Derivative of (sin8x*ln(2x+5))

1. Apply the product rule:

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

   $f{\left(x \right)} = \sin{\left(8 x \right)}$; to find $\frac{d}{d x} f{\left(x \right)}$:

   1. Let $u = 8 x$.

   2. The derivative of sine is cosine:

      $$\frac{d}{d u} \sin{\left(u \right)} = \cos{\left(u \right)}$$

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} 8 x$:

      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: $8$

      The result of the chain rule is:

      $$8 \cos{\left(8 x \right)}$$

   $g{\left(x \right)} = \log{\left(2 x + 5 \right)}$; to find $\frac{d}{d x} g{\left(x \right)}$:

   1. Let $u = 2 x + 5$.

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

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(2 x + 5\right)$:

      1. Differentiate $2 x + 5$ 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: $2$

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

         The result is: $2$

      The result of the chain rule is:

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

   The result is: $8 \log{\left(2 x + 5 \right)} \cos{\left(8 x \right)} + \frac{2 \sin{\left(8 x \right)}}{2 x + 5}$

2. Now simplify:

   $$\frac{2 \left(4 \left(2 x + 5\right) \log{\left(2 x + 5 \right)} \cos{\left(8 x \right)} + \sin{\left(8 x \right)}\right)}{2 x + 5}$$

The answer is: