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:

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

### References

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

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