Derivative of y=(x^2-3x+9)(x-3)

1. Apply the product rule:

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

   $f{\left(x \right)} = \left(x^{2} - 3 x\right) + 9$; to find $\frac{d}{d x} f{\left(x \right)}$:

   1. Differentiate $\left(x^{2} - 3 x\right) + 9$ term by term:

      1. Differentiate $x^{2} - 3 x$ term by term:

         1. Apply the power rule: $x^{2}$ goes to $2 x$

         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: $-3$

         The result is: $2 x - 3$

      2. The derivative of the constant $9$ is zero.

      The result is: $2 x - 3$

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

   1. Differentiate $x - 3$ term by term:

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

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

      The result is: $1$

   The result is: $x^{2} - 3 x + \left(x - 3\right) \left(2 x - 3\right) + 9$

2. Now simplify:

   $$3 x^{2} - 12 x + 18$$

The answer is:

$$3 x^{2} - 12 x + 18$$