Derivative of 9(sin(x))/16cos(x)

1. Apply the quotient rule, which is:

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

   $f{\left(x \right)} = 9 \sin{\left(x \right)} \cos{\left(x \right)}$ and $g{\left(x \right)} = 16$.

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

   1. The derivative of a constant times a function is the constant times the derivative of the function.

      1. Apply the product rule:

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

         $f{\left(x \right)} = \cos{\left(x \right)}$; to find $\frac{d}{d x} f{\left(x \right)}$:

         1. The derivative of cosine is negative sine:

            $$\frac{d}{d x} \cos{\left(x \right)} = - \sin{\left(x \right)}$$

         $g{\left(x \right)} = \sin{\left(x \right)}$; to find $\frac{d}{d x} g{\left(x \right)}$:

         1. The derivative of sine is cosine:

            $$\frac{d}{d x} \sin{\left(x \right)} = \cos{\left(x \right)}$$

         The result is: $- \sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}$

      So, the result is: $- 9 \sin^{2}{\left(x \right)} + 9 \cos^{2}{\left(x \right)}$

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

   1. The derivative of the constant $16$ is zero.

   Now plug in to the quotient rule:

   $- \frac{9 \sin^{2}{\left(x \right)}}{16} + \frac{9 \cos^{2}{\left(x \right)}}{16}$

2. Now simplify:

   $$\frac{9 \cos{\left(2 x \right)}}{16}$$

The answer is:

$$\frac{9 \cos{\left(2 x \right)}}{16}$$