Derivative of y=ln^5(sqrt(3x+5))

1. Let $u = \log{\left(\sqrt{3 x + 5} \right)}$.

2. Apply the power rule: $u^{5}$ goes to $5 u^{4}$

3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \log{\left(\sqrt{3 x + 5} \right)}$:

   1. Let $u = \sqrt{3 x + 5}$.

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

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \sqrt{3 x + 5}$:

      1. Let $u = 3 x + 5$.

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

      3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(3 x + 5\right)$:

         1. Differentiate $3 x + 5$ term by term:

            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: $3$

            2. The derivative of the constant $5$ is zero.

            The result is: $3$

         The result of the chain rule is:

         $$\frac{3}{2 \sqrt{3 x + 5}}$$

      The result of the chain rule is:

      $$\frac{3}{2 \left(3 x + 5\right)}$$

   The result of the chain rule is:

   $$\frac{15 \log{\left(\sqrt{3 x + 5} \right)}^{4}}{2 \left(3 x + 5\right)}$$

4. Now simplify:

   $$\frac{15 \log{\left(3 x + 5 \right)}^{4}}{32 \left(3 x + 5\right)}$$

The answer is: