Derivative of x*ln(x+1)-cos4x

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

   1. Apply the product rule:

      $$\frac{d}{d x} f{\left(x \right)} g{\left(x \right)} = f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}$$

      $f{\left(x \right)} = x$; to find $\frac{d}{d x} f{\left(x \right)}$:

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

      $g{\left(x \right)} = \log{\left(x + 1 \right)}$; to find $\frac{d}{d x} g{\left(x \right)}$:

      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:

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

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

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

      1. Let $u = 4 x$.

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

         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:

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

      So, the result is: $4 \sin{\left(4 x \right)}$

   The result is: $\frac{x}{x + 1} + \log{\left(x + 1 \right)} + 4 \sin{\left(4 x \right)}$

2. Now simplify:

   $$\frac{x + \left(x + 1\right) \left(\log{\left(x + 1 \right)} + 4 \sin{\left(4 x \right)}\right)}{x + 1}$$

The answer is:

$$\frac{x + \left(x + 1\right) \left(\log{\left(x + 1 \right)} + 4 \sin{\left(4 x \right)}\right)}{x + 1}$$