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To solve this problem, we need to differentiate the function:

\[
y = \sin^7(x)
\]

with respect to \( x \).

### Step 1: Rewrite the Function
We can rewrite \( y = \sin^7(x) \) as:
\[
y = (\sin(x))^7
\]

### Step 2: Apply the Chain Rule
To differentiate \( y = (\sin(x))^7 \), we'll use the chain rule. The chain rule states that if \( y = u^n \) and \( u = f(x) \), then:

\[
\frac{dy}{dx} = n \cdot u^{n-1} \cdot \frac{du}{dx}
\]

In this case:
- \( u = \sin(x) \)
- \( n = 7 \)

### Step 3: Differentiate
1. Differentiate the outer function with respect to \( u \):
   \[
   \frac{d}{dx}(\sin(x))^7 = 7(\sin(x))^{6} \cdot \frac{d}{dx}(\sin(x))
   \]
   
2. Now, differentiate \( \sin(x) \) with respect to \( x \):
   \[
   \frac{d}{dx}(\sin(x)) = \cos(x)
   \]

### Step 4: Combine the Results
Substitute \( \frac{d}{dx}(\sin(x)) = \cos(x) \) back into the equation:
\[
y' = 7(\sin(x))^6 \cdot \cos(x)
\]

### Final Answer
The derivative of \( y = \sin^7(x) \) is:
\[
y' = 7 \sin^6(x) \cos(x)
\]

### Conclusion
The correct answer is:
\[
y' = 7 \sin^6(x) \cos(x)
\]

This corresponds to the second option in the multiple-choice answers.

Would you like further details or have any questions?

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Here are 5 related questions to expand your understanding:

1. How would you apply the chain rule if the exponent were different, such as \( \sin^5(x) \)?
2. How would the derivative change if the function were \( \cos^7(x) \) instead of \( \sin^7(x) \)?
3. What is the second derivative of \( y = \sin^7(x) \)?
4. How would you approach the problem if the function was \( y = (\sin(x) + \cos(x))^7 \)?
5. What if the function was \( y = \sin^7(2x) \)? How would the derivative change?

**Tip:** Always break down complex functions into inner and outer functions for easier application of the chain rule.

What is the derivative of \( y = \sin^7(x) \)?

Learn how to find the derivative of y = sin^7(x) with a step-by-step solution. This calculus problem covers key concepts like the chain rule and differentiating trigonometric functions. Ideal for students in Grades 11-12.

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