We are asked to differentiate the function $y = \frac{4x^2 + 3}{x^2 + 1}$ with respect to $x$.

1. Problem Description

We are asked to differentiate the function $y = \frac{4x^2 + 3}{x^2 + 1}$ with respect to $x$.

2. Solution Steps

We will use the quotient rule to find the derivative. The quotient rule states that if $y = \frac{u}{v}$, then $\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$.

In our case, $u = 4x^2 + 3$ and $v = x^2 + 1$.

First, we find the derivatives of $u$ and $v$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(4x^2 + 3) = 8x$

$\frac{dv}{dx} = \frac{d}{dx}(x^2 + 1) = 2x$

Now, we can apply the quotient rule:

$\frac{dy}{dx} = \frac{(x^2 + 1)(8x) - (4x^2 + 3)(2x)}{(x^2 + 1)^2}$

$\frac{dy}{dx} = \frac{8x^3 + 8x - 8x^3 - 6x}{(x^2 + 1)^2}$

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

3. Final Answer

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