https://file.aiquickdraw.com/mathbot/file/1725165812943q1cjeda1.jpg

The expression in the image is \( f(x) = (x+1)(x+2)(x+3)\dots(x+100) \) and you are asked to calculate its derivative \( f'(x) \).

### Step-by-Step Solution:

The function \( f(x) \) is a product of linear factors:

\[
f(x) = (x+1)(x+2)(x+3)\dots(x+100)
\]

This function can be written as:

\[
f(x) = \prod_{k=1}^{100} (x+k)
\]

To find the derivative \( f'(x) \), we can use the product rule. The product rule for derivatives states that if you have a function that is the product of several functions \( u(x) = v(x) \cdot w(x) \), then its derivative is:

\[
u'(x) = v'(x) \cdot w(x) + v(x) \cdot w'(x)
\]

For a product of multiple functions, this generalizes to:

\[
f'(x) = \sum_{k=1}^{100} \left[ \prod_{\substack{j=1 \\ j \neq k}}^{100} (x+j) \right]
\]

This means that for each \( k \), we differentiate the factor \( (x+k) \), which simply gives 1, and then multiply by the product of all the remaining factors.

### Example for Small Case:
To clarify this approach, consider a simpler case with three factors:

\[
f(x) = (x+1)(x+2)(x+3)
\]

The derivative \( f'(x) \) would be:

\[
f'(x) = \frac{d}{dx}[(x+1)(x+2)(x+3)] = (x+2)(x+3) + (x+1)(x+3) + (x+1)(x+2)
\]

This approach applies similarly to the 100-factor case, though the full expression would be much larger.

### Conclusion:
The derivative \( f'(x) \) can be expressed as a sum of 100 terms, where each term is the product of 99 of the linear factors, omitting one at a time.

Would you like a more detailed step-by-step calculation or have any specific questions?

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Here are five related questions:
1. How does the product rule for derivatives apply to multiple factors?
2. What is the general formula for the derivative of a product of \( n \) functions?
3. How would you approximate \( f'(x) \) for large \( x \)?
4. What special properties might \( f'(x) \) have at specific points?
5. How could you extend this method to the product of nonlinear functions?

**Tip:** When dealing with products of many functions, breaking down the problem using the product rule and considering smaller cases can make the solution easier to understand.

Learn how to calculate the derivative of the function f(x) = (x+1)(x+2)(x+3)...(x+100) using the product rule for derivatives. This problem involves differentiating a product of linear factors, suitable for advanced high school students and college-level calculus learners.

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