Differentiate the following with respect to $x$: $\sqrt{e^{\sqrt{x}}}, x > 0$

Let $y = \sqrt{e^{\sqrt{x}}}$.
Then,$y^2 = e^{\sqrt{x}}$.
By differentiating this relationship with respect to $x$,we obtain:
$\frac{d}{dx}(y^2) = \frac{d}{dx}(e^{\sqrt{x}})$.
Using the chain rule,we get:
$2y \frac{dy}{dx} = e^{\sqrt{x}} \cdot \frac{d}{dx}(\sqrt{x})$.
Since $\frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}$,we have:
$2y \frac{dy}{dx} = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}}$.
$\frac{dy}{dx} = \frac{e^{\sqrt{x}}}{4y\sqrt{x}}$.
Substituting $y = \sqrt{e^{\sqrt{x}}}$ back into the equation:
$\frac{dy}{dx} = \frac{e^{\sqrt{x}}}{4\sqrt{e^{\sqrt{x}}}\sqrt{x}}$.
$\frac{dy}{dx} = \frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}}$.