Derivative of (x-1)^2*sinx

1. Apply the product rule:

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

   $f{\left(x \right)} = \left(x - 1\right)^{2}$; to find $\frac{d}{d x} f{\left(x \right)}$:

   1. Let $u = x - 1$.

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

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(x - 1\right)$:

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

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

         2. The derivative of the constant $-1$ is zero.

         The result is: $1$

      The result of the chain rule is:

      $$2 x - 2$$

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

   1. The derivative of sine is cosine:

      $$\frac{d}{d x} \sin{\left(x \right)} = \cos{\left(x \right)}$$

   The result is: $\left(x - 1\right)^{2} \cos{\left(x \right)} + \left(2 x - 2\right) \sin{\left(x \right)}$

2. Now simplify:

   $$\left(x - 1\right) \left(\left(x - 1\right) \cos{\left(x \right)} + 2 \sin{\left(x \right)}\right)$$

The answer is:

$$\left(x - 1\right) \left(\left(x - 1\right) \cos{\left(x \right)} + 2 \sin{\left(x \right)}\right)$$