Minor changes

source/calculus/notations.rst

@@ -5,17 +5,27 @@ Differentiation Notations
 Leibniz's Notation
 -------------------
 
-The derivative of a function :math:`f` at :math:`x` is given
-by :math:`\lim\limits_{h\to0} \frac{f(x+h)-f(x)}{h}`.
-
+The derivative of a function :math:`f` at :math:`x` is given by :math:`\lim\limits_{h\to0} \frac{f(x+h)-f(x)}{h}`.
 The Leibniz's notation expresses the derivative of :math:`f` as :math:`\frac{dy}{dx}` with :math:`y=f(x)`.
-
 More details `here <https://www.me.psu.edu/cimbala/me420/Homework/dydx_quotient_article.html>`__.
 
 .. note::
    The following :math:`\frac{dx}{dt}` means :math:`x` is a function of :math:`t` such as :math:`x=f(t)`.
    See explanations `here <https://www.khanacademy.org/math/ap-calculus-ab/ab-diff-contextual-applications-new/ab-4-4/v/differentiating-related-functions-intro>`__.
 
+:math:`\frac{d}{dx}` is an operator not a quotient!
+Although it behaves like a quotient. In fact it is a limit:
+
+.. math::
+
+   \lim\limits_{h\to0} \frac{f(x+h)-f(x)}{h} \ne \frac{\lim\limits_{h\to0} \left(f(x+h)-f(x)\right)}{\lim\limits_{h\to0} h}
+
+See, we cannot express this limit-of-a-quotient as a-quotient-of-the-limits, then the derivative is not a quotient.
+More details `here <https://www.me.psu.edu/cimbala/me420/Homework/dydx_quotient_article.html>`__.
+
+Can we perform operation without relying on this assumption?
+The answer is yes! Using the chain rule.
+
 Lagrange's Notation
 -------------------
 Also cal prime notation

source/electronics/maxwell/gauss.rst

@@ -0,0 +1,2 @@
+Gauss's Law
+============

source/electronics/maxwell/index.rst

@@ -0,0 +1,8 @@
+The Maxwell's Equations
+========================================
+
+.. toctree::
+   :maxdepth: 2
+   :caption: First equation
+
+   gauss.rst

source/index.rst

@@ -33,6 +33,7 @@ Welcome to ScienceNotes's documentation!
    :numbered:
    :caption: Electronics
 
+   electronics/maxwell/index.rst
    electronics/metrics.rst
    electronics/components.rst
    electronics/kirchhoff.rst