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This is a Chain Rule worksheet, where you need to find the derivative of each function. I can guide you through the solution of these problems step by step. Let's go through the first few problems:

### 1. \( f(x) = (2x^2 - 5x)^3 \)
We will use the chain rule for this function. Let \( u = 2x^2 - 5x \), so \( f(x) = u^3 \).
- First, differentiate \( u^3 \) with respect to \( u \): \( \frac{d}{du}(u^3) = 3u^2 \).
- Then, differentiate \( u = 2x^2 - 5x \) with respect to \( x \): \( \frac{d}{dx}(2x^2 - 5x) = 4x - 5 \).

Now, apply the chain rule:
\[
f'(x) = 3(2x^2 - 5x)^2 \cdot (4x - 5)
\]

### 2. \( f(x) = \sqrt{5x^3 - 2x} \)
We rewrite the square root as a power of \( \frac{1}{2} \): \( f(x) = (5x^3 - 2x)^{1/2} \).
- Let \( u = 5x^3 - 2x \), then \( f(x) = u^{1/2} \).
- The derivative of \( u^{1/2} \) with respect to \( u \) is \( \frac{1}{2}u^{-1/2} \).
- Differentiate \( u = 5x^3 - 2x \) with respect to \( x \): \( \frac{d}{dx}(5x^3 - 2x) = 15x^2 - 2 \).

Now apply the chain rule:
\[
f'(x) = \frac{1}{2}(5x^3 - 2x)^{-1/2} \cdot (15x^2 - 2)
\]

### 3. \( y = 3\sin(x - 3) \)
For this one, use the chain rule combined with the derivative of \( \sin(x) \).
- Differentiate \( 3\sin(x - 3) \) with respect to \( (x - 3) \): The derivative of \( \sin(x) \) is \( \cos(x) \), so we get \( 3\cos(x - 3) \).

Thus, \( y' = 3\cos(x - 3) \).

### 4. \( y = -2\cos(x^2 + 2) \)
- The derivative of \( \cos(x) \) is \( -\sin(x) \), so we differentiate \( -2\cos(x^2 + 2) \) with respect to \( (x^2 + 2) \).
- First, differentiate \( \cos(x^2 + 2) \), which gives \( -\sin(x^2 + 2) \), then apply the chain rule:
- Derivative of \( x^2 + 2 \) with respect to \( x \) is \( 2x \).

Now apply the chain rule:
\[
y' = -2 \cdot (-\sin(x^2 + 2)) \cdot (2x) = 4x \sin(x^2 + 2)
\]

### 5. \( g(x) = \sin^2(3x^2) \)
This is the chain rule applied twice.
- Let \( u = \sin(3x^2) \), so \( g(x) = u^2 \).
- First, differentiate \( u^2 \): \( \frac{d}{du}(u^2) = 2u \).
- Now differentiate \( u = \sin(3x^2) \): The derivative of \( \sin(x) \) is \( \cos(x) \), so \( \frac{d}{dx}\sin(3x^2) = \cos(3x^2) \cdot 6x \) (by applying the chain rule again to \( 3x^2 \)).

Now combine everything:
\[
g'(x) = 2\sin(3x^2) \cdot (6x \cos(3x^2)) = 12x \sin(3x^2) \cos(3x^2)
\]

These are the solutions for the first five problems. Would you like detailed explanations for more problems or do you have any specific questions?

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Here are five related questions for further understanding:

1. How is the chain rule different from the product rule?
2. What would happen if a function includes multiple inner functions? How would the chain rule apply?
3. Can the chain rule be applied to implicit differentiation problems? How?
4. How do you differentiate trigonometric functions that have nested functions inside?
5. In what cases might you need to apply the chain rule more than once?

**Tip**: Always identify the "inner" and "outer" functions when applying the chain rule. This makes the process clearer and more structured.

Find the derivative of each function using the chain rule.

Explore how to find the derivative of composite functions using the chain rule. This worksheet includes problems with trigonometric, exponential, and power functions. Perfect for college or university students studying calculus.

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