Derivative of (20cos20x)/(sin20x)

1. Apply the quotient rule, which is:

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

   $f{\left(x \right)} = 20 \cos{\left(20 x \right)}$ and $g{\left(x \right)} = \sin{\left(20 x \right)}$.

   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. Let $u = 20 x$.

      2. The derivative of cosine is negative sine:

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

      3. Then, apply the chain rule. Multiply by $\frac{d}{d x} 20 x$:

         1. 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: $20$

         The result of the chain rule is:

         $$- 20 \sin{\left(20 x \right)}$$

      So, the result is: $- 400 \sin{\left(20 x \right)}$

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

   1. Let $u = 20 x$.

   2. The derivative of sine is cosine:

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

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} 20 x$:

      1. 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: $20$

      The result of the chain rule is:

      $$20 \cos{\left(20 x \right)}$$

   Now plug in to the quotient rule:

   $\frac{- 400 \sin^{2}{\left(20 x \right)} - 400 \cos^{2}{\left(20 x \right)}}{\sin^{2}{\left(20 x \right)}}$

2. Now simplify:

   $$- \frac{400}{\sin^{2}{\left(20 x \right)}}$$

The answer is:

$$- \frac{400}{\sin^{2}{\left(20 x \right)}}$$