Derivative of sin4x+cos(3x-3)

1. Differentiate $\sin{\left(4 x \right)} + \cos{\left(3 x - 3 \right)}$ term by term:

   1. Let $u = 4 x$.

   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 $\frac{d}{d x} 4 x$:

      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: $4$

      The result of the chain rule is:

      $$4 \cos{\left(4 x \right)}$$

   4. Let $u = 3 x - 3$.

   5. The derivative of cosine is negative sine:

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

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

      1. Differentiate $3 x - 3$ 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: $3$

         2. The derivative of the constant $-3$ is zero.

         The result is: $3$

      The result of the chain rule is:

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

   The result is: $- 3 \sin{\left(3 x - 3 \right)} + 4 \cos{\left(4 x \right)}$

2. Now simplify:

   $$- 3 \sin{\left(3 x - 3 \right)} + 4 \cos{\left(4 x \right)}$$

The answer is: