Derivative of cos6x-sin3x

1. Differentiate $- \sin{\left(3 x \right)} + \cos{\left(6 x \right)}$ term by term:

   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:

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

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

      1. Let $u = 3 x$.

      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 $\frac{d}{d x} 3 x$:

         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:

         $$3 \cos{\left(3 x \right)}$$

      So, the result is: $- 3 \cos{\left(3 x \right)}$

   The result is: $- 6 \sin{\left(6 x \right)} - 3 \cos{\left(3 x \right)}$

The answer is:

$$- 6 \sin{\left(6 x \right)} - 3 \cos{\left(3 x \right)}$$