# Derivative of Natural Logarithmic Functions

A function defined by $y = {\log _a}x,\,\,\,x > 0$, where $x = {a^y},\,\,\,a > 0$, $a \ne 1$ is called the logarithm of $x$ to the base $a$. The natural logarithmic function is written as $y = {\log _e}x$ or $y = \ln x$.

We shall prove the formula for the derivative of the natural logarithm function using definition or the first principle method.

Let us suppose that the function is of the form $y = f\left( x \right) = \ln x$

First we take the increment or small change in the function:
$\begin{gathered} y + \Delta y = \ln \left( {x + \Delta x} \right) \\ \Delta y = \ln \left( {x + \Delta x} \right) – y \\ \end{gathered}$

Putting the value of function $y = \ln x$ in the above equation, we get
$\begin{gathered} \Delta y = \ln \left( {x + \Delta x} \right) – \ln x \\ \Delta y = \ln \left( {\frac{{x + \Delta x}}{x}} \right) \\ \Delta y = \ln \left( {1 + \frac{{\Delta x}}{x}} \right) \\ \end{gathered}$

Dividing both sides by $\Delta x$, we get
$\frac{{\Delta y}}{{\Delta x}} = \frac{1}{{\Delta x}}\ln \left( {1 + \frac{{\Delta x}}{x}} \right)$

Multiplying and dividing the right hand side by $x$, we have
$\begin{gathered} \Rightarrow \frac{{\Delta y}}{{\Delta x}} = \frac{1}{x}\frac{x}{{\Delta x}}\ln \left( {1 + \frac{{\Delta x}}{x}} \right) \\ \Rightarrow \frac{{\Delta y}}{{\Delta x}} = \frac{1}{x}\ln {\left( {1 + \frac{{\Delta x}}{x}} \right)^{\frac{x}{{\Delta x}}}} \\ \end{gathered}$

Taking the limit of both sides as $\Delta x \to 0$, we have
$\begin{gathered} \Rightarrow \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{1}{x}\ln {\left( {1 + \frac{{\Delta x}}{x}} \right)^{\frac{x}{{\Delta x}}}} \\ \Rightarrow \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \frac{1}{x}\mathop {\lim }\limits_{\Delta x \to 0} \ln {\left( {1 + \frac{{\Delta x}}{x}} \right)^{\frac{x}{{\Delta x}}}} \\ \end{gathered}$

Consider $\frac{{\Delta x}}{x} = u \Rightarrow \frac{x}{{\Delta x}} = \frac{1}{u}$, as $\Delta x \to 0$ then $u \to 0$, we get
$\Rightarrow \frac{{dy}}{{dx}} = \frac{1}{x}\mathop {\lim }\limits_{u \to 0} \ln {\left( {1 + u} \right)^{\frac{1}{u}}}$

Using the relation from the limit $\mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{\frac{1}{x}}} = e$, we have
$\begin{gathered} \frac{{dy}}{{dx}} = \frac{1}{x}\ln \left( e \right) \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{x} \\ \Rightarrow \frac{d}{{dx}}\left( {\ln x} \right) = \frac{1}{x} \\ \end{gathered}$

Example: Find the derivative of $y = f\left( x \right) = \ln {x^2}$

We have the given function as
$y = \ln {x^2}$

Differentiating with respect to variable $x$, we get
$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\ln {x^2}$

Using the rule, $\frac{d}{{dx}}\left( {\ln x} \right) = \frac{1}{x}$, we get
$\begin{gathered} \frac{{dy}}{{dx}} = \frac{1}{{{x^2}}}\frac{d}{{dx}}\left( {{x^2}} \right) \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{{2x}}{{{x^2}}} \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{2}{x} \\ \end{gathered}$