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Hint: Find the 1st, 2nd, 3rd… order derivatives at x=0 and substitute in general form of Maclaurin’s series expansion.

Let us suppose the given function as \[f\left( x \right) = {\log _e}\left( {1 + x} \right)\]

According to Maclaurin’s series expansion for any function,

\[f\left( x \right) = f\left( 0 \right) + \dfrac{x}{{1!}}f'\left( 0 \right) + \dfrac{{{x^2}}}{{2!}}f''\left( 0 \right) + \dfrac{{{x^3}}}{{3!}}f'''\left( 0 \right) + ......{\text{ }} \to {\text{(1)}}\]

Now, let us find \[f'\left( x \right)\] by differentiating \[f\left( x \right) = {\log _e}\left( {1 + x} \right)\] with respect to $x$, we get

\[{\text{ }}f'\left( x \right) = \dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} = \dfrac{{d\left( {{{\log }_e}\left( {1 + x} \right)} \right)}}{{dx}} = \left( {\dfrac{1}{{1 + x}}} \right)\left( {\dfrac{{d\left( {1 + x} \right)}}{{dx}}} \right) = \dfrac{1}{{1 + x}}\]

For \[f''\left( x \right)\], differentiate \[f'\left( x \right)\] with respect to $x$, we get

\[f''\left( x \right) = \dfrac{{d\left( {f'\left( x \right)} \right)}}{{dx}} = \dfrac{{d\left( {\dfrac{1}{{1 + x}}} \right)}}{{dx}} = \dfrac{{d\left[ {{{\left( {1 + x} \right)}^{ - 1}}} \right]}}{{dx}} = - \dfrac{1}{{{{\left( {1 + x} \right)}^2}}}\left( {\dfrac{{d\left( {1 + x} \right)}}{{dx}}} \right) = - \dfrac{1}{{{{\left( {1 + x} \right)}^2}}}\]

Similarly, \[f'''\left( x \right) = \dfrac{{d\left( {f''\left( x \right)} \right)}}{{dx}} = \dfrac{{d\left( { - \dfrac{1}{{{{\left( {1 + x} \right)}^2}}}} \right)}}{{dx}} = - \dfrac{{d\left( {{{\left( {1 + x} \right)}^{ - 2}}} \right)}}{{dx}} = \dfrac{2}{{{{\left( {1 + x} \right)}^3}}}\]

Now, put \[x = 0\] in the expressions of \[f\left( x \right),{\text{ }}f'\left( x \right),{\text{ }}f''\left( x \right),f'''\left( x \right)\], etc.

Therefore, \[f\left( 0 \right) = {\log _e}\left( {1 + 0} \right) = {\log _e}\left( 1 \right) = 0\]

\[f'\left( 0 \right) = \dfrac{1}{{1 + 0}} = 1\], \[f''\left( 0 \right) = - \dfrac{1}{{{{\left( {1 + 0} \right)}^2}}} = - 1\], \[f'''\left( 0 \right) = \dfrac{2}{{{{\left( {1 + 0} \right)}^3}}} = 2\], etc.

Now substitute all the above values in equation (1), we get

\[

f\left( x \right) = 0 + \dfrac{x}{1} \times 1 + \dfrac{{{x^2}}}{2} \times \left( { - 1} \right) + \dfrac{{{x^3}}}{6} \times 2 + ....... \\

\Rightarrow f\left( x \right) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + ....... \\

\]

The above equation represents the Maclaurin’s series expansion for \[f\left( x \right) = {\log _e}\left( {1 + x} \right)\].

Note: These types of problems are solved by finding the first derivative, second derivative, third derivative and so on of the given function and then finally substituting the value of the variable as 0. In this particular question, we have calculated only up to the third derivative just in order to avoid unnecessary computations. If we observe carefully the next term can be predicted easily.

Let us suppose the given function as \[f\left( x \right) = {\log _e}\left( {1 + x} \right)\]

According to Maclaurin’s series expansion for any function,

\[f\left( x \right) = f\left( 0 \right) + \dfrac{x}{{1!}}f'\left( 0 \right) + \dfrac{{{x^2}}}{{2!}}f''\left( 0 \right) + \dfrac{{{x^3}}}{{3!}}f'''\left( 0 \right) + ......{\text{ }} \to {\text{(1)}}\]

Now, let us find \[f'\left( x \right)\] by differentiating \[f\left( x \right) = {\log _e}\left( {1 + x} \right)\] with respect to $x$, we get

\[{\text{ }}f'\left( x \right) = \dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} = \dfrac{{d\left( {{{\log }_e}\left( {1 + x} \right)} \right)}}{{dx}} = \left( {\dfrac{1}{{1 + x}}} \right)\left( {\dfrac{{d\left( {1 + x} \right)}}{{dx}}} \right) = \dfrac{1}{{1 + x}}\]

For \[f''\left( x \right)\], differentiate \[f'\left( x \right)\] with respect to $x$, we get

\[f''\left( x \right) = \dfrac{{d\left( {f'\left( x \right)} \right)}}{{dx}} = \dfrac{{d\left( {\dfrac{1}{{1 + x}}} \right)}}{{dx}} = \dfrac{{d\left[ {{{\left( {1 + x} \right)}^{ - 1}}} \right]}}{{dx}} = - \dfrac{1}{{{{\left( {1 + x} \right)}^2}}}\left( {\dfrac{{d\left( {1 + x} \right)}}{{dx}}} \right) = - \dfrac{1}{{{{\left( {1 + x} \right)}^2}}}\]

Similarly, \[f'''\left( x \right) = \dfrac{{d\left( {f''\left( x \right)} \right)}}{{dx}} = \dfrac{{d\left( { - \dfrac{1}{{{{\left( {1 + x} \right)}^2}}}} \right)}}{{dx}} = - \dfrac{{d\left( {{{\left( {1 + x} \right)}^{ - 2}}} \right)}}{{dx}} = \dfrac{2}{{{{\left( {1 + x} \right)}^3}}}\]

Now, put \[x = 0\] in the expressions of \[f\left( x \right),{\text{ }}f'\left( x \right),{\text{ }}f''\left( x \right),f'''\left( x \right)\], etc.

Therefore, \[f\left( 0 \right) = {\log _e}\left( {1 + 0} \right) = {\log _e}\left( 1 \right) = 0\]

\[f'\left( 0 \right) = \dfrac{1}{{1 + 0}} = 1\], \[f''\left( 0 \right) = - \dfrac{1}{{{{\left( {1 + 0} \right)}^2}}} = - 1\], \[f'''\left( 0 \right) = \dfrac{2}{{{{\left( {1 + 0} \right)}^3}}} = 2\], etc.

Now substitute all the above values in equation (1), we get

\[

f\left( x \right) = 0 + \dfrac{x}{1} \times 1 + \dfrac{{{x^2}}}{2} \times \left( { - 1} \right) + \dfrac{{{x^3}}}{6} \times 2 + ....... \\

\Rightarrow f\left( x \right) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + ....... \\

\]

The above equation represents the Maclaurin’s series expansion for \[f\left( x \right) = {\log _e}\left( {1 + x} \right)\].

Note: These types of problems are solved by finding the first derivative, second derivative, third derivative and so on of the given function and then finally substituting the value of the variable as 0. In this particular question, we have calculated only up to the third derivative just in order to avoid unnecessary computations. If we observe carefully the next term can be predicted easily.

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