Derivative of 1-x+sin(x)-ln(1+x)

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

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

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

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

         2. 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: $-1$

         The result is: $-1$

      2. The derivative of sine is cosine:

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

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

   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 $\log{\left(u \right)}$ 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. The derivative of the constant $1$ is zero.

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

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

2. Now simplify:

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

The answer is:

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