The problem consists of two parts. The first part asks to differentiate $y = 3x^2 + 2x + 5$ from first principles. The second part asks to differentiate the following functions with respect to $x$: (a) $(3x^2)(x^2 + 2x + 1)$ (b) $\frac{x^2}{3x + 1}$

1. Problem Description

The problem consists of two parts.

The first part asks to differentiate $y = 3x^2 + 2x + 5$ from first principles.

The second part asks to differentiate the following functions with respect to $x$:

(a) $(3x^2)(x^2 + 2x + 1)$

(b) $\frac{x^2}{3x + 1}$

2. Solution Steps

Part 1: Differentiate $y = 3x^2 + 2x + 5$ from first principles.

The first principle of differentiation is defined as:

$\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Here, $f(x) = 3x^2 + 2x + 5$.

Therefore, $f(x+h) = 3(x+h)^2 + 2(x+h) + 5 = 3(x^2 + 2xh + h^2) + 2x + 2h + 5 = 3x^2 + 6xh + 3h^2 + 2x + 2h + 5$.

$\frac{f(x+h) - f(x)}{h} = \frac{(3x^2 + 6xh + 3h^2 + 2x + 2h + 5) - (3x^2 + 2x + 5)}{h} = \frac{6xh + 3h^2 + 2h}{h} = 6x + 3h + 2$.

Now, we take the limit as $h \to 0$:

$\frac{dy}{dx} = \lim_{h \to 0} (6x + 3h + 2) = 6x + 3(0) + 2 = 6x + 2$.

Part 2(a): Differentiate $(3x^2)(x^2 + 2x + 1)$ with respect to $x$.

Let $y = (3x^2)(x^2 + 2x + 1) = 3x^4 + 6x^3 + 3x^2$.

$\frac{dy}{dx} = 12x^3 + 18x^2 + 6x$.

Part 2(b): Differentiate $\frac{x^2}{3x + 1}$ with respect to $x$.

Let $y = \frac{x^2}{3x + 1}$. We use the quotient rule:

$\frac{d}{dx}(\frac{u}{v}) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$

Here, $u = x^2$ and $v = 3x + 1$.

$\frac{du}{dx} = 2x$ and $\frac{dv}{dx} = 3$.

$\frac{dy}{dx} = \frac{(3x + 1)(2x) - (x^2)(3)}{(3x + 1)^2} = \frac{6x^2 + 2x - 3x^2}{(3x + 1)^2} = \frac{3x^2 + 2x}{(3x + 1)^2}$.

3. Final Answer

1. The derivative of $y = 3x^2 + 2x + 5$ from first principles is $6x + 2$.

2. (a) The derivative of $(3x^2)(x^2 + 2x + 1)$ is $12x^3 + 18x^2 + 6x$.

(b) The derivative of $\frac{x^2}{3x + 1}$ is $\frac{3x^2 + 2x}{(3x + 1)^2}$.