Derivative of -4sin(2x)-cos(x+1)

1. Differentiate $- 4 \sin{\left(2 x \right)} - \cos{\left(x + 1 \right)}$ term by term:

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

      1. Let $u = 2 x$.

      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 $\frac{d}{d x} 2 x$:

         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: $2$

         The result of the chain rule is:

         $$2 \cos{\left(2 x \right)}$$

      So, the result is: $- 8 \cos{\left(2 x \right)}$

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

      1. Let $u = x + 1$.

      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 $\frac{d}{d x} \left(x + 1\right)$:

         1. Differentiate $x + 1$ term by term:

            1. Apply the power rule: $x$ goes to $1$

            2. The derivative of the constant $1$ is zero.

            The result is: $1$

         The result of the chain rule is:

         $$- \sin{\left(x + 1 \right)}$$

      So, the result is: $\sin{\left(x + 1 \right)}$

   The result is: $\sin{\left(x + 1 \right)} - 8 \cos{\left(2 x \right)}$

2. Now simplify:

   $$\sin{\left(x + 1 \right)} - 8 \cos{\left(2 x \right)}$$

The answer is:

$$\sin{\left(x + 1 \right)} - 8 \cos{\left(2 x \right)}$$