Derivative of (21-x^2)/(7x+9)

1. Apply the quotient rule, which is:

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

   $f{\left(x \right)} = 21 - x^{2}$ and $g{\left(x \right)} = 7 x + 9$.

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

   1. Differentiate $21 - x^{2}$ term by term:

      1. The derivative of the constant $21$ 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: $x^{2}$ goes to $2 x$

         So, the result is: $- 2 x$

      The result is: $- 2 x$

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

   1. Differentiate $7 x + 9$ term by term:

      1. The derivative of the constant $9$ 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: $7$

      The result is: $7$

   Now plug in to the quotient rule:

   $\frac{7 x^{2} - 2 x \left(7 x + 9\right) - 147}{\left(7 x + 9\right)^{2}}$

The answer is:

$$\frac{7 x^{2} - 2 x \left(7 x + 9\right) - 147}{\left(7 x + 9\right)^{2}}$$