derivative of limited summation what's the simple way to find $$\frac{d}{dx}$$ of limited summation ? 
$$\frac{d}{dx}\sum_{n=1}^{13}\left({x^n}\right)$$
Is there a general formula for this?
what's the sum when x=1
 A: \begin{equation}
\dfrac{d}{dx} \sum_{n = 1}^{13}x^{n} = \sum_{n = 1}^{13}\dfrac{d}{dx} x^{n} = \sum_{n = 1}^{13} n x^{n - 1}
\end{equation}
A: HINT You can do termwise differentiation which is valid within the annulus of convergence or if you do not want a sum as an answer you can for example use the factorization $$(1+x+\cdots + x^n)(x-1) = x^{n+1}-1$$ followed by differentiating the function this gives you.
A: We can write :
$\tag{1}f(x):=x^1 + x^2 + x^3 + \dots +x^{13}= x(1+x^1 + x^2 + x^3 + \dots +x^{12})=x \dfrac{\ \ 1-x^{13}}{1-x}$
by using the formula for a (finite) sum of geometric series.
Differentiation of (1) (using formula $(uv)'=u'v+uv'$, then $(\dfrac{u}{v})'=\dfrac{u'v-uv'}{v^2}$) gives the closed form expression:
$$f'(x)=\dfrac{1-x^{13}}{1-x}+x \dfrac{-13x^{12}(1-x)+(1-x^{13})}{(1-x)^2}$$
which can be even further simplified into:
$$=\dfrac{(1-x^{13})(1-x)+x\left(-13x^{12}(1-x)+(1-x^{13})\right)}{(1-x)^2}$$
$$=\dfrac{(1-x^{13})-13x^{13}(1-x)}{(1-x)^2}$$

$$\tag{2}f'(x)=\begin{cases}\dfrac{13x^{14}-14x^{13}+1}{(1-x)^2}& \text{if} \ x \neq 1 & \ \ (a)\\\dfrac{13 \times 14}{2}& \text{if} \ x = 1 & \ \ (b)\end{cases}$$

(see Remark below for the case (2b).)
The method employed here is clearly general : instead of 13 in (1), one could take a general $N$ and obtain :

$$\tag{2}f(x)=\sum_{n=1}^N x^n \ \implies \ f'(x)=\begin{cases}\dfrac{Nx^{N+1}-(N+1)x^{N}+1}{(1-x)^2}& \text{if} \ x \neq 1 & \ \ (a)\\\dfrac{N(N+1)}{2}& \text{if} \ x = 1 & \ \ (b)\end{cases}$$

Remark (explanation of formula (2b)): According to (2a), $f'(x)$ looks undefined for $x=1$ : it appears as a limit $\dfrac{\infty}{\infty}$, but we can overcome this difficulty. Setting $f(x)=\dfrac{N(x)}{D(x)}$, let us apply l'Hospital's rule twice, and get :
$$f'(1):=\lim_{x \to 1}f'(x)=\lim_{x \to 1} \dfrac{N''(x)}{D''(x)}=\dfrac{13 \times 14}{2}$$
explainig the value in formula (2b).
A: there is a formula for the derivative of $a_0+a_1x+a_2x^2...a_nx^n=\sum\limits_{j=0}^na_jx^j$.
$$\boxed{f^{(k)}=\sum_{j=k}^{n}a_j\frac{j!}{(j-k)!}x^{j-k}}$$ where $k$ is the number of derivatives you take. in this case we have $a_0=0,a_j=1,k=1,n=13$ hence$${f'=\sum_{j=1}^{13}\frac{j!}{(j-1)!}x^{j-1}}=\sum_{j=1}^{13}jx^{j-1}$$
you can find the proof for this formula here
