To find the derivative of the function \( y = 8x^3 e^{-x} \), we will need to apply the **product rule** and the **chain rule**.

### Steps:

1. **Recall the product rule**:  
   If \( y = f(x) \cdot g(x) \), then:
   \[
   y' = f'(x)g(x) + f(x)g'(x)
   \]

2. Let \( f(x) = 8x^3 \) and \( g(x) = e^{-x} \).  
   Now, find the derivatives of \( f(x) \) and \( g(x) \).

   - \( f'(x) = \frac{d}{dx}(8x^3) = 24x^2 \)
   - \( g'(x) = \frac{d}{dx}(e^{-x}) = -e^{-x} \) (by the chain rule)

3. Now apply the product rule:
   \[
   y' = f'(x)g(x) + f(x)g'(x)
   \]
   Substituting the values:
   \[
   y' = (24x^2)(e^{-x}) + (8x^3)(-e^{-x})
   \]

4. Simplify the expression:
   \[
   y' = 24x^2 e^{-x} - 8x^3 e^{-x}
   \]

5. Factor out \( e^{-x} \):
   \[
   y' = e^{-x}(24x^2 - 8x^3)
   \]

Thus, the derivative is:
\[
y' = e^{-x}(24x^2 - 8x^3)
\]

Would you like further explanation or clarification on any step?

### 5 Related Questions:
1. How is the product rule applied to differentiate two functions?
2. What is the chain rule and how does it apply to exponential functions?
3. Can you find the second derivative of \( y = 8x^3 e^{-x} \)?
4. How does factoring simplify the final derivative expression?
5. What are the general rules for differentiating exponential functions?

### Tip:
When using the product rule, it’s helpful to differentiate each part separately and then combine them, ensuring all steps are clear and organized.

Learn how to find the derivative of the function y = 8x^3 e^(-x) using the product rule and chain rule. This calculus problem involves key differentiation techniques and provides a detailed step-by-step solution. Suitable for students in Grades 11-12 or at the college level.

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