Derivative of Log X

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 common logarithmic function is written as $$y = {\log _{10}}x$$.

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 \[y = f\left( x \right) = {\log _a}x\]

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

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

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

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

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

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
\[ \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{x}\mathop {\lim }\limits_{u \to 0} {\log _a}{\left( {1 + u} \right)^{\frac{1}{u}}}\]

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

Example: Find the derivative of \[y = f\left( x \right) = {\log _{10}}{x^2}\]

We have the given function as
\[y = {\log _{10}}{x^2}\]

Differentiating with respect to variable $$x$$, we get
\[\frac{{dy}}{{dx}} = \frac{d}{{dx}}{\log _{10}}{x^2}\]

Using the rule, $$\frac{d}{{dx}}\left( {{{\log }_a}x} \right) = \frac{1}{{x\ln a}}$$, we get
\[\begin{gathered} \frac{{dy}}{{dx}} = \frac{1}{{{x^2}\ln 10}}\frac{d}{{dx}}\left( {{x^2}} \right) \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{{2x}}{{{x^2}\ln 10}} \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{2}{{x\ln 10}} \\ \end{gathered} \]