We are asked to differentiate the given expressions with respect to $x$. The expressions are: a) $(3x^2)(x^2)(x^2 + 2x + 1)$ b) $\frac{x^2}{3x+1}$

1. Problem Description

We are asked to differentiate the given expressions with respect to $x$. The expressions are:

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

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

2. Solution Steps

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

First, simplify the expression:

$y = 3x^4(x^2 + 2x + 1) = 3x^6 + 6x^5 + 3x^4$.

Now, differentiate with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(3x^6 + 6x^5 + 3x^4)$.

Using the power rule, $\frac{d}{dx}(x^n) = nx^{n-1}$, we get:

$\frac{dy}{dx} = 3(6x^5) + 6(5x^4) + 3(4x^3) = 18x^5 + 30x^4 + 12x^3$.

b) Let $y = \frac{x^2}{3x+1}$.

We need to use the quotient rule:

If $y = \frac{u}{v}$, then $\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$.

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

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

Applying the quotient rule:

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

3. Final Answer

a) $\frac{dy}{dx} = 18x^5 + 30x^4 + 12x^3$

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