Derivative of sin2x-ln*(x+1)

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

   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:

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

   4. 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 $\log{\left(u \right)}$ is $\frac{1}{u}$.

      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:

         $$\frac{1}{x + 1}$$

      So, the result is: $- \frac{1}{x + 1}$

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

2. Now simplify:

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

The answer is:

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