Question

How do I use the limit definition of derivative on $f\left( x \right)=\ln \left( x \right)$?

Answer 1

Assume ‘h’ as the small change in x and hence find the small change in the function f (x) given as f(x + h). Now, use the limit definition of derivative and apply the formula: $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{f\left( x+h \right)-f\left( x \right)}{h} \right)$, substitute the value of the given functions. Now, use the property of log given as: $\log m-\log n=\log \left( \dfrac{m}{n} \right)$ and apply the formula of expansion of logarithm given as: $\log \left( 1+a \right)=a-\dfrac{{{a}^{2}}}{2}+\dfrac{{{a}^{3}}}{3}-\dfrac{{{a}^{4}}}{4}+.........$ to simplify the limit to get the answer.

Complete step by step solution:
Here, we have been provided with the function $f\left( x \right)=\ln \left( x \right)$ and we are asked to find its derivative using the limit definition.
We know that derivative of a function is defined as the rate of change of function. In other words we can say that it is the measure of change in the value of the function with respect to the change in the value of the variable on which it depends. This change is infinitesimally small, that means tending to zero. Mathematically we have,
$\Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{f\left( x+h \right)-f\left( x \right)}{h} \right)$
In the above formula we have ‘h’ as the infinitesimally small change in the variable x.
Now, let us come to the question. We have the function $f\left( x \right)=\ln \left( x \right)$, so substituting (x + h) in place of x in the function, we get,
$\Rightarrow f\left( x+h \right)=\ln \left( x+h \right)$
Now, substituting the value of the functions f (x) and f (x + h) in the limit formula, we get,
$\Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{\ln \left( x+h \right)-\ln x}{h} \right)$
Using the formula: $\log m-\log n=\log \left( \dfrac{m}{n} \right)$, we get,
$\begin{aligned} & \Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{\ln \left( \dfrac{x+h}{x} \right)}{h} \right) \\\ & \Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{\ln \left( 1+\dfrac{h}{x} \right)}{h} \right) \\\ \end{aligned}$
Now, using the expansion formula of log given as: $\log \left( 1+a \right)=a-\dfrac{{{a}^{2}}}{2}+\dfrac{{{a}^{3}}}{3}-\dfrac{{{a}^{4}}}{4}+.........$, we get,
$\begin{aligned} & \Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{\dfrac{h}{x}-\dfrac{1}{2}{{\left( \dfrac{h}{x} \right)}^{2}}+\dfrac{1}{3}{{\left( \dfrac{h}{x} \right)}^{3}}-\dfrac{1}{4}{{\left( \dfrac{h}{x} \right)}^{4}}+......}{h} \right) \\\ & \Rightarrow f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{\dfrac{1}{x}-\dfrac{h}{2}{{\left( \dfrac{1}{x} \right)}^{2}}+\dfrac{{{h}^{2}}}{3}{{\left( \dfrac{1}{x} \right)}^{3}}-\dfrac{{{h}^{3}}}{4}{{\left( \dfrac{1}{x} \right)}^{4}}+......}{h} \right) \\\ \end{aligned}$
Substituting the value h = 0, we get,
$\Rightarrow f'\left( x \right)=\dfrac{1}{x}$
Hence, the above relation is our answer.

Note:
One may note that the limit definition of derivative is also known as the first principle of differentiation. It is the basic definition of derivative and all the formulas of derivatives of different functions are derived from the first principle. You must remember the expansion formula of logarithm to solve the above question. One of the important relations you may derive from here is: $\underset{a\to 0}{\mathop{\lim }}\,\left( \dfrac{\ln \left( 1+a \right)}{a} \right)=1$.