The problem is to find the derivative of the function $y = e^{\frac{1}{2}x} \sin^2(x)$ with respect to $x$.

1. Problem Description

The problem is to find the derivative of the function $y = e^{\frac{1}{2}x} \sin^2(x)$ with respect to $x$.

2. Solution Steps

We need to find $\frac{dy}{dx}$. We will use the product rule and the chain rule. The product rule states that if $y = u(x)v(x)$, then $\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)$. In this case, let $u(x) = e^{\frac{1}{2}x}$ and $v(x) = \sin^2(x)$.

First, let's find the derivative of $u(x)$ with respect to $x$:

$u(x) = e^{\frac{1}{2}x}$

$u'(x) = \frac{d}{dx}(e^{\frac{1}{2}x})$

Using the chain rule:

$u'(x) = e^{\frac{1}{2}x} \cdot \frac{d}{dx}(\frac{1}{2}x)$

$u'(x) = e^{\frac{1}{2}x} \cdot \frac{1}{2}$

$u'(x) = \frac{1}{2}e^{\frac{1}{2}x}$

Next, let's find the derivative of $v(x)$ with respect to $x$:

$v(x) = \sin^2(x)$

$v'(x) = \frac{d}{dx}(\sin^2(x))$

Using the chain rule:

$v'(x) = 2\sin(x) \cdot \frac{d}{dx}(\sin(x))$

$v'(x) = 2\sin(x) \cdot \cos(x)$

$v'(x) = 2\sin(x)\cos(x)$

We can also write this as $v'(x) = \sin(2x)$ using the double angle formula.

Now, we apply the product rule:

$\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)$

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

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

We can factor out $e^{\frac{1}{2}x}$:

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

We can also use the double angle formula $\sin(2x) = 2\sin(x)\cos(x)$:

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

3. Final Answer

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