How do you derive the taylor polynomial for $ \dfrac{x}{1+x}$ ?

Hint: In order to do this problem, you can use the differentiation of the above term and then continue differentiating till we get a known series. Then you have to put all these resulting values sin the taylor formula to get the required answer.

Complete step by step answer:
According to the problem, we are asked to derive the taylor polynomial for $ \dfrac{x}{1+x}$.
The taylor’s formula is given by $f\left(x\right)= \sum\limits_{n=0}^{\infty }{\dfrac{{{f}^{\left( n \right)}}\left( a \right)}{n!}{{\left( x-a \right)}^{n}}}$------ (1)
First, we take $ f\left( x \right)=\dfrac{x}{1+x}$. So first we find the derivative of f(x)
Therefore, after differentiating we get:
We use the $\dfrac{u}{v}$ rule.
$ \Rightarrow f'\left( x \right)=\dfrac{{\left(1+x\right).1}{-x\left(1\right)}}{{{\left( 1+x \right)}^{2}}}=\dfrac{1}{{{\left( 1+x \right)}^{2}}}$
Then differentiating again the f’(x), we get
Here, we use the formula $f\left(x\right)=\dfrac{1}{{{x}^{n}}} \Rightarrow f'\left(x\right)=\dfrac{-n}{{{x}^{n+1}}}$. Therefore, we get:
$ \Rightarrow f'\left( x \right)=-\dfrac{2}{{{\left( 1+x \right)}^{3}}}$
Then differentiating again the f’’(x), we get
Here, we use the formula $f\left(x\right)=\dfrac{1}{{{x}^{n}}} \Rightarrow f'\left(x\right)=\dfrac{-n}{{{x}^{n+1}}}$. Therefore, we get:
$ \Rightarrow f'\left( x \right)=\dfrac{6}{{{\left( 1+x \right)}^{4}}}$
Then differentiating again the f’’’(x), we get
Here, we use the formula $f\left(x\right)=\dfrac{1}{{{x}^{n}}} \Rightarrow f'\left(x\right)=\dfrac{-n}{{{x}^{n+1}}}$. Therefore, we get:
$ \Rightarrow f'\left( x \right)=-\dfrac{24}{{{\left( 1+x \right)}^{5}}}$
We have to differentiate them continually to infinity. But now, we have established a series.
In the next step, we have to find all the derivatives of f(x) in equation 1. After substituting all the values, we get:
$\Rightarrow \dfrac{f\left( a \right)}{0!}{{\left( x-a \right)}^{0}}+\dfrac{f'\left( a \right)}{1!}{{\left( x-a \right)}^{1}}+\dfrac{f''\left( a \right)}{2!}{{\left( x-a \right)}^{2}}+\dfrac{f'''\left( a \right)}{3!}{{\left( x-a \right)}^{3}}+\dfrac{f''''\left( a \right)}{4!}{{\left( x-a \right)}^{4}}...........$
Now we have to substitute the values
$\Rightarrow \dfrac{a}{1+a}.1+\dfrac{1}{{{\left( 1+a \right)}^{2}}}\left( x-a \right)-\dfrac{2}{2.{{\left( 1+a \right)}^{3}}}{{\left( x-a \right)}^{2}}+\dfrac{6}{6.{{\left( 1+a \right)}^{4}}}{{\left( x-a \right)}^{3}}-\dfrac{24}{24.{{\left( 1+a \right)}^{5}}}{{\left( x-a \right)}^{4}}...........$
Further simplifying the equation, we get
$\Rightarrow \dfrac{a}{1+a}.+\dfrac{1}{{{\left( 1+a \right)}^{2}}}\left( x-a \right)-\dfrac{1}{{{\left( 1+a \right)}^{3}}}{{\left( x-a \right)}^{2}}+\dfrac{1}{{{\left( 1+a \right)}^{4}}}{{\left( x-a \right)}^{3}}-\dfrac{1}{{{\left( 1+a \right)}^{5}}}{{\left( x-a \right)}^{4}}...........$
Finally, we have derived the taylor series for the term $\dfrac{x}{1+x}$.
Therefore, after all the derivation, we get the taylor polynomial for the term $\dfrac{x}{1+x}$ as,
$\dfrac{a}{1+a}.1+\dfrac{1}{{{\left( 1+a \right)}^{2}}}\left( x-a \right)-\dfrac{2}{2.{{\left( 1+a \right)}^{2}}}{{\left( x-a \right)}^{2}}+\dfrac{6}{6.{{\left( 1+a \right)}^{3}}}{{\left( x-a \right)}^{3}}-\dfrac{24}{24.{{\left( 1+a \right)}^{4}}}{{\left( x-a \right)}^{4}}...........$.

Note:
In these kinds of questions, correct substitution is important. If one substitution goes wrong, the whole answer may go wrong. In this question, you could verify the derivatives by using integration. If after integration of the answer, you get the question, then the derivative is correct.