How a Multivariable Function Changes with Respect to an Azimuthal Coordinate

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

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

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

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

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

This is a significant assumption.

Then, using the product rule,

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

Evaluate $\frac{\partial [g(y) h(z)]}{\partial \phi}$ using the product rule again:

$ \frac{\partial [g(y) h(z)]}{\partial \phi} = h(z) \frac{\partial g(y) }{\partial \phi} + g(y) \frac{\partial h(z)}{\partial \phi} $.

Substituting,

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

Distribute.

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

Substitute:

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

Rearranging:

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

Since multiplication is commutative,

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

Rewriting in a slightly different form,

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

Rewriting with a distributive property:

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

References

[1] http://www.thphys.nuim.ie/Notes/MP469/Laplace.pdf

Christina Daniel

I am a research assistant in theoretical physics. This website, derive-it.com, serves to organize my ongoing learning and research as well as to provide a resource to other learners around the world.

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