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}