Derivative of (2x-4)/(5+x)

1. Apply the quotient rule, which is:

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

   $f{\left(x \right)} = 2 x - 4$ and $g{\left(x \right)} = x + 5$.

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

   1. Differentiate $2 x - 4$ term by term:

      1. The derivative of the constant $-4$ is zero.

      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: $2$

      The result is: $2$

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

   1. Differentiate $x + 5$ term by term:

      1. The derivative of the constant $5$ is zero.

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

      The result is: $1$

   Now plug in to the quotient rule:

   $\frac{14}{\left(x + 5\right)^{2}}$

The answer is:

$$\frac{14}{\left(x + 5\right)^{2}}$$