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For part (iii) of the question, the function is:

\[
y = \ln(2x^2) + \sin(x^2)
\]

We need to find \(\frac{dy}{dx}\).

### Step-by-step solution:

1. **Differentiate \( \ln(2x^2) \):**
   Using the chain rule, the derivative of \(\ln(u)\) is \(\frac{1}{u} \cdot \frac{du}{dx}\), where \(u = 2x^2\). So:
   \[
   \frac{d}{dx}[\ln(2x^2)] = \frac{1}{2x^2} \cdot \frac{d}{dx}[2x^2] = \frac{1}{2x^2} \cdot 4x = \frac{2}{x}.
   \]

2. **Differentiate \(\sin(x^2)\):**
   Again, using the chain rule, the derivative of \(\sin(u)\) is \(\cos(u) \cdot \frac{du}{dx}\), where \(u = x^2\). So:
   \[
   \frac{d}{dx}[\sin(x^2)] = \cos(x^2) \cdot \frac{d}{dx}[x^2] = \cos(x^2) \cdot 2x.
   \]

3. **Combine the results:**
   Adding the derivatives of the two parts:
   \[
   \frac{dy}{dx} = \frac{2}{x} + 2x\cos(x^2).
   \]

### Final Answer:
\[
\frac{dy}{dx} = \frac{2}{x} + 2x\cos(x^2).
\]

Would you like a detailed explanation of any step, or assistance with the other parts? 

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### Relative Questions:
1. How does the chain rule apply to logarithmic and trigonometric functions?
2. What are common mistakes to avoid when differentiating composite functions?
3. How would the answer change if \( \sin(x^2) \) was replaced by \( \cos(x^2) \)?
4. Why is it important to simplify expressions like \(\ln(2x^2)\) before differentiating?
5. Can you use these differentiation techniques to solve related rates problems?

**Tip:** Always verify your results by simplifying before and after applying differentiation rules to avoid errors!

Find the derivative of ln(2x^2) + sin(x^2) using the chain rule. Learn how to differentiate logarithmic and trigonometric functions step by step. This problem covers key calculus concepts, suitable for advanced high school or undergraduate math students.

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