If \[log3 = 1.0986\] then the approximate value of \[log\left( {9.01} \right)\] is
A) \[2.1983\]
B) \[2.1893\]
C) \[2.1389\]
D) \[2.1789\]

Hint: The derivative of a logarithmic function is the reciprocal of the argument. As always, the chain rule tells us to also multiply by the derivative of the argument. So if \[f\left( x \right) = \log \left( u \right)\] then
\[f'\left( x \right) = \dfrac{1}{u}.u'\]; \[\dfrac{d}{{dx}}(\log x) = \dfrac{1}{x}\]; \[\dfrac{d}{{dx}}\left( {lo{g_b}x} \right) = \dfrac{1}{{\left( {\log b} \right)x}}\]
Basic Idea: the derivative of a logarithmic function is that of the reciprocal of the things inside.
Using the properties of logarithms will sometimes make the differentiation process easier.
The derivative of a logarithmic function has been proved by using first principles.

Complete step-by-step answer:
It is given within the problem \[log3 = 1.0986\] and finds the approximate value of \[log\left( {9.01} \right)\].
For this, we will assume that \[y = logx\]
When differentiating the equation, we get then, \[\vartriangle x = \dfrac{{dy}}{{dx}} = \dfrac{1}{x}\]
And when \[x{\text{ = }}9\]so we get \[f'(x) = \vartriangle x = \dfrac{1}{9}\]
Now, we will assume that \[x{\text{ = }}9\],\[\vartriangle x{\text{ = }}0.01\]and \[f(x + \vartriangle x) = log\left( {9.01} \right)\]
Now, for finding the approximate value of \[f(x + \vartriangle x) = log\left( {9.01} \right)\].
\[
   \Rightarrow f(x){\text{ }} = {\text{ }}log\left( {{\text{9 }} + {\text{ 0}}{\text{.01}}} \right) \\
   \Rightarrow f(x){\text{ }} = {\text{ }}log\left( {{\text{x }} + {\text{ }}\vartriangle {\text{x}}} \right){\text{ }}
 \]
Now on differentiating the equation \[f(x) = log\left( {9.01} \right)\] with respect to $x$
Then the derivative of f(x) is given by:
\[ \Rightarrow {f'}(x){\text{ }} = {\text{ }}\mathop {\lim }\limits_{\vartriangle x \to 0} \dfrac{{f(x + \vartriangle x) - f(x)}}{{\vartriangle x}}\]
We write this as:
\[ \Rightarrow f(x + \vartriangle x) = \vartriangle x{f'}(x) + f(x)\]
\[ \Rightarrow f(x + \vartriangle x) = 0.01{f'}(9) + f(9)\]
Substituting the value of x =9 in the above equation.
\[ \Rightarrow f(x + \vartriangle x) = 0.01{f'}(9) + \log 9\]
Replacing 9 with 32 as nine is equal to square three.
\[ \Rightarrow f(x + \vartriangle x) = 0.01{f'}(9) + \log {3^2}\]
Substituting the value of \[f'(x) = \vartriangle x = \dfrac{1}{9}\] in the above equation.
\[ \Rightarrow f(x + \vartriangle x) = 0.01\left( {\dfrac{1}{9}} \right) + 2\log 3\]
Here, we will substitute the value of \[log3 = 1.0986\] in the above equation.
\[ \Rightarrow f(x + \vartriangle x) = \dfrac{{0.01}}{9} + 2.1972\]
So the final answer is :
\[ \Rightarrow f(x + \vartriangle x) = 0.0011 + 2.1972 = 2.1983\]
So, the approximate value of \[log\left( {9.01} \right)\] is \[2.1983\]

Therefore, the option (A) is the correct answer.

Note: In this question the value of \[log\left( {9.01} \right)\] has to be calculated. Unfortunately, we can only use the logarithm laws to help us in a limited number of logarithm differentiation question types. Most often, we like to seek out the derivative of a logarithm of some function of x. Differentiate the logarithmic functions.