Let's work through the derivatives of each function given in the image.

The functions are:

1. $g(x) = x^3 - 8x$
2. $h(x) = 3x^2 + 2x$
3. $f(x) = (x^3 - 8x)(3x^2 + 2x)$

Step-by-step solutions:

1. $g'(x)$ (Derivative of $g(x) = x^3 - 8x$)

• Apply the power rule: $\frac{d}{dx} [x^n] = nx^{n-1}$
• Derivative of $x^3$ is $3x^2$
• Derivative of $-8x$ is $-8$

Thus, $g'(x) = 3x^2 - 8$

2. $h'(x)$ (Derivative of $h(x) = 3x^2 + 2x$)

• Derivative of $3x^2$ is $6x$
• Derivative of $2x$ is $2$

Thus, $h'(x) = 6x + 2$

3. $f'(x)$ (Derivative of $f(x) = (x^3 - 8x)(3x^2 + 2x)$)

• Use the product rule: $\frac{d}{dx} [u(x) \cdot v(x)] = u'(x)v(x) + u(x)v'(x)$
• Let $u(x) = x^3 - 8x$ and $v(x) = 3x^2 + 2x$

First, find the derivatives of $u(x)$ and $v(x)$:

• $u'(x) = 3x^2 - 8$
• $v'(x) = 6x + 2$

Now apply the product rule: $f'(x) = u'(x) v(x) + u(x) v'(x)$ $f'(x) = (3x^2 - 8)(3x^2 + 2x) + (x^3 - 8x)(6x + 2)$ Now expand both terms:

1. $(3x^2 - 8)(3x^2 + 2x) = 9x^4 + 6x^3 - 24x^2 - 16x$
2. $(x^3 - 8x)(6x + 2) = 6x^4 + 2x^3 - 48x^2 - 16x$

Combine like terms: $f'(x) = 9x^4 + 6x^3 - 24x^2 - 16x + 6x^4 + 2x^3 - 48x^2 - 16x$ $f'(x) = 15x^4 + 8x^3 - 72x^2 - 32x$

Thus, $f'(x) = 15x^4 + 8x^3 - 72x^2 - 32x$.

Would you like more details on any of these steps?

Here are 5 related questions to help expand your understanding:

1. What is the power rule for differentiation?
2. How does the product rule work, and when should you use it?
3. Can you apply the chain rule to composite functions during differentiation?
4. How would you find the second derivative of the function $g(x)$?
5. How do you factor a polynomial after differentiating?

Tip: Always simplify your expression after applying the product or chain rule to avoid complex terms.