How a Multivariable Function Changes with Respect to a Polar Coordinate

For a functions $f(x)$, $g(y)$ and $h(z)$, the chain rule yields

$ \frac{\partial f(x)}{\partial \theta} = ( \frac{\partial f}{\partial x} ) ( \frac{\partial x}{\partial \theta} )$.

$ \frac{\partial g(y)}{\partial \theta}  = (\frac{\partial g}{\partial y}) (\frac{\partial y}{\partial \theta} )$.

$ \frac{\partial h(z)}{\partial \theta}  = (\frac{\partial h}{\partial z})( \frac{\partial z}{\partial \theta} )$.

Suppose $F(x,y,z) \equiv f(x) g(y) h(z)$.

This is a significant assumption.

Then, using the product rule twice,

$ \frac{\partial F(x,y,z)}{\partial \theta} = (\frac{\partial f(x)}{\partial \theta}) g(y) h(z) + f(x) \bigg( (\frac{\partial g(y)  }{\partial \theta}) h(z) + g(y) (\frac{\partial h(z)  }{\partial \theta}) \bigg) $.

Simplifying,

$ \frac{\partial F(x,y,z) }{\partial \theta}  = (\frac{\partial f(x)}{\partial \theta}) g(y) h(z) + f(x) ( \frac{\partial g(y)  }{\partial \theta} ) h(z) + f(x) g(y) ( \frac{\partial h(z)  }{\partial \theta} ) $.

Substituting,

$ \frac{\partial F(x,y,z)}{\partial \theta}  = ( ( \frac{\partial f}{\partial x} ) ( \frac{\partial x}{\partial \theta} ) ) g(y) h(z) + f(x) ( (\frac{\partial g}{\partial y}) (\frac{\partial y}{\partial \theta} ) ) h(z) + f(x) g(y) ( (\frac{\partial h}{\partial z})( \frac{\partial z}{\partial \theta} ) ) $.

Rearranging,

$ \frac{\partial F(x,y,z)}{\partial \theta}  =  ( \frac{\partial F(x,y,z) }{\partial x} ) ( \frac{\partial x}{\partial \theta} )  + (\frac{\partial F(x,y,z) }{\partial y}) (\frac{\partial y}{\partial \theta} )  +  (\frac{\partial F(x,y,z) }{\partial z})( \frac{\partial z}{\partial \theta} ) $.

Since multiplication is commutative,

$ \frac{\partial F(x,y,z)}{\partial \theta}  =   ( \frac{\partial x}{\partial \theta} ) ( \frac{\partial F(x,y,z) }{\partial x} )  +  (\frac{\partial y}{\partial \theta} ) (\frac{\partial F(x,y,z) }{\partial y}) +  ( \frac{\partial z}{\partial \theta} ) (\frac{\partial F(x,y,z) }{\partial z}) $.

Rewriting in a slightly different form,

$ \frac{\partial }{\partial \theta} F(x,y,z) =   ( \frac{\partial x}{\partial \theta} ) ( \frac{\partial }{\partial x} F(x,y,z) ) +  (\frac{\partial y}{\partial \theta} ) (\frac{\partial  }{\partial y} F(x,y,z) ) +  ( \frac{\partial z}{\partial \theta} ) (\frac{\partial  }{\partial z} F(x,y,z) ) $.

Removing parentheses so that the rightmost partial derivatives do not look like isolated factors,

$ \frac{\partial }{\partial \theta} F(x,y,z) =   ( \frac{\partial x}{\partial \theta} )  \frac{\partial }{\partial x} F(x,y,z)  +  (\frac{\partial y}{\partial \theta} ) \frac{\partial  }{\partial y} F(x,y,z)  +  ( \frac{\partial z}{\partial \theta} ) \frac{\partial  }{\partial z} F(x,y,z)  $.

Rewriting,

$ \frac{\partial }{\partial \theta} F(x,y,z) =   \bigg( ( \frac{\partial x}{\partial \theta} )  \frac{\partial }{\partial x}  +  (\frac{\partial y}{\partial \theta} ) \frac{\partial  }{\partial y}  +  ( \frac{\partial z}{\partial \theta} ) \frac{\partial  }{\partial z}  \bigg) F(x,y,z)$.

Assuming the only function to the right of the differential operators is $F(x,y,z) \equiv f(x) g(y) h(z)$, one can write:

$ \frac{\partial }{\partial \theta}  =   ( \frac{\partial x}{\partial \theta} )  \frac{\partial }{\partial x}  +  (\frac{\partial y}{\partial \theta} ) \frac{\partial  }{\partial y}  +  ( \frac{\partial z}{\partial \theta} ) \frac{\partial  }{\partial z} $.

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