Step 1:

We notice that the given function is a composite function, or a function within another function. To find the derivative of composite functions, we will use the chain rule. First, we can rewrite the function as: 

[latex]\large{h(x) = (e^{2x})^{\frac{1}{2}}}[/latex]

We can see that the inner function is [latex]e^{2x}[/latex] and the outer function is the square root function or, the power of [latex]\frac{1}{2}[/latex].

The chain rule states:

[latex]\Large{\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)}[/latex]

Step 2:

Applying the chain rule, we have:

[latex]\Large{\frac{d}{dx}(e^{2x})^{\frac{1}{2}} = \frac{1}{2}(e^{2x})^{-\frac{1}{2}} \cdot \frac{d}{dx}(e^{2x})}[/latex]

To find [latex]\frac{d}{dx}e^{2x}[/latex], we can use the chain rule, as follows:

[latex]\large{\frac{d}{dx}e^{2x} = e^{2x} \cdot \frac{d}{dx}2x = 2e^{2x}}[/latex]

Combining everything, we have:

[latex]\large{\frac{d}{dx}(e^{2x})^{\frac{1}{2}} = \frac{2e^{2x}}{2\sqrt{e^{2x}}} = \frac{e^{2x}}{\sqrt{e^{2x}}}}[/latex]

Step 3:

Simplifying, we have:

[latex]\Large{ = e^{2x}(e^{2x})^{-\frac{1}{2}}}[/latex]

[latex]\Large{ = e^{2x}(e^{-x}) = e^x}[/latex]

Thus, this is the final answer for the derivative. So:

[latex]\Large{h'(x) = e^x}[/latex]

Solution Complete!