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
\[\begin{gathered} \frac{{dy}}{{dx}} = \frac{1}{{{x^2}}}\frac{d}{{dx}}\left( {{x^2}} \right) \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{{2x}}{{{x^2}}} \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{2}{x} \\ \end{gathered} \]