Derivative of y=ln(cos(6x))+9x

1. Differentiate $9 x + \log{\left(\cos{\left(6 x \right)} \right)}$ term by term:

   1. Let $u = \cos{\left(6 x \right)}$.

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

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \cos{\left(6 x \right)}$:

      1. Let $u = 6 x$.

      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 $\frac{d}{d x} 6 x$:

         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: $6$

         The result of the chain rule is:

         $$- 6 \sin{\left(6 x \right)}$$

      The result of the chain rule is:

      $$- \frac{6 \sin{\left(6 x \right)}}{\cos{\left(6 x \right)}}$$

   4. 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: $9$

   The result is: $- \frac{6 \sin{\left(6 x \right)}}{\cos{\left(6 x \right)}} + 9$

2. Now simplify:

   $$9 - 6 \tan{\left(6 x \right)}$$

The answer is:

$$9 - 6 \tan{\left(6 x \right)}$$