Derivative of y=(sinx-3x)(x+1)

1. Apply the product rule:

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

   $f{\left(x \right)} = - 3 x + \sin{\left(x \right)}$; to find $\frac{d}{d x} f{\left(x \right)}$:

   1. Differentiate $- 3 x + \sin{\left(x \right)}$ term by term:

      1. The derivative of sine is cosine:

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

      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: $-3$

      The result is: $\cos{\left(x \right)} - 3$

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

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

2. Now simplify:

   $$- 3 x + \left(x + 1\right) \left(\cos{\left(x \right)} - 3\right) + \sin{\left(x \right)}$$

The answer is:

$$- 3 x + \left(x + 1\right) \left(\cos{\left(x \right)} - 3\right) + \sin{\left(x \right)}$$