Let $p=f(v-w, v-u, u-w)$ be a differentiable function. Show that $$\frac{\partial p}{\partial u}+\frac{\partial p}{\partial v}+\frac{\partial p}{\partial w}=0$$

Solution Let $x=v-w$, $y=v-u$, and $z=u-w$. Then $p=f(x,y,z)$. We use an extension of Chain Rule II. Since $\dfrac{\partial x}{\partial u}=0,$ $\dfrac{\partial y}{\partial u} =-1,$ and $\dfrac{\partial z}{\partial u}=1,$ we have $$\frac{\partial p}{\partial u}=\frac{\partial p}{\partial x}\frac{\partial x}{ \partial u}+\frac{\partial p}{\partial y}\frac{\partial y}{\partial u}+\frac{ \partial p}{\partial z}\frac{\partial z}{\partial u}=\frac{\partial p}{ \partial x}(0)+\frac{\partial p}{\partial y}(-1)+\frac{\partial p}{\partial z }(1)=-\frac{\partial p}{\partial y}+\frac{\partial p}{\partial z}\hbox{ }$$

856

Since $\dfrac{\partial x}{\partial v}=1,$ $\dfrac{\partial y}{\partial v}=1,$ and $\dfrac{\partial z}{\partial v}=0,$ we have $$\frac{\partial p}{\partial v}=\frac{\partial p}{\partial x}\frac{\partial x}{ \partial v}+\frac{\partial p}{\partial y}\frac{\partial y}{\partial v}+\frac{ \partial p}{\partial z}\frac{\partial z}{\partial v}=\frac{\partial p}{ \partial x}(1)+\frac{\partial p}{\partial y}(1)+\frac{\partial p}{\partial z} (0)=\frac{\partial p}{\partial x}+\frac{\partial p}{\partial y}$$

Since $\dfrac{\partial x}{\partial w}=-1,$ $\dfrac{\partial y}{\partial w} =0,$ and $\dfrac{\partial z}{\partial w}=-1,$ we have $$\frac{\partial p}{\partial w}=\frac{\partial p}{\partial x}\frac{\partial x}{ \partial w}+\frac{\partial p}{\partial y}\frac{\partial y}{\partial w}+\frac{ \partial p}{\partial z}\frac{\partial z}{\partial w}=\frac{\partial p}{ \partial x}(-1)+\frac{\partial p}{\partial y}(0)+\frac{\partial p}{\partial z }(-1)=-\frac{\partial p}{\partial x}-\frac{\partial p}{\partial z}$$

Adding these, we get $$\underset{\color{#0066A7}{\hbox{ }\hspace{5pc}\dfrac{{\partial p}}{{\partial u}}}}{\dfrac{ \partial p}{\partial u}+\dfrac{\partial p}{\partial v}+\dfrac{\partial p}{ \partial w}=\underbrace{\left( -\frac{\partial p}{\partial y}+\frac{\partial p}{\partial z}\right) }}+\underset{\color{#0066A7}{\dfrac{{\partial p}}{{ \partial v}}}}{\underbrace{\left( \frac{\partial p}{\partial x}+\frac{ \partial p}{\partial y}\right) }}+\underset{\color{#0066A7}{\dfrac{{\partial p}}{ {\partial w}}}}{\underbrace{\left( -\frac{\partial p}{\partial x}- \frac{\partial p}{\partial z}\right) \,\,\,}}=0$$