Derivative of 1/sqrt(3x+1)

1. Apply the quotient rule, which is:

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

   $f{\left(x \right)} = 1$ and $g{\left(x \right)} = \sqrt{3 x + 1}$.

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

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

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

   1. Let $u = 3 x + 1$.

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

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(3 x + 1\right)$:

      1. Differentiate $3 x + 1$ 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: $3$

         The result is: $3$

      The result of the chain rule is:

      $$\frac{3}{2 \sqrt{3 x + 1}}$$

   Now plug in to the quotient rule:

   $- \frac{3}{2 \left(3 x + 1\right)^{\frac{3}{2}}}$

The answer is:

$$- \frac{3}{2 \left(3 x + 1\right)^{\frac{3}{2}}}$$