Part 1: Using the Product Rule

The product rule for differentiation states that if we have two functions $f(z)$ and $g(z)$, then the derivative of their product is:

$\frac{d}{dz} \left( f(z) \cdot g(z) \right) = f'(z) \cdot g(z) + f(z) \cdot g'(z)$

In this case, we have:

• $f(z) = 6 - z^2$
• $g(z) = z^3 - z + 3$

Now, let's compute the derivatives of these functions:

• $f'(z) = \frac{d}{dz} (6 - z^2) = -2z$
• $g'(z) = \frac{d}{dz} (z^3 - z + 3) = 3z^2 - 1$

Using the product rule:

$h'(z) = f'(z) \cdot g(z) + f(z) \cdot g'(z)$

Substitute the values:

$h'(z) = (-2z) \cdot (z^3 - z + 3) + (6 - z^2) \cdot (3z^2 - 1)$

Part 2: Expanding the Product

We now expand the product $(6 - z^2)(z^3 - z + 3)$. Use the distributive property:

$(6 - z^2)(z^3 - z + 3) = 6(z^3 - z + 3) - z^2(z^3 - z + 3)$

First, expand $6(z^3 - z + 3)$:

$6z^3 - 6z + 18$

Now expand $-z^2(z^3 - z + 3)$:

$-z^2 \cdot z^3 = -z^5$ $-z^2 \cdot (-z) = z^3$ $-z^2 \cdot 3 = -3z^2$

Putting it all together:

$6z^3 - 6z + 18 - z^5 + z^3 - 3z^2$

Simplify the expression:

$h(z) = -z^5 + 7z^3 - 3z^2 - 6z + 18$

Part 3: Taking the Derivative of the Expanded Product

Now, differentiate the expanded product:

$\frac{d}{dz} \left( -z^5 + 7z^3 - 3z^2 - 6z + 18 \right)$

The derivative is:

$h'(z) = -5z^4 + 21z^2 - 6z - 6$

Final Answer:

$h'(z) = -5z^4 + 21z^2 - 6z - 6$

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Do you want further details or clarifications on any step?

Relative Questions:

1. How is the product rule applied in differentiation?
2. What is the general formula for the product rule?
3. How do you compute derivatives of polynomials?
4. How do you simplify expressions after expanding products?
5. What is the importance of simplifying before differentiating?

Tip:

When dealing with products, applying the product rule before expanding often simplifies complex differentiation problems.