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We are asked to find the derivative of the function:

\[
y = \sin\left( \sqrt[3]{x} \right) + \sqrt[3]{\sin(6x)}
\]

Let’s break this into two parts for differentiation using the chain rule.

### Step 1: Differentiate \( y = \sin\left( \sqrt[3]{x} \right) \)

Let \( u = \sqrt[3]{x} = x^{1/3} \), so the function becomes:

\[
y_1 = \sin(u)
\]

The derivative of \( y_1 \) with respect to \( x \) is:

\[
\frac{dy_1}{dx} = \cos(u) \cdot \frac{du}{dx}
\]

Now, differentiate \( u = x^{1/3} \):

\[
\frac{du}{dx} = \frac{1}{3}x^{-2/3}
\]

So, the derivative of the first part is:

\[
\frac{dy_1}{dx} = \cos\left( \sqrt[3]{x} \right) \cdot \frac{1}{3}x^{-2/3}
\]

### Step 2: Differentiate \( y = \sqrt[3]{\sin(6x)} \)

Let \( v = \sin(6x) \), so the function becomes:

\[
y_2 = v^{1/3}
\]

The derivative of \( y_2 \) with respect to \( x \) is:

\[
\frac{dy_2}{dx} = \frac{1}{3}v^{-2/3} \cdot \frac{dv}{dx}
\]

Now, differentiate \( v = \sin(6x) \):

\[
\frac{dv}{dx} = 6\cos(6x)
\]

So, the derivative of the second part is:

\[
\frac{dy_2}{dx} = \frac{1}{3} \left( \sin(6x) \right)^{-2/3} \cdot 6\cos(6x)
\]

### Final Derivative

Now we combine both derivatives:

\[
\frac{dy}{dx} = \cos\left( \sqrt[3]{x} \right) \cdot \frac{1}{3}x^{-2/3} + \frac{6}{3} \cdot \left( \sin(6x) \right)^{-2/3} \cdot \cos(6x)
\]

Simplifying the second term:

\[
\frac{dy}{dx} = \cos\left( \sqrt[3]{x} \right) \cdot \frac{1}{3}x^{-2/3} + 2 \cdot \left( \sin(6x) \right)^{-2/3} \cdot \cos(6x)
\]

This is the derivative of the given function.

Would you like further clarifications on any step or details?

### Related Questions:
1. How can we apply the chain rule to more complex trigonometric functions?
2. What are the properties of fractional exponents in derivatives?
3. How does the chain rule work with composite functions?
4. What are common pitfalls when differentiating functions involving cube roots?
5. How can we simplify derivatives involving trigonometric functions with higher-order roots?

**Tip:** When differentiating composite functions, always identify the inner function clearly for easier application of the chain rule.

Find the derivative of the trigonometric function y = sin(cube root of x) + cube root of sin(6x).

Learn how to find the derivative of the function y = sin(cube root of x) + cube root of sin(6x). This solution applies the chain rule and fractional exponents to solve trigonometric functions involving roots. Suitable for high school and college-level calculus students.

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