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.