Derivative of log2(sqrt3x+4)

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

   1. Let $u = \sqrt{3 x} + 4$.

   2. The derivative of $\log{\left(u \right)}$ is $\frac{1}{u}$.

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(\sqrt{3 x} + 4\right)$:

      1. Differentiate $\sqrt{3 x} + 4$ term by term:

         1. Let $u = 3 x$.

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

         3. Then, apply the chain rule. Multiply by $\frac{d}{d x} 3 x$:

            1. 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 of the chain rule is:

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

         4. The derivative of the constant $4$ is zero.

         The result is: $\frac{\sqrt{3}}{2 \sqrt{x}}$

      The result of the chain rule is:

      $$\frac{\sqrt{3}}{2 \sqrt{x} \left(\sqrt{3 x} + 4\right)}$$

   So, the result is: $\frac{\sqrt{3}}{2 \sqrt{x} \left(\sqrt{3 x} + 4\right) \log{\left(2 \right)}}$

2. Now simplify:

   $$\frac{\sqrt{3}}{2 \left(4 \sqrt{x} + \sqrt{3} x\right) \log{\left(2 \right)}}$$

The answer is: