Derivative of -sqrt(2x+1^-1)/(2x+1)

1. Apply the quotient rule, which is:

   $$\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$$

   $f{\left(x \right)} = - \sqrt{2} \sqrt{x + 0.5}$ and $g{\left(x \right)} = 2 x + 1$.

   To find $\frac{d}{d x} f{\left(x \right)}$:

   1. The derivative of a constant times a function is the constant times the derivative of the function.

      1. Let $u = x + 0.5$.

      2. Apply the power rule: $\sqrt{u}$ goes to $\frac{1}{2 \sqrt{u}}$

      3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(x + 0.5\right)$:

         1. Differentiate $x + 0.5$ term by term:

            1. The derivative of the constant $0.5$ is zero.

            2. Apply the power rule: $x$ goes to $1$

            The result is: $1$

         The result of the chain rule is:

         $$\frac{1}{2 \sqrt{x + 0.5}}$$

      So, the result is: $- \frac{\sqrt{2}}{2 \sqrt{x + 0.5}}$

   To find $\frac{d}{d x} g{\left(x \right)}$:

   1. Differentiate $2 x + 1$ term by term:

      1. The derivative of the constant $1$ is zero.

      2. The derivative of a constant times a function is the constant times the derivative of the function.

         1. Apply the power rule: $x$ goes to $1$

         So, the result is: $2$

      The result is: $2$

   Now plug in to the quotient rule:

   $\frac{2 \sqrt{2} \sqrt{x + 0.5} - \frac{\sqrt{2} \left(2 x + 1\right)}{2 \sqrt{x + 0.5}}}{\left(2 x + 1\right)^{2}}$

2. Now simplify:

   $$\frac{2.0 \sqrt{2}}{\sqrt{x + 0.5} \left(8.0 x + 4.0\right)}$$

The answer is:

$$\frac{2.0 \sqrt{2}}{\sqrt{x + 0.5} \left(8.0 x + 4.0\right)}$$