# 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
$\begin{gathered} \frac{{dy}}{{dx}} = \frac{1}{{{x^2}\ln 10}}\frac{d}{{dx}}\left( {{x^2}} \right) \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{{2x}}{{{x^2}\ln 10}} \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{2}{{x\ln 10}} \\ \end{gathered}$