Question: > 
> Find the Maclaurin series for the function
>  $$f(x)=x\ln(x+1)$$
> 
> 
> 

So finding the derivatives is the first step. How many derivatives I need to find is explicitly said so I'll just go till the $4^{th}$ derivative.

$$\begin{align}
f'(x)&=\frac{x}{x+1}+\ln(x+1) & f'(0) &=0 \\
f''(x)&=\frac{1}{(x+1)^2}+\frac{1}{x+1} & f''(0)&=2 \\
f^{(3)}(x)&=\frac{-x-3}{(x+1)^3} & f^{(3)}(0)&=0 \\
f^{(4)}(x)&=\frac{2x+8}{(x+1)^4} & f^{(4)}(0)&=8
\end{align}$$

So now plugging in for the Maclaurin form 

$$P\_n(x)=0+x^2-\frac{1}{2}x^3+\frac{1}{3}x^4+....$$

Is this correct?

Answer: 
Adding to @Dr Zafar Ahmed's answer, A simpler approach is, if $b>1$ and $b^x = \log\_b(x)$, we automatically have $b^x = \log\_b(x) = x$. From the first and the last bit, we get $b = x^{1/x}$ (It is crucial here that b>1. See if you can think why).

The [Graph of x^(1/x)](https://i.stack.imgur.com/G0wbc.png) looks like this. A bit of basic calculus tells us that the function reaches its maximum at $e^{1/e}$. But notice also, that the function is one-one at that exact point only. 
So, given any "$y$" value, the function always takes 2 $x$ values. So, this directly means that in our case, we obtain a unique $b$ only when $b = e^{1/e} \sim 1.444667861$.
Number of real roots of
$$b^x=\log\_b x, b>1, x>0~~~~(1)$$
$$f(x)=b^x-\log\_b x \implies f(0)= +\infty,~ f(\infty)= +\infty.$$ So $f(x)=0$ will have no real roots or even number of real roots or one root (two coincident roots) critically if $b=b\_0$ such that $f(x\_0)=0$, when ${b\_0}^x=\frac{1}{x\_0}$ (when $y=b^x$ touches $y=\log\_b x$).

Next $$f'(x)=b^{x} \ln b- \frac{1}{x} \ln b>0 ~\text{for}~ b>1, x>0.$$
This means $f'(x)$ will have atmost one real root and this inturn will mean that $f(x)=0$ will have atmost two real roots.
For exampole for $b=\sqrt{2}$, $b^x=\log\_b x$ has two roots $x=2,4$.
Eq.(1) can have exactly one root only when $y=b^x$ touches $y=\log\_b x$ at $x=x\_0$, when $b=b\_0.$
such that ${b\_0}^x=\frac{1}{x\_0}$.

Finlly Eq. (1) will have one real root if $b=b\_0$, no real root if $b>b\_0$ and two reak roots if $1 <b <b\_0$. the value of $b\_0$ turns out to be a little more than $\sqrt{2}$, namely $b\_0=1.444667861009766.$ See the Fig. below for the critical stuation when the curves touch at $x=e$. $y=b^x$ is blue, $y=\log\_b x$ is read and $b=b\_0$

[![Fig.[1]](https://i.stack.imgur.com/hUsut.png)