Derivative of x^4/(x+1)^3

1. Apply the quotient rule, which is:

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

   $f{\left(x \right)} = x^{4}$ and $g{\left(x \right)} = \left(x + 1\right)^{3}$.

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

   1. Apply the power rule: $x^{4}$ goes to $4 x^{3}$

   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:

   $\frac{- 3 x^{4} \left(x + 1\right)^{2} + 4 x^{3} \left(x + 1\right)^{3}}{\left(x + 1\right)^{6}}$

2. Now simplify:

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

The answer is:

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