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The given expression is:

\[
f(x) = 7 \sqrt[3]{x} - \frac{3}{x^3} - 3x + 6
\]

We need to find the derivative \( f'(x) \).

### Step-by-step solution:

1. **Rewrite the terms in exponent form**:
   - \( 7 \sqrt[3]{x} \) becomes \( 7x^{1/3} \)
   - \( \frac{3}{x^3} \) becomes \( 3x^{-3} \)
   - The term \( -3x \) remains as is.
   - The constant \( 6 \) remains unchanged (its derivative will be 0).

So the function now looks like:
\[
f(x) = 7x^{1/3} - 3x^{-3} - 3x + 6
\]

2. **Differentiate each term**:
   - For \( 7x^{1/3} \), apply the power rule: \( \frac{d}{dx}(x^n) = n x^{n-1} \)
     \[
     \frac{d}{dx}(7x^{1/3}) = 7 \cdot \frac{1}{3} x^{1/3 - 1} = \frac{7}{3}x^{-2/3}
     \]
   - For \( -3x^{-3} \), apply the power rule again:
     \[
     \frac{d}{dx}(-3x^{-3}) = -3 \cdot (-3) x^{-4} = 9x^{-4}
     \]
   - For \( -3x \), the derivative is simply:
     \[
     \frac{d}{dx}(-3x) = -3
     \]
   - The derivative of the constant \( 6 \) is 0.

3. **Combine the results**:
   \[
   f'(x) = \frac{7}{3}x^{-2/3} + 9x^{-4} - 3
   \]

Thus, the derivative of the given function is:

\[
f'(x) = \frac{7}{3}x^{-2/3} + 9x^{-4} - 3
\]

Would you like more details on any specific part of this solution?

### 5 Related Questions:
1. How do you apply the power rule to expressions with fractional exponents?
2. What is the derivative of a cube root function like \( \sqrt[3]{x} \)?
3. How can you differentiate rational expressions like \( \frac{1}{x^3} \)?
4. Why is the derivative of a constant always zero?
5. Can you explain the general rules for differentiating terms with negative exponents?

### Tip:
When differentiating functions with roots or fractional exponents, always rewrite them in exponent form to simplify the application of the power rule.

Find the derivative of: 7√x − 3/x³ − 3x + 6

This page explains how to find the derivative of the function 7√x − 3/x³ − 3x + 6 using the power rule. The solution covers differentiation of terms with fractional and negative exponents, providing a detailed step-by-step process. Suitable for students in Grades 11-12 studying calculus.

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