Derivative of cot(2*x)/cos(6*x)

1. Apply the quotient rule, which is:

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

   $f{\left(x \right)} = \cot{\left(2 x \right)}$ and $g{\left(x \right)} = \cos{\left(6 x \right)}$.

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

   1. There are multiple ways to do this derivative.

      Method #1

      1. Rewrite the function to be differentiated:

         $$\cot{\left(2 x \right)} = \frac{1}{\tan{\left(2 x \right)}}$$

      2. Let $u = \tan{\left(2 x \right)}$.

      3. Apply the power rule: $\frac{1}{u}$ goes to $- \frac{1}{u^{2}}$

      4. Then, apply the chain rule. Multiply by $\frac{d}{d x} \tan{\left(2 x \right)}$:

         1. Rewrite the function to be differentiated:

            $$\tan{\left(2 x \right)} = \frac{\sin{\left(2 x \right)}}{\cos{\left(2 x \right)}}$$

         2. Apply the quotient rule, which is:

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

            $f{\left(x \right)} = \sin{\left(2 x \right)}$ and $g{\left(x \right)} = \cos{\left(2 x \right)}$.

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

            1. Let $u = 2 x$.

            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 $\frac{d}{d x} 2 x$:

               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: $2$

               The result of the chain rule is:

               $$2 \cos{\left(2 x \right)}$$

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

            1. Let $u = 2 x$.

            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 $\frac{d}{d x} 2 x$:

               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: $2$

               The result of the chain rule is:

               $$- 2 \sin{\left(2 x \right)}$$

            Now plug in to the quotient rule:

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

         The result of the chain rule is:

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

      Method #2

      1. Rewrite the function to be differentiated:

         $$\cot{\left(2 x \right)} = \frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}}$$

      2. Apply the quotient rule, which is:

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

         $f{\left(x \right)} = \cos{\left(2 x \right)}$ and $g{\left(x \right)} = \sin{\left(2 x \right)}$.

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

         1. Let $u = 2 x$.

         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 $\frac{d}{d x} 2 x$:

            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: $2$

            The result of the chain rule is:

            $$- 2 \sin{\left(2 x \right)}$$

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

         1. Let $u = 2 x$.

         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 $\frac{d}{d x} 2 x$:

            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: $2$

            The result of the chain rule is:

            $$2 \cos{\left(2 x \right)}$$

         Now plug in to the quotient rule:

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

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

   1. Let $u = 6 x$.

   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 $\frac{d}{d x} 6 x$:

      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: $6$

      The result of the chain rule is:

      $$- 6 \sin{\left(6 x \right)}$$

   Now plug in to the quotient rule:

   $\frac{- \frac{\left(2 \sin^{2}{\left(2 x \right)} + 2 \cos^{2}{\left(2 x \right)}\right) \cos{\left(6 x \right)}}{\cos^{2}{\left(2 x \right)} \tan^{2}{\left(2 x \right)}} + 6 \sin{\left(6 x \right)} \cot{\left(2 x \right)}}{\cos^{2}{\left(6 x \right)}}$

2. Now simplify:

   $$\frac{2 \left(\frac{3 \cos{\left(2 x \right)}}{4} - \cos{\left(6 x \right)} - \frac{3 \cos{\left(10 x \right)}}{4}\right)}{\sin^{2}{\left(2 x \right)} \cos^{2}{\left(6 x \right)}}$$

The answer is: