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

1. Apply the product rule:

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

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

   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: $1$

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

   1. Let $u = x + 3$.

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

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(x + 3\right)$:

      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 of the chain rule is:

      $$2 x + 6$$

   The result is: $\left(x - 3\right) \left(2 x + 6\right) + \left(x + 3\right)^{2}$

2. Now simplify:

   $$3 \left(x - 1\right) \left(x + 3\right)$$

The answer is:

$$3 \left(x - 1\right) \left(x + 3\right)$$