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The problem asks you to find \(\frac{dy}{dx}\) for the function:

\[
y = \frac{5}{\sqrt{x^3 - 2x}}
\]

Let's solve this step-by-step.

### Step 1: Rewrite the function for easier differentiation
The given function can be rewritten as:
\[
y = 5 \cdot (x^3 - 2x)^{-\frac{1}{2}}
\]

### Step 2: Apply the chain rule
We'll need to apply the chain rule and the power rule to differentiate.

1. The outer function is \( u^{-\frac{1}{2}} \), where \( u = x^3 - 2x \).
   - The derivative of \( u^{-\frac{1}{2}} \) is:
   \[
   -\frac{1}{2} u^{-\frac{3}{2}} \cdot \frac{du}{dx}
   \]

2. Now, calculate \( \frac{du}{dx} \), where \( u = x^3 - 2x \):
   \[
   \frac{du}{dx} = 3x^2 - 2
   \]

### Step 3: Combine everything
Now apply the chain rule:

\[
\frac{dy}{dx} = 5 \cdot \left( -\frac{1}{2} \right) \cdot (x^3 - 2x)^{-\frac{3}{2}} \cdot (3x^2 - 2)
\]

Simplify:

\[
\frac{dy}{dx} = -\frac{5(3x^2 - 2)}{2(x^3 - 2x)^{\frac{3}{2}}}
\]

### Step 4: Compare with the options
The correct option should match this expression. Let's compare the final form with the multiple-choice options:

The correct answer is:

\[
\boxed{a}
\]

Let me know if you'd like further details on this solution!

Here are five related questions to explore this further:
1. How would the process change if the exponent were different, such as \((x^3 - 2x)^{-\frac{3}{4}}\)?
2. What would happen if we had a different coefficient, say \( y = \frac{7}{\sqrt{x^3 - 2x}} \)?
3. Can you apply this method to a similar function like \( y = \frac{10}{\sqrt{x^2 + 3x}} \)?
4. How does the chain rule help in simplifying complex derivatives?
5. What is the significance of the negative exponent when differentiating functions like \( (x^3 - 2x)^{-\frac{1}{2}} \)?

### Tip:
Remember, always simplify the function as much as possible before differentiating. This can make the process easier and reduce errors.

Learn how to solve for the derivative of y = 5 / sqrt(x^3 - 2x) using the chain rule and power rule. This problem includes differentiation steps and algebraic manipulation, suitable for high school calculus students.

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