Taylor Polynomial $\ln(1-x)$ Given $f(x) = \ln(1-x)$ centered at $0$, with $n = 0,1,2$:
Why is it that the first derivative (for $n=1$): $$\frac{d}{dx}(p(x)) = -x$$ in a Taylor Polynomial? The answer that I got was $$\frac{d}{dx}(p(x)) = 1$$
I got the right answer  for the second derivative $$\frac{d^2}{d^2x}(p(x)) = -\frac{x^2}{2}$$
Overall the final Taylor polynomial is: 
$$-x-\frac{x^2}{2}$$ 
I got: $$x-\frac{x^2}{2}$$
Where exactly is that negative coming from?
 A: You can derive this without the use of derivatives, 
$$
\frac{1}{1-x}=\sum_{n=0}^\infty x^n
$$
for $|x|<1$. 
Then integrating 
we find 
$$
-\ln(1-x)+C=\sum_{n=0}^\infty \frac{x^{n+1}}{n+1}
$$
where the negative came from the substitution $u=1-x$ in the integral. Plugging in $x=0$ we find 
$$
C=0
$$
so putting it all together we have 
$$
\ln(1-x)=-\sum_{n=0}^\infty \frac{x^{n+1}}{n+1}
$$
A: Considering $y = 1-x$ by the chain rule we have
$$ f'(x) = \frac{d}{dx}[\ln(y)] = \frac{d}{dy}[\ln(y)] \frac{dy}{dx} = \frac{1}{y} (-1) = - \frac{1}{1-x} = \frac{1}{x-1} $$
That's where the negative comes from.
A: \begin{align}
\int\frac{du}{u-1} = {} & \log\left| u-1 \right| + C \\[10pt]
= {} & \log(1-u) + C \text{ if $ u$ is near $1$} \\
& \text{since in that case, $u-1$ is negative.} \\[10pt]
\text{So } \int_0^x \frac{du}{u-1} & = \log(1-x) - \log(1-0) = \log(1-x).
\end{align}
Thus we have
\begin{align}
\log(1-x) & = \int_0^x \frac{du}{u-1} = \int_0^x \left( -1-u - u^2 -u^3 -u^4 - \cdots \right) \,du \\[10pt]
& = \left[ -u -\frac {u^2} 2 - \frac{u^3} 3 - \frac {u^4} 4 - \cdots \right]_{u\,:=\,0}^{u\,:=\,x} = -x-\frac{x^2} 2 - \frac{x^3} 3 - \frac {x^4} 4 - \cdots
\end{align}
The series on the right above, and the function on the left above, should be equal in the values of their $n$th derivatives for $n=0,1,2,3,\ldots\,.$
