y)=0

$ y=g_y(g_z(x,y),x)

$ \implies 1=\left.\frac{\partial g_y}{\partial z}\right|_{z=g_z}\frac{\partial g_z}{\partial y}

$ \iff {\frac{\partial g_z}{\partial y}}^{-1}=\left.\frac{\partial g_y}{\partial z}\right|_{z=g_z}

$ \implies 0=\left.\frac{\partial g_y}{\partial z}\right|_{z=g_z}\frac{\partial g_z}{\partial x}+\left.\frac{\partial g_y}{\partial x}\right|_{z=g_z}

$ \iff {\frac{\partial g_z}{\partial x}}^{-1}=-\left.\left({\frac{\partial g_y}{\partial x}}^{-1}\frac{\partial g_y}{\partial z}\right)\right|_{z=g_z}

$ z=g_z(x,g_y(z,x))

$ \implies0=\left.\left(\frac{\partial g_z}{\partial y}\frac{\partial g_y}{\partial x}+\frac{\partial g_z}{\partial x}\right)\right|_{y=g_y}

$ \iff\left.\frac{\partial g_z}{\partial x}\right|_{y=g_y}=-\left.\left(\frac{\partial g_z}{\partial y}\frac{\partial g_y}{\partial x}\right)\right|_{y=g_y}

$ \iff\left.\frac{\partial g_z}{\partial x}\right|_{y=g_y(g_z(x,y),x)}=-\left.\left(\frac{\partial g_z}{\partial y}\frac{\partial g_y(g_z(x,y),x)}{\partial x}\right)\right|_{y=g_y(g_z(x,y),x)}

$ \iff \frac{\partial g_z}{\partial x}=-\frac{\partial g_z}{\partial y}\left.\frac{\partial g_y}{\partial x}\right|_{z=g_z}

$ \iff {\frac{\partial g_z}{\partial y}}^{-1}=-{\frac{\partial g_z}{\partial x}}^{-1}\left.\frac{\partial g_y}{\partial x}\right|_{z=g_z}

2通りで表せることがわかった

$ {\frac{\partial g_z}{\partial y}}^{-1}=\left.\frac{\partial g_y}{\partial z}\right|_{z=g_z}

$ {\frac{\partial g_z}{\partial y}}^{-1}=-{\frac{\partial g_z}{\partial x}}^{-1}\left.\frac{\partial g_y}{\partial x}\right|_{z=g_z}=-\left.\left(\frac{\partial g_x}{\partial z}\frac{\partial g_y}{\partial x}\right)\right|_{z=g_z}

予想

変数をひっくりかえすだけなら全微分と同じ

↑だと、xが増えている

$ \therefore \left.\left(\frac{\partial g_x}{\partial z}\frac{\partial g_z}{\partial y}\frac{\partial g_y}{\partial x}\right)\right|_{z=g_z}=-1

それぞれ他方の変数の函数と考えて偏微分した後、特定の2変数の函数に書き換えたものは常に-1になる