We are given the function $f(x) = (x+1)(x+2)(x+3)...(x+100)$. We need to find the derivative of $f(x)$ and evaluate it at $x = -1$.

1. Problem Description

We are given the function $f(x) = (x+1)(x+2)(x+3)...(x+100)$. We need to find the derivative of $f(x)$ and evaluate it at $x = -1$.

2. Solution Steps

First, we take the natural logarithm of both sides of the equation:

$ln(f(x)) = ln((x+1)(x+2)(x+3)...(x+100))$

$ln(f(x)) = ln(x+1) + ln(x+2) + ln(x+3) + ... + ln(x+100)$

Now, we differentiate both sides with respect to x:

$\frac{f'(x)}{f(x)} = \frac{1}{x+1} + \frac{1}{x+2} + \frac{1}{x+3} + ... + \frac{1}{x+100}$

Then we have:

$f'(x) = f(x) \left(\frac{1}{x+1} + \frac{1}{x+2} + \frac{1}{x+3} + ... + \frac{1}{x+100}\right)$

$f'(x) = (x+1)(x+2)(x+3)...(x+100) \left(\frac{1}{x+1} + \frac{1}{x+2} + \frac{1}{x+3} + ... + \frac{1}{x+100}\right)$

We need to evaluate $f'(-1)$. When $x = -1$, $f(x) = (-1+1)(-1+2)(-1+3)...(-1+100) = 0 * (1)(2)...(99) = 0$. So $f(-1) = 0$.

Then, we can examine the limit of the expression for $f'(x)$ as $x$ approaches $-1$.

$f'(-1) = \lim_{x \to -1} (x+1)(x+2)(x+3)...(x+100) \left(\frac{1}{x+1} + \frac{1}{x+2} + \frac{1}{x+3} + ... + \frac{1}{x+100}\right)$

We can rewrite this as:

$f'(-1) = \lim_{x \to -1} (x+2)(x+3)...(x+100) + (x+1)(x+2)(x+3)...(x+100) \left(\frac{1}{x+2} + \frac{1}{x+3} + ... + \frac{1}{x+100}\right)$

Then we substitute $x = -1$ into the first term:

$(-1+2)(-1+3)...(-1+100) = (1)(2)(3)...(99) = 99!$

In the second term, when $x = -1$, the expression $(x+1)(x+2)...(x+100)$ becomes 0, so the entire term becomes

0. Therefore:

$f'(-1) = 99! + 0 = 99!$

3. Final Answer

$f'(-1) = 99!$