Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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DERIVATIVES

**Problem 7: Derivative Calculation**

Calculate the derivative of the function \( f(x) = 10x^4 - 32x + e^2 \).

**Solution:**

To find the derivative \( f'(x) \), we'll apply the rules of differentiation:

1. **Power Rule**: If \( f(x) = ax^n \), then \( f'(x) = nax^{n-1} \).
2. **Constant Rule**: The derivative of a constant is zero.

Applying these rules to each term in the function:

- For \( 10x^4 \): 
  - The derivative is \( 4 \times 10x^{4-1} = 40x^3 \).

- For \( -32x \): 
  - The derivative is \(-32 \).

- For \( e^2 \): 
  - Since \( e^2 \) is a constant, its derivative is \( 0 \).

**Final Derivative**: 

\[ f'(x) = 40x^3 - 32 \]
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Transcribed Image Text:**Problem 7: Derivative Calculation** Calculate the derivative of the function \( f(x) = 10x^4 - 32x + e^2 \). **Solution:** To find the derivative \( f'(x) \), we'll apply the rules of differentiation: 1. **Power Rule**: If \( f(x) = ax^n \), then \( f'(x) = nax^{n-1} \). 2. **Constant Rule**: The derivative of a constant is zero. Applying these rules to each term in the function: - For \( 10x^4 \): - The derivative is \( 4 \times 10x^{4-1} = 40x^3 \). - For \( -32x \): - The derivative is \(-32 \). - For \( e^2 \): - Since \( e^2 \) is a constant, its derivative is \( 0 \). **Final Derivative**: \[ f'(x) = 40x^3 - 32 \]
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