2 beginner calculus questions - Help needed I am in an online bridge course starting my masters after not doing calculus for 12 years.  My professor does not reply to emails, online "office hours" ive been told to email him since we have moved past this, but he still doesn't reply.  So I am posting homework that I simply do not understand and have failed in the internet to find answers.
1st: Find the derivative of 
$$f_k(x)=\sum_{n=0}^k\frac{x^n}{n!}$$
2nd: Find the derivative of 
$$f(x)=\sum_{n=0}^\infty\frac{x^n}{n!}$$
Assuming that you can differentiate an infinite sum term-wise.
Based on its derivative, can you guess what function $f(x)$ is?
3rd: Find  the derivitive of $$f(p) = \ln\left(\prod_{i=1}^np^{x_i}(1-p)^{1-x_i} \right)$$
Between 1 and 2: I don't understand how the derivative would change based upon the summation end changing from a real number to infinity.
For the 3rd, I feel like my basic algebra is failing me because I get $$\ln\left(\prod(p-p)^1\right)$$  which I am very certain is not correct.  
Any help would be appreciated to understand what is going on here and what I'm missing.  Thanks in advance for my help on my first post.
 A: $(1) $First, note that the first term in this series is a constant real number with no variable in it. The remaining terms will be a function of $x$...
$$\sum_{n=0}^{k} \frac{x^n}{n!} = \frac{x^0}{0!} + \sum_{n=1}^{k} \frac{x^n}{n!} = \frac{1}{1} + \sum_{n=1}^{k} \frac{x^n}{n!} = 1 + \sum_{n=1}^{k} \frac{x^n}{n!}$$
Second, recall that the derivative of sums or differences is the sums or differences of the derivatives. Hence,
$$\frac{d}{dx} \sum_{n=0}^{k} \frac{x^n}{n!} = \frac{d}{dx} [1 + \sum_{n=1}^{k} \frac{x^n}{n!}] = \frac{d}{dx} [1] + \frac{d}{dx}[\sum_{n=1}^{k} \frac{x^n}{n!}] = \frac{d}{dx}\sum_{n=1}^{k} \frac{x^n}{n!} = \sum_{n=1}^{k} \frac{d}{dx} \frac{x^n}{n!}$$
Now you differentiate with respect to $x$. You may be wondering "What do I do with $n$?" The answer is that you treat $n$ as though it is a constant. Note that $n$ is an index that tells you how many terms are in the series as well as how the constants in each term differ from one term to the next. So,
$$\sum_{n=1}^{k} \frac{d}{dx}\frac{x^n}{n!} = \sum_{n=1}^{k} \frac{1}{n!} \frac{d}{dx}x^n = \sum_{n=1}^{k} \frac{1}{n!} n x^{(n-1)} = \sum_{n=1}^{k} \frac{n x^{(n-1)}}{n!}$$
Now, note that there is an opportunity to simplify this fraction if we expand the factorial some...
$$\sum_{n=1}^{k} \frac{n x^{(n-1)}}{n!} = \sum_{n=1}^{k} \frac{n x^{(n-1)}}{n(n-1)!} = \sum_{n=1}^{k} \frac{x^{(n-1)}}{(n-1)!}$$
Thus, 
$$\frac{d}{dx} \sum_{n=0}^{k} \frac{x^n}{n!} = \sum_{n=1}^{k} \frac{x^{(n-1)}}{(n-1)!}$$

$(2)$ You are correct about nothing changing between $1$ and $2$. The reasoning is the same, only now it translates to a sum over infinite terms. Nice intuition! 
