Step 1:

We notice that the given function is a quotient of two other functions. To find the derivative of this, we can use the quotient rule. First, let’s label the two functions as:

[latex]\large{f(x) = \sqrt{x}}[/latex] and [latex]\large{g(x) = 2x+1}[/latex]

To use the quotient rule, we also need the derivative of these functions. We can easily find the derivatives as:

[latex]\large{f'(x) = \frac{1}{2\sqrt{x}}}[/latex] and [latex]\large{g'(x) = 2}[/latex]

Step 2:

The quotient rule formula states that:

[latex]\large{\frac{d}{dx}\frac{f(x)}{g(x)} = \frac{g(x)f'(x) – f(x)g'(x)}{g(x)^2}}[/latex]

Thus, substituting into the quotient rule formula, we have:

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

Step 3:

Simplifying, we have:

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

Now, we can simplify the numerator as shown:

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

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

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

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

[latex]\Large{\frac{dy}{dx}= \frac{-2x+1}{2\sqrt{x}(2x+1)^2}}[/latex]

Solution Complete!