Derivative of y=зx^3+3x^2-3x-3-2(x+1)^3/(x-1)^3

1. Differentiate $3 x^{3} + 3 x^{2} - 3 x - \frac{2 \left(x + 1\right)^{3}}{\left(x - 1\right)^{3}} - 3$ 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: $x^{3}$ goes to $3 x^{2}$

      So, the result is: $9 x^{2}$

   2. The derivative of a constant times a function is the constant times the derivative of the function.

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

      So, the result is: $6 x$

   3. The derivative of a constant times a function is the constant times the derivative of the function.

      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: $3$

      So, the result is: $-3$

   4. The derivative of the constant $\left(-1\right) 3$ is zero.

   5. The derivative of a constant times a function is the constant times the derivative of the function.

      1. The derivative of a constant times a function is the constant times the derivative of the function.

         1. Apply the quotient rule, which is:

            $$\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$$

            $f{\left(x \right)} = \left(x + 1\right)^{3}$ and $g{\left(x \right)} = \left(x - 1\right)^{3}$.

            To find $\frac{d}{d x} f{\left(x \right)}$:

            1. Let $u = x + 1$.

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

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

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

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

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

                  The result is: $1$

               The result of the chain rule is:

               $$3 \left(x + 1\right)^{2}$$

            To find $\frac{d}{d x} g{\left(x \right)}$:

            1. Let $u = x - 1$.

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

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

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

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

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

                  The result is: $1$

               The result of the chain rule is:

               $$3 \left(x - 1\right)^{2}$$

            Now plug in to the quotient rule:

            $\frac{3 \left(x - 1\right)^{3} \left(x + 1\right)^{2} - 3 \left(x - 1\right)^{2} \left(x + 1\right)^{3}}{\left(x - 1\right)^{6}}$

         So, the result is: $\frac{2 \cdot \left(3 \left(x - 1\right)^{3} \left(x + 1\right)^{2} - 3 \left(x - 1\right)^{2} \left(x + 1\right)^{3}\right)}{\left(x - 1\right)^{6}}$

      So, the result is: $- \frac{2 \cdot \left(3 \left(x - 1\right)^{3} \left(x + 1\right)^{2} - 3 \left(x - 1\right)^{2} \left(x + 1\right)^{3}\right)}{\left(x - 1\right)^{6}}$

   The result is: $9 x^{2} + 6 x - 3 - \frac{2 \cdot \left(3 \left(x - 1\right)^{3} \left(x + 1\right)^{2} - 3 \left(x - 1\right)^{2} \left(x + 1\right)^{3}\right)}{\left(x - 1\right)^{6}}$

2. Now simplify:

   $$\frac{3 \left(\left(x - 1\right)^{4} \cdot \left(3 x^{2} + 2 x - 1\right) - 2 \left(x - 1\right) \left(x + 1\right)^{2} + 2 \left(x + 1\right)^{3}\right)}{\left(x - 1\right)^{4}}$$

The answer is:

$$\frac{3 \left(\left(x - 1\right)^{4} \cdot \left(3 x^{2} + 2 x - 1\right) - 2 \left(x - 1\right) \left(x + 1\right)^{2} + 2 \left(x + 1\right)^{3}\right)}{\left(x - 1\right)^{4}}$$