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**Hint**: Assume the function = $f\left( x \right)={{\log }_{\text{e}}}\left( x \right)$.Since the function to be calculated consists of a small change, convert the given logarithmic function in the form of \[{{\log }_{\text{e}}}\left( x+dx \right)\] where ‘x’ represents the original value and ‘dx’ represents the small change in the original value. In this question; x= 4 and dx=0.01

**:**

__Complete step-by-step answer__Then, differentiate the function with respect to x and substitute the values of ‘x’ and ‘dx’ to get the solution.

Let a function \[f\left( x \right)=y={{\log }_{\text{e}}}\left( x \right)......(1)\]

Differentiate both sides of equation (1) with respect to ‘x’, we get:

\[f'\left( x \right)=\dfrac{dy}{dx}\]

Therefore, we can write: \[dy=f'\left( x \right)dx\]

So, from equation (1), we can say: \[dy=f'\left( {{\log }_{\text{e}}}\left( x \right) \right)dx\]

Since, \[f'\left( {{\log }_{\text{e}}}\left( x \right) \right)=\dfrac{1}{x}\]

So, we can write: \[dy=\dfrac{1}{x}dx......(2)\]

Now by increasing f(x) by an element ‘dx’, we get:

\[f\left( x+dx \right)={{\log }_{\text{e}}}\left( x+dx \right)\]

Comparing with equation (1), we can write:

\[y+dy={{\log }_{\text{e}}}\left( x+dx \right)\]

Therefore, \[dy={{\log }_{\text{e}}}\left( x+dx \right)-y\]

\[\Rightarrow dy={{\log }_{\text{e}}}\left( x+dx \right)-{{\log }_{\text{e}}}\left( x \right)......(3)\]

Substitute the value of ‘dy’ from equation (2) in equation (4):

\[\dfrac{1}{x}dx={{\log }_{\text{e}}}\left( x+dx \right)-{{\log }_{\text{e}}}\left( x \right)......(4)\]

Now, compare equation (4) with the function given in the question i.e. \[{{\log }_{\text{e}}}\left( 4.01 \right)\]

Consider x = 4 and dx = 0.01 and put the values in equation (4).

We get:

\[\left( \dfrac{1}{4}\times 0.01 \right)={{\log }_{\text{e}}}\left( 4.01 \right)-{{\log }_{\text{e}}}\left( 4 \right)\]

Therefore, \[{{\log }_{\text{e}}}\left( 4.01 \right)={{\log }_{\text{e}}}\left( 4 \right)+\left( \dfrac{1}{4}\times 0.01 \right)\]

\[\begin{align}

& \Rightarrow {{\log }_{\text{e}}}\left( 4.01 \right)=1.3868-0.0025 \\

& \Rightarrow {{\log }_{\text{e}}}\left( 4.01 \right)=1.3893 \\

\end{align}\]

**So, the correct answer is “Option C”.**

**Note**: As we know that, \[\begin{align}

& \underset{\Delta x\to 0}{\mathop{\lim }}\,\dfrac{\Delta y}{\Delta x}=\dfrac{dy}{dx}=f'\left( x \right) \\

& \Delta y=f'\left( x \right)\Delta x \\

& dy=f'\left( x \right)dx \\

\end{align}\] \[\underset{\Delta x\to 0}{\mathop{\lim }}\,\dfrac{\Delta y}{\Delta x}=\dfrac{dy}{dx}=f'\left( x \right)\]

Therefore, \[\Delta y=f'\left( x \right)\Delta x\]

Also \[dy=f'\left( x \right)dx\]

Hence, we can use differentials to calculate small changes in the dependent variable (dy) of a function corresponding to small changes in the independent variable f(x).

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