To find the derivative of \((n+1)(-x)^n\) with respect to \(x\), we can apply the product rule, since we have a product of two functions: \(f(x) = n+1\) and \(g(x) = (-x)^n\).

The product rule states:
\[
(fg)' = f'g + fg'
\]

### Step 1: Differentiate each part
1. **Differentiate \(f(x) = n+1\):**
   \[
   f'(x) = 0
   \]
   since \(n+1\) is constant.

2. **Differentiate \(g(x) = (-x)^n\):**
   Using the chain rule:
   \[
   g'(x) = n(-x)^{n-1} \cdot (-1) = -n(-x)^{n-1}
   \]

### Step 2: Apply the product rule
Substitute these results into the product rule:
\[
\frac{d}{dx} \left((n+1)(-x)^n\right) = f'(x)g(x) + f(x)g'(x)
\]
\[
= 0 \cdot (-x)^n + (n+1)(-n)(-x)^{n-1}
\]
\[
= -(n+1)n(-x)^{n-1}
\]

### Final Answer
\[
\frac{d}{dx} \left((n+1)(-x)^n\right) = -(n+1)n(-x)^{n-1}
\]

Would you like further explanation or have any questions?

---

1. How does the chain rule apply in derivatives of functions with powers?
2. What is the importance of the product rule in calculus?
3. How would the answer change if \(n\) were a variable rather than a constant?
4. What is the derivative of \(x^n\) when \(n\) is a negative integer?
5. How can we extend this derivative to multivariable functions?

**Tip:** When differentiating powers of \(x\), remember to apply the chain rule if the base is more complex than \(x\) itself.

This page explains how to find the derivative of (n+1)(-x)^n using calculus techniques such as the product rule and chain rule. Detailed steps are provided, covering key concepts in differentiation for high school students in Grades 11-12.

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