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.