What is a Derivative and Partial Derivative? How to find derivative of an arbitrary function?

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
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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.

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