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Find the derivative using chain rule and product rule

The given function from the image is:

\[
h(t) = t^3 \sqrt{t^2 - t}
\]

To find the derivative \( h'(t) \), we need to apply both the **product rule** and the **chain rule**.

### Steps:

1. **Product Rule**: The product rule states that if you have a function that is the product of two functions, say \( u(t) \cdot v(t) \), then the derivative is:
   \[
   (uv)' = u'v + uv'
   \]
   Here, \( u(t) = t^3 \) and \( v(t) = \sqrt{t^2 - t} \).

2. **Chain Rule for \( v(t) \)**: 
   The function \( v(t) = \sqrt{t^2 - t} \) can be rewritten as \( (t^2 - t)^{1/2} \). The chain rule for differentiating \( (f(g(t))) \) is:
   \[
   \frac{d}{dt} [f(g(t))] = f'(g(t)) \cdot g'(t)
   \]

Let's now differentiate step by step.

### Step 1: Derivative of \( u(t) = t^3 \)
\[
u'(t) = 3t^2
\]

### Step 2: Derivative of \( v(t) = (t^2 - t)^{1/2} \)
Using the chain rule:
\[
v'(t) = \frac{1}{2} (t^2 - t)^{-1/2} \cdot (2t - 1)
\]
Simplifying:
\[
v'(t) = \frac{2t - 1}{2 \sqrt{t^2 - t}}
\]

### Step 3: Apply the Product Rule
Using the product rule:
\[
h'(t) = u'(t) \cdot v(t) + u(t) \cdot v'(t)
\]
Substituting the expressions for \( u'(t) \), \( u(t) \), \( v(t) \), and \( v'(t) \):
\[
h'(t) = (3t^2) \cdot \sqrt{t^2 - t} + (t^3) \cdot \frac{2t - 1}{2 \sqrt{t^2 - t}}
\]

### Final Expression for \( h'(t) \)
Thus, the derivative of the function is:
\[
h'(t) = 3t^2 \sqrt{t^2 - t} + \frac{t^3 (2t - 1)}{2 \sqrt{t^2 - t}}
\]

This is the derivative of the function \( h(t) = t^3 \sqrt{t^2 - t} \).

Would you like to further simplify this expression or have any specific questions? 

Here are five related questions for practice:
1. How would you apply the chain rule to differentiate a composite function like \( \sqrt{g(t)} \)?
2. What is the general form of the product rule for three functions multiplied together?
3. Can you find the second derivative \( h''(t) \) using the quotient rule?
4. What happens to \( h'(t) \) when \( t \to 0 \)?
5. How can the derivative of \( h(t) \) help analyze the behavior of the function in terms of maxima or minima?

**Tip:** When applying both product and chain rules, ensure each component is carefully differentiated before substituting into the product rule.

Learn how to find the derivative using both the chain rule and product rule with a detailed step-by-step solution. This problem covers differentiation techniques, including the product rule and chain rule formulas. Suitable for advanced high school students in Grades 11-12.

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