Derivative of (2x+1)*cos(3x-5)

1. Apply the product rule:

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

   $f{\left(x \right)} = 2 x + 1$; to find $\frac{d}{d x} f{\left(x \right)}$:

   1. Differentiate $2 x + 1$ 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: $2$

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

      The result is: $2$

   $g{\left(x \right)} = \cos{\left(3 x - 5 \right)}$; to find $\frac{d}{d x} g{\left(x \right)}$:

   1. Let $u = 3 x - 5$.

   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 $\frac{d}{d x} \left(3 x - 5\right)$:

      1. Differentiate $3 x - 5$ 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 $-5$ is zero.

         The result is: $3$

      The result of the chain rule is:

      $$- 3 \sin{\left(3 x - 5 \right)}$$

   The result is: $- 3 \left(2 x + 1\right) \sin{\left(3 x - 5 \right)} + 2 \cos{\left(3 x - 5 \right)}$

2. Now simplify:

   $$\left(- 6 x - 3\right) \sin{\left(3 x - 5 \right)} + 2 \cos{\left(3 x - 5 \right)}$$

The answer is:

$$\left(- 6 x - 3\right) \sin{\left(3 x - 5 \right)} + 2 \cos{\left(3 x - 5 \right)}$$