Derivative of (x-5)^2(2x+8)

1. Apply the product rule:

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

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

   1. Let $u = x - 5$.

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

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(x - 5\right)$:

      1. Differentiate $x - 5$ term by term:

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

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

         The result is: $1$

      The result of the chain rule is:

      $$2 x - 10$$

   $g{\left(x \right)} = 2 x + 8$; to find $\frac{d}{d x} g{\left(x \right)}$:

   1. Differentiate $2 x + 8$ 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: $x$ goes to $1$

         So, the result is: $2$

      2. The derivative of the constant $8$ is zero.

      The result is: $2$

   The result is: $2 \left(x - 5\right)^{2} + \left(2 x - 10\right) \left(2 x + 8\right)$

2. Now simplify:

   $$6 \left(x - 5\right) \left(x + 1\right)$$

The answer is:

$$6 \left(x - 5\right) \left(x + 1\right)$$