The problem asks to find the derivative of the function $y = \sqrt{xe^{2x^2+3}}$.

1. Problem Description

The problem asks to find the derivative of the function $y = \sqrt{xe^{2x^2+3}}$.

2. Solution Steps

First, rewrite the function using exponents:

$y = (xe^{2x^2+3})^{\frac{1}{2}}$.

Now we will use the chain rule to find the derivative.

The chain rule states that if $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$.

Also, we will use the product rule: If $y = u(x)v(x)$, then $\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)$.

And, the derivative of $e^{f(x)}$ is $e^{f(x)} f'(x)$.

Applying the chain rule:

$\frac{dy}{dx} = \frac{1}{2} (xe^{2x^2+3})^{-\frac{1}{2}} \cdot \frac{d}{dx} (xe^{2x^2+3})$

Now we need to find the derivative of $xe^{2x^2+3}$ using the product rule.

Let $u(x) = x$ and $v(x) = e^{2x^2+3}$.

Then $u'(x) = 1$ and $v'(x) = e^{2x^2+3} \cdot \frac{d}{dx} (2x^2+3) = e^{2x^2+3} \cdot (4x)$.

So, $\frac{d}{dx} (xe^{2x^2+3}) = (1) e^{2x^2+3} + x(4x e^{2x^2+3}) = e^{2x^2+3} + 4x^2 e^{2x^2+3} = e^{2x^2+3}(1+4x^2)$.

Substituting this back into the original derivative:

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

$\frac{dy}{dx} = \frac{e^{2x^2+3} (1+4x^2)}{2 \sqrt{xe^{2x^2+3}}}$

3. Final Answer

$\frac{dy}{dx} = \frac{e^{2x^2+3}(1+4x^2)}{2\sqrt{xe^{2x^2+3}}}$