To differentiate \( f(x) = \sin^5(x) \), we will apply the **chain rule**. The function can be rewritten as \( f(x) = (\sin(x))^5 \), and we'll treat it as a composition of two functions.

Let \( u = \sin(x) \), so that \( f(x) = u^5 \).

### Step 1: Differentiate \( f(x) = u^5 \) with respect to \( u \):
\[
\frac{d}{du}(u^5) = 5u^4
\]

### Step 2: Differentiate \( u = \sin(x) \) with respect to \( x \):
\[
\frac{du}{dx} = \cos(x)
\]

### Step 3: Apply the chain rule:
\[
f'(x) = \frac{d}{dx} [(\sin(x))^5] = 5(\sin(x))^4 \cdot \cos(x)
\]

Thus, the derivative is:
\[
f'(x) = 5\sin^4(x) \cos(x)
\]

### Step 4: Find \( f'(4) \)
Now, substitute \( x = 4 \) into the derivative:
\[
f'(4) = 5\sin^4(4) \cos(4)
\]

Using approximate values:
- \( \sin(4) \approx -0.7568 \)
- \( \cos(4) \approx -0.6536 \)

Now, compute:
\[
f'(4) = 5(-0.7568)^4 (-0.6536)
\]

Let's compute this expression to get the final value.The value of \( f'(4) \) is approximately \( -1.072 \).

### Summary:
- The derivative is \( f'(x) = 5\sin^4(x) \cos(x) \).
- The value of the derivative at \( x = 4 \) is approximately \( f'(4) \approx -1.072 \).

Would you like further details or have any questions? Here are 5 related questions to explore further:

1. How do you apply the chain rule in general cases?
2. What are the derivatives of other trigonometric functions?
3. How would you differentiate \( f(x) = \cos^3(x) \)?
4. Can you apply the product rule in combination with the chain rule?
5. What is the geometric meaning of the derivative of \( \sin(x) \) raised to a power?

**Tip:** When differentiating composite functions, always identify the inner and outer functions clearly to apply the chain rule correctly.

f(x) = sin^5(x), find f'(x) and find f'(4)

Explore the step-by-step differentiation of f(x) = sin^5(x) using the chain rule and find the derivative at x = 4. This problem involves key calculus concepts such as trigonometric functions and differentiation. Ideal for students in Grades 11-12.

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