Derivative of (0,5-x)cosx+sinx

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

   1. Apply the quotient rule, which is:

      $$\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$$

      $f{\left(x \right)} = \left(1 - 2 x\right) \cos{\left(x \right)}$ and $g{\left(x \right)} = 2$.

      To find $\frac{d}{d x} f{\left(x \right)}$:

      1. Apply the product rule:

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

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

         1. Differentiate $1 - 2 x$ 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: $-2$

            The result is: $-2$

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

         1. The derivative of cosine is negative sine:

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

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

      To find $\frac{d}{d x} g{\left(x \right)}$:

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

      Now plug in to the quotient rule:

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

   2. The derivative of sine is cosine:

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

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

2. Now simplify:

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

The answer is: