The function given is:

$f(x) = \sqrt{-7x^3 + 7}$

We are tasked with finding the derivative of $f(x)$. Let's calculate it step by step.

Step 1: Rewrite the function

The square root can be rewritten as a power of $1/2$:

$f(x) = \left( -7x^3 + 7 \right)^{1/2}$

Step 2: Differentiate using the chain rule

The derivative of $f(x)$ is calculated as:

$f'(x) = \frac{1}{2} \left( -7x^3 + 7 \right)^{-1/2} \cdot \frac{d}{dx}(-7x^3 + 7)$

Step 3: Differentiate the inner function

The derivative of the inner function, $-7x^3 + 7$, is:

$\frac{d}{dx}(-7x^3 + 7) = -21x^2$

Step 4: Combine the results

Substitute the derivative of the inner function into the chain rule expression:

$f'(x) = \frac{1}{2} \left( -7x^3 + 7 \right)^{-1/2} \cdot (-21x^2)$

Simplify:

$f'(x) = \frac{-21x^2}{2\sqrt{-7x^3 + 7}}$

Final Answer:

$f'(x) = \frac{-21x^2}{2\sqrt{-7x^3 + 7}}$

This is the derivative of the given function.

Would you like further clarification or a graphical representation of this? Here are some related questions to deepen understanding:

1. What is the domain of $f(x)$ where the derivative exists?
2. How would the behavior of $f'(x)$ change for values of $x$ close to the boundaries of the domain?
3. Can you explain why the chain rule is applied in this problem?
4. How would you verify the derivative using a numerical approach?
5. What happens to $f'(x)$ as $x \to \pm\infty$, if applicable?

Tip: Always check the domain of the square root to ensure the expression inside is non-negative.