Derivative of -(x+1)/(4x+7)

1. Apply the quotient rule, which is:

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

   $f{\left(x \right)} = - x - 1$ and $g{\left(x \right)} = 4 x + 7$.

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

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

      1. The derivative of the constant $-1$ 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: $-1$

      The result is: $-1$

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

   1. Differentiate $4 x + 7$ term by term:

      1. The derivative of the constant $7$ 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: $4$

      The result is: $4$

   Now plug in to the quotient rule:

   $- \frac{3}{\left(4 x + 7\right)^{2}}$

The answer is:

$$- \frac{3}{\left(4 x + 7\right)^{2}}$$