Derivative of y=x^5(x-3)^3

1. Apply the product rule:

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

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

   1. Apply the power rule: $x^{5}$ goes to $5 x^{4}$

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

   1. Let $u = x - 3$.

   2. Apply the power rule: $u^{3}$ goes to $3 u^{2}$

   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 $\left(-1\right) 3$ is zero.

         The result is: $1$

      The result of the chain rule is:

      $$3 \left(x - 3\right)^{2}$$

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

2. Now simplify:

   $$x^{4} \left(x - 3\right)^{2} \cdot \left(8 x - 15\right)$$

The answer is:

$$x^{4} \left(x - 3\right)^{2} \cdot \left(8 x - 15\right)$$