Answer :

To find [tex]\(\frac{dy}{dx}\)[/tex] for the function [tex]\(y = (3x^3 + 2x^2 + 1)(3x^2 + 4)\)[/tex], we will apply the product rule. The product rule states that if you have two functions [tex]\(u(x)\)[/tex] and [tex]\(v(x)\)[/tex], then the derivative of their product is:

[tex]\[ \frac{d}{dx} [u(x) v(x)] = u'(x) v(x) + u(x) v'(x) \][/tex]

Here, let's define:
[tex]\[ u(x) = 3x^3 + 2x^2 + 1 \][/tex]
[tex]\[ v(x) = 3x^2 + 4 \][/tex]

We will first find the derivatives [tex]\(u'(x)\)[/tex] and [tex]\(v'(x)\)[/tex]:

1. Differentiate [tex]\(u(x)\)[/tex]:
[tex]\[ u(x) = 3x^3 + 2x^2 + 1 \][/tex]
[tex]\[ u'(x) = \frac{d}{dx} (3x^3) + \frac{d}{dx} (2x^2) + \frac{d}{dx} (1) \][/tex]
[tex]\[ u'(x) = 9x^2 + 4x + 0 \][/tex]
[tex]\[ u'(x) = 9x^2 + 4x \][/tex]

2. Differentiate [tex]\(v(x)\)[/tex]:
[tex]\[ v(x) = 3x^2 + 4 \][/tex]
[tex]\[ v'(x) = \frac{d}{dx} (3x^2) + \frac{d}{dx} (4) \][/tex]
[tex]\[ v'(x) = 6x + 0 \][/tex]
[tex]\[ v'(x) = 6x \][/tex]

Now we can apply the product rule:
[tex]\[ \frac{dy}{dx} = u'(x) v(x) + u(x) v'(x) \][/tex]

Substitute [tex]\(u(x)\)[/tex], [tex]\(u'(x)\)[/tex], [tex]\(v(x)\)[/tex], and [tex]\(v'(x)\)[/tex]:

[tex]\[ \frac{dy}{dx} = (9x^2 + 4x)(3x^2 + 4) + (3x^3 + 2x^2 + 1)(6x) \][/tex]

Simplify each term separately:

1. First term:
[tex]\[ (9x^2 + 4x)(3x^2 + 4) \][/tex]
[tex]\[ = 9x^2(3x^2) + 9x^2(4) + 4x(3x^2) + 4x(4) \][/tex]
[tex]\[ = 27x^4 + 36x^2 + 12x^3 + 16x \][/tex]
[tex]\[ = 27x^4 + 12x^3 + 36x^2 + 16x \][/tex]

2. Second term:
[tex]\[ (3x^3 + 2x^2 + 1)(6x) \][/tex]
[tex]\[ = (3x^3)(6x) + (2x^2)(6x) + (1)(6x) \][/tex]
[tex]\[ = 18x^4 + 12x^3 + 6x \][/tex]

Now, combine the two terms:

[tex]\[ \frac{dy}{dx} = (27x^4 + 12x^3 + 36x^2 + 16x) + (18x^4 + 12x^3 + 6x) \][/tex]
[tex]\[ = 27x^4 + 18x^4 + 12x^3 + 12x^3 + 36x^2 + 16x + 6x \][/tex]
[tex]\[ = 45x^4 + 24x^3 + 36x^2 + 22x \][/tex]

Thus, the derivative [tex]\(\frac{dy}{dx}\)[/tex] is:

[tex]\[ \frac{dy}{dx} = 6x(3x^3 + 2x^2 + 1) + (3x^2 + 4)(9x^2 + 4x) \][/tex]