Derivative of (4cosx-3)(2x²-4x+5)

1. Apply the product rule:

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

   $f{\left(x \right)} = 4 \cos{\left(x \right)} - 3$; to find $\frac{d}{d x} f{\left(x \right)}$:

   1. Differentiate $4 \cos{\left(x \right)} - 3$ term by term:

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

         1. The derivative of cosine is negative sine:

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

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

      2. The derivative of the constant $-3$ is zero.

      The result is: $- 4 \sin{\left(x \right)}$

   $g{\left(x \right)} = \left(2 x^{2} - 4 x\right) + 5$; to find $\frac{d}{d x} g{\left(x \right)}$:

   1. Differentiate $\left(2 x^{2} - 4 x\right) + 5$ term by term:

      1. Differentiate $2 x^{2} - 4 x$ 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: $x^{2}$ goes to $2 x$

            So, the result is: $4 x$

         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: $-4$

         The result is: $4 x - 4$

      2. The derivative of the constant $5$ is zero.

      The result is: $4 x - 4$

   The result is: $\left(4 x - 4\right) \left(4 \cos{\left(x \right)} - 3\right) - 4 \left(\left(2 x^{2} - 4 x\right) + 5\right) \sin{\left(x \right)}$

2. Now simplify:

   $$4 \left(x - 1\right) \left(4 \cos{\left(x \right)} - 3\right) + 4 \left(- 2 x^{2} + 4 x - 5\right) \sin{\left(x \right)}$$

The answer is:

$$4 \left(x - 1\right) \left(4 \cos{\left(x \right)} - 3\right) + 4 \left(- 2 x^{2} + 4 x - 5\right) \sin{\left(x \right)}$$