Math Problem Statement

y= 8x^3 e^-x find the derivative

Solution

To find the derivative of the function y=8x3exy = 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)g(x)y = f(x) \cdot g(x), then: y=f(x)g(x)+f(x)g(x)y' = f'(x)g(x) + f(x)g'(x)

  2. Let f(x)=8x3f(x) = 8x^3 and g(x)=exg(x) = e^{-x}.
    Now, find the derivatives of f(x)f(x) and g(x)g(x).

    • f(x)=ddx(8x3)=24x2f'(x) = \frac{d}{dx}(8x^3) = 24x^2
    • g(x)=ddx(ex)=exg'(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)y' = f'(x)g(x) + f(x)g'(x) Substituting the values: y=(24x2)(ex)+(8x3)(ex)y' = (24x^2)(e^{-x}) + (8x^3)(-e^{-x})

  4. Simplify the expression: y=24x2ex8x3exy' = 24x^2 e^{-x} - 8x^3 e^{-x}

  5. Factor out exe^{-x}: y=ex(24x28x3)y' = e^{-x}(24x^2 - 8x^3)

Thus, the derivative is: y=ex(24x28x3)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=8x3exy = 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.

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Product Rule
Chain Rule

Formulas

Product rule: (f(x)g(x))' = f'(x)g(x) + f(x)g'(x)
Chain rule: (e^(-x))' = -e^(-x)

Theorems

Product Rule
Chain Rule

Suitable Grade Level

Grades 11-12 or College Level