Partial derivative with respect to intermediate variable

Question

Suppose we have $f(x,y,z)$, where $x=g(r,\theta,\phi)$, $y=h(r,\theta,\phi)$, $z=t(r,\theta,\phi)$. How do we find partial derivative with respect to $x$ and express it as a "function" of $r$, $\theta$ and $\phi$. I am worried because the other variables are also functions of $r$, $\theta$ and $\phi$. EDIT: And if it is possible to be explained fundamentaly.

Answer 1

Using the chain rule for partial derivatives, you probably already got something like:

$\begin{align} \dfrac{\partial f}{\partial r} &= \dfrac{\partial f}{\partial x}\dfrac{\partial g}{\partial r} + \dfrac{\partial f}{\partial y}\dfrac{\partial h}{\partial r} + \dfrac{\partial f}{\partial z}\dfrac{\partial t}{\partial r} \\ \dfrac{\partial f}{\partial \theta} &= \dfrac{\partial f}{\partial x}\dfrac{\partial g}{\partial \theta} + \dfrac{\partial f}{\partial y}\dfrac{\partial h}{\partial \theta} + \dfrac{\partial f}{\partial z}\dfrac{\partial t}{\partial \theta} \\ \dfrac{\partial f}{\partial \phi} &= \dfrac{\partial f}{\partial x}\dfrac{\partial g}{\partial \phi} + \dfrac{\partial f}{\partial y}\dfrac{\partial h}{\partial \phi} + \dfrac{\partial f}{\partial z}\dfrac{\partial t}{\partial \phi} \end{align}$

which is a set of simultaneous equations and hence can be expressed in matrix form.

$\left[ \begin{array}{c} \dfrac{\partial f}{\partial r} \\ \dfrac{\partial f}{\partial \theta} \\ \dfrac{\partial f}{\partial \phi} \end{array} \right] = \begin{bmatrix}\dfrac{\partial g}{\partial r} & \dfrac{\partial h}{\partial r} & \dfrac{\partial t}{\partial r} \\ \dfrac{\partial g}{\partial \theta} & \dfrac{\partial h}{\partial \theta} & \dfrac{\partial t}{\partial \theta} \\ \dfrac{\partial g}{\partial \phi} & \dfrac{\partial h}{\partial \phi} & \dfrac{\partial t}{\partial \phi}\end{bmatrix} \left[ \begin{array}{c} \dfrac{\partial f}{\partial x} \\ \dfrac{\partial f}{\partial y} \\ \dfrac{\partial f}{\partial z} \end{array} \right]$

Whenever this matrix will be invertible you will always be able to solve for $\partial f / \partial x$ and the like. If you are working in spherical coordinates the matrix is always invertible (see http://mathworld.wolfram.com/SphericalCoordinates.html for more information).