Derivative of sin(2cos3x)

1. Let $u = 2 \cos{\left(3 x \right)}$.

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 $\frac{d}{d x} 2 \cos{\left(3 x \right)}$:

   1. 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 cosine is negative sine:

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

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

         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:

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

      So, the result is: $- 6 \sin{\left(3 x \right)}$

   The result of the chain rule is:

   $$- 6 \sin{\left(3 x \right)} \cos{\left(2 \cos{\left(3 x \right)} \right)}$$

The answer is:

$$- 6 \sin{\left(3 x \right)} \cos{\left(2 \cos{\left(3 x \right)} \right)}$$