Derivative of y=(x-9)^2*e^(x-9)

1. Apply the product rule:

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

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

   1. Let $u = x - 9$.

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

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(x - 9\right)$:

      1. Differentiate $x - 9$ term by term:

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

         2. The derivative of the constant $\left(-1\right) 9$ is zero.

         The result is: $1$

      The result of the chain rule is:

      $$2 x - 18$$

   $g{\left(x \right)} = e^{x - 9}$; to find $\frac{d}{d x} g{\left(x \right)}$:

   1. Let $u = x - 9$.

   2. The derivative of $e^{u}$ is itself.

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(x - 9\right)$:

      1. Differentiate $x - 9$ term by term:

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

         2. The derivative of the constant $\left(-1\right) 9$ is zero.

         The result is: $1$

      The result of the chain rule is:

      $$e^{x - 9}$$

   The result is: $\left(x - 9\right)^{2} e^{x - 9} + \left(2 x - 18\right) e^{x - 9}$

2. Now simplify:

   $$\left(x - 9\right) \left(x - 7\right) e^{x - 9}$$

The answer is:

$$\left(x - 9\right) \left(x - 7\right) e^{x - 9}$$