MCQ The nth derivative of $f(x)=\frac{1+x}{1-x}$ Let $f(x)=\dfrac{1+x}{1-x}$ The nth derivative of f is equal to:


*

*$\dfrac{2n}{(1-x)^{n+1}} $

*$\dfrac{2(n!)}{(1-x)^{2n}} $

*$\dfrac{2(n!)}{(1-x)^{n+1}} $


by Leibniz formula 
$$ {\displaystyle \left( \dfrac{1+x}{1-x}\right)^{(n)}=\sum _{k=0}^{n}{\binom {n}{k}}\ (1+x)^{(k)}\ \left(\dfrac{1}{1-x}\right)^{(n-k)}}$$
using the hint 


*

*$\dfrac{1+x}{1-x}=\dfrac{2-(1-x)}{1-x}=\dfrac2{1-x}-1$ and 

*$\left(\dfrac{1}{x}\right)^{n}=\dfrac{(-1)^{n}n!}{x^{n+1}}$


so 
$${\displaystyle \left( \dfrac{1+x}{1-x} \right)^{(n)} = \left( \dfrac{2}{1-x}-1 \right)^{(n)}=2\dfrac{ (-1)^{n}n! }{ (1-x)^{n+1} }   }  $$ but this result isn't apear in any proposed answers
what about the method of Lord Shark the Unknown
tell me please this way holds for any mqc question contain find the n th derivative so it's suffice to check each answer in y case i will start with first   


*

*let $f_n(x)=\dfrac{2n}{(1-x)^{n+1}}$ then $f_{n+1}(x)=\dfrac{2(n+1)}{(1-x)^{n+2}}$ do i have $f'_{n}=f_{n+1}$  let calculate $$ f'_n=\dfrac{-2n(n+1)}{(1-x)^{n+2}}\neq f_{n+1}$$

*let $f_n(x)=\dfrac{2(n!)}{(1-x)^{2n}}$ then $f_{n+1}(x)=\dfrac{2((n+1)!)}{(1-x)^{2(n+1)}}$ do i have $f'_{n}=f_{n+1}$  let calculate $$ f'_n=\dfrac{-2(n!)(2n)}{(1-x)^{4n}}\neq f_{n+1}$$

*let $f_n(x)=\dfrac{2(n!)}{(1-x)^{n+1}}$ then $f_{n+1}(x)=\dfrac{2((n+1)!)}{(1-x)^{n+2}}$ do i have $f'_{n}=f_{n+1}$  let calculate $$ f'_n=\dfrac{2(n!)(n+1)}{((1-x)^{n+1})^{2}}=\dfrac{2((n+1)!)}{(1-x)^{2n+2}}\neq f_{n+1}$$


You can do it with Leibniz's formula (not that it's easier than without it); just consider that
$$
(1+x)^{(k)}=
\begin{cases}
1+x & k=0 \\
1 & k=1 \\
0 & k>1
\end{cases}
$$
so the formula gives
$$
\left( \frac{1+x}{1-x}\right)^{\!(n)}=
(1+x)\left(\dfrac{1}{1-x}\right)^{\!(n)}+
n\left(\frac{1}{1-x}\right)^{\!(n-1)}
$$
Now make a conjecture about
$$
\left(\frac{1}{1-x}\right)^{\!(n)}
$$
