-
How a Multivariable Function Changes with Respect to the Cartesian Coordinates
Suppose $F(x,y,z) \equiv f(x) g(y) h(z)$. This post shows how to calculate a partial derivative of a multivariable function $F(x,y,z)$ with respect to each Cartesian coordinates. The resulting expressions are in spherical coordinates. In previous posts (see references [2], [3], and [4]), the following equations were written: $ \frac{\partial }{\partial r} F(x,y,z) = \bigg( ( \frac{\partial x}{\partial r}…
-
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…
-
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…
-
How a Multivariable Function Changes with Respect to a Radial Coordinate
First focus on the Cartesian coordinate $x$, which depends on the spherical coordinates $r,\theta,$ and $\phi$. For a function $f(x)$, the chain rule yields $ \frac{\partial f(x)}{\partial r} = ( \frac{\partial f}{\partial x} ) ( \frac{\partial x}{\partial r} )$. Similarly for functions $g(y)$ and $h(z)$: $ \frac{\partial g(y)}{\partial r} = (\frac{\partial g}{\partial y}) (\frac{\partial y}{\partial r}…