Part 2: ‘Second Order Differentiation‘
Just as a function has a rate of change, the rate of change itself can also change. How? Let’s dig into that.
2.1 Second Order Exact Differentiation
If we again consider infinite small algebra, for example, let’s take:
\begin{equation} \label{eq:myequ}
y=f(x)
\end{equation}
as our function. Here, we previously found that:
\begin{equation} \label{eq:exactdiff}
dy=\frac{\partial{y}}{\partial{x}}dx
\end{equation}
and \(\frac{\partial{y}}{\partial{x}}\) as a function of some variable. In this case, function of a single variable \(x\). Let it be equal to function \(g(x)\) . So;
\begin{equation} \label{eq:diff}
y^\prime=\frac{\partial{y}}{\partial{x}}=g(x)
\end{equation}
Now, by using the limit again
\begin{equation}\label{eq:limitsequation2}\begin{aligned} y^\prime+ \lim_{\Delta y^\prime \to 0}\Delta y^\prime=\lim_{\Delta x \to 0}g(x+\Delta x)\end{aligned}\end{equation}
is found. Here, \(y^\prime\) is the derivative of \(y\) as defined above. If we write the equation \eqref{eq:limitsequation2} in terms of infinitesmall changes:
\begin{equation} \label{eq:df1}
y^\prime+dy^\prime=g(x+dx)
\end{equation}
is obtained. This is called the second order infinitesmall changes equality.
Now, if we take the equation \(y^\prime=2x\), which we found for \(y=x^2\), as an example, using the equality \eqref{eq:df1}:
\begin{equation} \label{eq:df2}
y^\prime+dy^\prime=2(x+dx)
\end{equation}
is obtained. Using this equation
\begin{equation} \label{eq:df3}
dy^\prime=2dx
\end{equation}
is found. Then, exact differentiation ratio is as :
\begin{equation} \label{eq:df4}
\frac{dy^\prime}{dx}=2
\end{equation}
found. This equation is the second order derivation of \(y \).
2.2 Second Order Partial Differentiation
Let’s dig into second order partial differentiation. Again assuming that;
\begin{equation} \label{eq:df5}y=f(x,z)\end{equation}
as our function. If we write the exact differentiation:
\begin{equation} \label{eq:df6}dy=\frac{\partial y}{\partial x}dx+\frac{\partial y}{\partial z}dz\end{equation}
is found. If we divide both sides by \(dx\):
\begin{equation} \label{eq:df7}\frac{ dy}{ dx}=\frac{\partial y}{\partial x}+\frac{\partial y}{\partial z}\frac{ dz}{ dx}\end{equation}
is obtained Here, if \(x\) ve \(z\) are independent \(\frac{ dz}{ dx}\) will be \(0\). In other words if the variables are independent then exact differentiation ratio of them will be \(0\). For independent variables:
\begin{equation} \label{eq:df8}\frac{dy}{ dx}=\frac{\partial y}{\partial x}\end{equation}
is obtained Lets then take both sides exact differentiation;
\begin{equation} \label{eq:df9}d(\frac{ dy}{ dx})=d(\frac{\partial y}{\partial x})\end{equation}
\begin{equation} \label{eq:df10}d(\frac{ dy}{ dx})=\frac{\partial(\frac{\partial y}{\partial x})}{\partial x}dx+\frac{\partial{(\frac{\partial y}{\partial x})}}{\partial z}dz\end{equation}
again by dividing both sides by \(dx\);
\begin{equation} \label{eq:df11}\frac{ d(\frac{ dy}{ dx})}{dx}=\frac{\partial(\frac{\partial y}{\partial x})}{\partial x}+\frac{\partial{(\frac{\partial y}{\partial x})}}{\partial z}\frac{dz}{dx}\end{equation}
is obtained. Again using independency of variables;
\begin{equation} \label{eq:df14}\frac{ d(\frac{ dy}{ dx})}{dx}=\frac{\partial(\frac{\partial y}{\partial x})}{\partial x}\end{equation}
is obtained. If we write it in a well known notation;
\begin{equation} \label{eq:df15}\frac{ d^2 y}{ d x^2}=\frac{\partial^2 y}{\partial x^2}\end{equation}
is found.
Here we described second order exact and partial differentiation. If you have further question and comments please describe it below.