The Crux of Calculus

Define $\Delta x \equiv x_2 – x_1$, to be consistent with this post. Similarly, define $\Delta y \equiv y_2 – y_1$ and $\Delta z \equiv z_2 – z_1$.

The Cartesian coordinates are $x$, $y$, & $z$. In contrast, the spherical coordinates are $r$, $\theta$, & $\phi$. Here, $\phi$ is the azimuthal angle in the $xy$-plane.

Next, use this post to obtain the equations relating Cartesian coordinates to spherical coordinates. In particular:

$x = r \sin\theta \cos \phi $

$y = r \sin \theta \sin \phi $

$z = r \cos\theta $

Note that $x$ changes if $r$ changes, $\theta$ changes, and/or $\phi$ changes. Similarly, $y$ changes if $r$ changes, $\theta$ changes, and/or $\phi$ changes. Finally, $z$ changes if $r$ changes and/or $\theta$ changes. To express all of these options, it is possible to construct a matrix [1]:

Some of the functional dependencies have been indicated so that it is clear how the ratios of differences can eventually become derivatives. 

As done in previous posts, I constrain each angle to be nonnegative. 

Note that it is possible to multiply each element by the quantity in the denominator:


From algebra, the previous matrix is equal to:

These two matrices will be useful later.

Next, from this post, one type of derivative is

$ \lim_{\Delta x \rightarrow 0^+} \frac{f(a+\Delta x) – f(a)}{\Delta x} \equiv \frac{d f(x)}{dx}\big|_{a^+}$.


  • $\Delta f(\Delta x) \equiv f_2(\Delta x) – f_1(\Delta x)$
  • $f_2(\Delta x) \equiv f(a+\Delta x)$
  • $f_1(\Delta x) \equiv f(a)$

Then $\Delta f(\Delta x) = f(a+\Delta x) – f(a)$, and the previous definition indicates that

$ \lim_{\Delta x \rightarrow 0^+} \frac{ \Delta f( \Delta x) }{\Delta x} = \frac{d f(x)}{dx}\big|_{a^+}$.

Given the above information, perhaps it is useful to construct a matrix that is identical to the previous matrix except for an additional $\lim_{\Delta x \rightarrow 0^+}$ applied to each element of the matrix. This new matrix is:

These are derivatives, and rules based on limit properties are needed to evaluate each element. This will be done in a separate post.

Recall that:

Apply a limit to both sides of each pair of elements:

Using this property, the matrix on the right becomes:


Using this property, the previous matrix is equal to

Comparing each pair of elements from the previous two matrices, it is valid to write 

$\lim_{\Delta x \rightarrow 0^+} \frac{ \Delta f( \Delta x) }{\Delta x} = \frac{ \lim_{\Delta x \rightarrow 0^+} \Delta f( \Delta x) }{\lim_{\Delta x \rightarrow 0^+} \Delta x} $

Using $ \lim_{\Delta x \rightarrow 0^+} \frac{ \Delta f( \Delta x) }{\Delta x} = \frac{d f(x)}{dx}\big|_{a^+}$, the left side is equal to $\frac{d f(x)}{dx}\big|_{a^+}$,


$\frac{d f(x)}{dx}\big|_{a^+} = \lim_{\Delta x \rightarrow 0^+} \frac{ \Delta f( \Delta x) }{\Delta x} = \frac{ \lim_{\Delta x \rightarrow 0^+} \Delta f( \Delta x) }{\lim_{\Delta x \rightarrow 0^+} \Delta x} $

For convenience, define 

$ \boxed{ dx |_{a^+} \equiv \lim_{\Delta x \rightarrow 0^+} \Delta x }$.


$\frac{d f(x)}{dx}\big|_{a^+} = \frac{ \lim_{\Delta x \rightarrow 0^+} \Delta f( \Delta x) }{dx|_{a^+} } $.

Thinking of a derivative to be most fundamentally a ratio of two quantities, moving the $ |_{a^+}$ to the numerator and to the denominator, and multiplying both sides by the quantity $dx|_{a^+} $, one has

$ \boxed{ d f(x)\big|_{a^+} =  \lim_{\Delta x \rightarrow 0^+} \Delta f( \Delta x) } $.

Now dummy variables are defined in terms of limits. In fact, it appears that all three types of derivatives can be defined by starting with this equation as a definition for a dummy variable, and using this property. With this notation, $dx |_{a^+} $ is just a name for a limit.

Recall that


$ d f(x)\big|_{a^+} =  \lim_{\Delta x \rightarrow 0^+} \Delta f( \Delta x) $

$ \lim_{\Delta x \rightarrow 0^+} \frac{ \Delta f( \Delta x) }{\Delta x} = \frac{d f(x)}{dx}\big|_{a^+}$


$ dx |_{a^+} \equiv \lim_{\Delta x \rightarrow 0^+} \Delta x $

one has:

Note there are minor typos in that each $0^+$ should be replaced by $a^+$.

Now it is possible to convert dummy variables in terms of Cartesian coordinates into dummy variables in terms of Spherical Coordinates. To do this, one must evaluate each derivative in the matrix on the right side. This will be done in subsequent posts. Note that $a$ can be replaced by the corresponding independent variable, if needed.



Christina Daniel

I am a research assistant in theoretical physics. This website,, 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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