Power series for derivatives a. Differentiate the Taylor series centered at 0 for the following functions. b. Identify the function represented by the differentiated series. c. Give the interval of convergence of the power series for the derivative. f(x)=ln(1+x)

Taylor series of function f(x) at a is defined as:
$f(x)=f(a)+\frac{f^{\prime }(a)}{1!}(x-a)+\frac{f^{″}(a)}{2!}(x-a{)}^2+\frac{f^{‴}(a)}{3!}(x-a{)}^3+...$
Consider the given:
$f(x)=\ln (1+x)$
$f(x)\approx 0+\frac{\frac{d}{dx}(\ln (1+x))(0)}{1!}x+\frac{\frac{d^2}{d{x}^2}(\ln (1+x))(0)}{2!}{x}^2+\frac{\frac{d^3}{d{x}^3}(\ln (1+x))(0)}{3!}{x}^3+...$
$f(x)\approx 0+\frac{1}{1!}x+\frac{-1}{2!}{x}^2+\frac{2}{3!}{x}^3+\frac{-6}{4!}{x}^4+\frac{24}{5!}{x}^5+...$
$f(x)\approx x-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3-\frac{1}{4}{x}^4+\frac{1}{5}{x}^5+...$
b) The series sum representation-
$\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n+1}\frac{x^n}{n}$
Apply ratio test and find interval of convergence.
If $|\frac{a_{n+1}}{a_n}|\le q$ eventually for some $\sum _{n=1}^{\mathrm{\infty }}|a_n|$ converges
If $|\frac{a_{n+1}}{a_n}|>1$ eventually then $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ diverges.
Consider the given series:
$f(x)=\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n+1}\frac{x^n}{n}$
$a_n=(-1{)}^{n+1}\frac{x^n}{n}$
$a_{k+1}=(-1{)}^{n+2}\frac{x^{n+1}}{n+1}$
$\frac{a_{n+1}}{a_n}=\frac{(-1{)}^{n+2}\frac{x^{n+1}}{n+1}}{(-1{)}^{n+2}\frac{x^n}{n}}$
$=(-1)\frac{nx}{(n+1)}$
$L=\underset{n\to \mathrm{\infty }}{lim}|\frac{a_{n+1}}{a_n}|$
$=\underset{n\to \mathrm{\infty }}{lim}|(-1)\frac{x}{(1+\frac{1}{n})}|$
$=|-x|$
From ratio test, if L > 1 series diverges.
L< 1 series converges.
||x||<1 series converges.
Find the interval of convergence. $-1<x\le 1$
Interval of convergence $=\{x:x=-1<x\le 1\}$
Interval of convergence =(-1,1]