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]
