Derivative of y=(x^2-3x)÷(x+1)

1. Apply the quotient rule, which is:

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

   $f{\left(x \right)} = x^{2} - 3 x$ and $g{\left(x \right)} = x + 1$.

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

   1. Differentiate $x^{2} - 3 x$ term by term:

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

      2. 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: $-3$

      The result is: $2 x - 3$

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

   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: $1$

   Now plug in to the quotient rule:

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

2. Now simplify:

   $$\frac{x^{2} + 2 x - 3}{x^{2} + 2 x + 1}$$

The answer is:

$$\frac{x^{2} + 2 x - 3}{x^{2} + 2 x + 1}$$