Let's work through the derivatives $\frac{dy}{dx}$ for both parts (a) and (b).

(a) $y = \sin(x^3) \left(3x^5 - x + \tan(x)\right)^4$

We need to use the product rule and the chain rule. The product rule states: $\frac{d}{dx}(u \cdot v) = u'v + uv'$ Where $u = \sin(x^3)$ and $v = \left(3x^5 - x + \tan(x)\right)^4$.

Step 1: Differentiate $u = \sin(x^3)$

We apply the chain rule here: $u' = \cos(x^3) \cdot \frac{d}{dx}(x^3) = \cos(x^3) \cdot 3x^2$ So, $u' = 3x^2 \cos(x^3)$.

Step 2: Differentiate $v = \left(3x^5 - x + \tan(x)\right)^4$

Again, we apply the chain rule: $v' = 4 \left(3x^5 - x + \tan(x)\right)^3 \cdot \frac{d}{dx}(3x^5 - x + \tan(x))$ Now, we compute the derivative of the inner term: $\frac{d}{dx}(3x^5 - x + \tan(x)) = 15x^4 - 1 + \sec^2(x)$ Thus, $v' = 4 \left(3x^5 - x + \tan(x)\right)^3 (15x^4 - 1 + \sec^2(x))$.

Step 3: Apply the product rule

Now, apply the product rule: $\frac{dy}{dx} = u'v + uv'$ Substitute in the values for $u'$, $u$, $v'$, and $v$: $\frac{dy}{dx} = \left(3x^2 \cos(x^3)\right) \cdot \left(3x^5 - x + \tan(x)\right)^4 + \sin(x^3) \cdot 4 \left(3x^5 - x + \tan(x)\right)^3 (15x^4 - 1 + \sec^2(x))$ This is the final derivative for part (a).

(b) $y = \left( \frac{4\sqrt{x} - x^3 + 1}{\sec(3x) + 5} \right)^3$

For this part, we will again apply the chain rule since we have a function raised to the third power, along with the quotient rule for the inner function.

Step 1: Differentiate the outer function

Using the chain rule: $\frac{d}{dx} \left(f(x)^3\right) = 3 \cdot f(x)^2 \cdot f'(x)$ Here, $f(x) = \frac{4\sqrt{x} - x^3 + 1}{\sec(3x) + 5}$, so the derivative of the outer part is: $\frac{dy}{dx} = 3 \left( \frac{4\sqrt{x} - x^3 + 1}{\sec(3x) + 5} \right)^2 \cdot \frac{d}{dx} \left( \frac{4\sqrt{x} - x^3 + 1}{\sec(3x) + 5} \right)$

Step 2: Differentiate the inner function (quotient rule)

Now we use the quotient rule: $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$ Where $u = 4\sqrt{x} - x^3 + 1$ and $v = \sec(3x) + 5$.

• $u' = \frac{d}{dx} (4\sqrt{x} - x^3 + 1) = 2x^{-1/2} - 3x^2$
• $v' = \frac{d}{dx} (\sec(3x) + 5) = \sec(3x) \cdot \tan(3x) \cdot 3$

Now apply the quotient rule: $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{(2x^{-1/2} - 3x^2)(\sec(3x) + 5) - (4\sqrt{x} - x^3 + 1)(\sec(3x) \cdot \tan(3x) \cdot 3)}{(\sec(3x) + 5)^2}$

Step 3: Combine the results

Substitute the quotient rule result back into the chain rule: $\frac{dy}{dx} = 3 \left( \frac{4\sqrt{x} - x^3 + 1}{\sec(3x) + 5} \right)^2 \cdot \frac{(2x^{-1/2} - 3x^2)(\sec(3x) + 5) - (4\sqrt{x} - x^3 + 1)(\sec(3x) \cdot \tan(3x) \cdot 3)}{(\sec(3x) + 5)^2}$ This is the final derivative for part (b).

Would you like more details on any specific step, or have any further questions? Here are some related questions to further explore these concepts:

1. How do you apply the product rule to functions with more than two terms?
2. What are some tips for identifying when to use the chain rule in composite functions?
3. Could we simplify the expression $(3x^5 - x + \tan(x))^4$ before differentiating to make the process easier?
4. How would the quotient rule change if the denominator was a constant?
5. What is the general process for differentiating functions involving trigonometric expressions like $\tan(x)$ and $\sec(3x)$?

Tip: Always try to simplify complex expressions before differentiating, especially when applying the product or quotient rules, to avoid unnecessary complications.