Derivative of y=ln(cos7x+8)

1. Let $u = \cos{\left(7 x \right)} + 8$.

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

3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(\cos{\left(7 x \right)} + 8\right)$:

   1. Differentiate $\cos{\left(7 x \right)} + 8$ term by term:

      1. Let $u = 7 x$.

      2. The derivative of cosine is negative sine:

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

      3. Then, apply the chain rule. Multiply by $\frac{d}{d x} 7 x$:

         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: $7$

         The result of the chain rule is:

         $$- 7 \sin{\left(7 x \right)}$$

      4. The derivative of the constant $8$ is zero.

      The result is: $- 7 \sin{\left(7 x \right)}$

   The result of the chain rule is:

   $$- \frac{7 \sin{\left(7 x \right)}}{\cos{\left(7 x \right)} + 8}$$

4. Now simplify:

   $$- \frac{7 \sin{\left(7 x \right)}}{\cos{\left(7 x \right)} + 8}$$

The answer is:

$$- \frac{7 \sin{\left(7 x \right)}}{\cos{\left(7 x \right)} + 8}$$