Derivative of tan(2*x)/(1-cot(2*x))

1. Apply the quotient rule, which is:

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

   $f{\left(x \right)} = \tan{\left(2 x \right)}$ and $g{\left(x \right)} = 1 - \cot{\left(2 x \right)}$.

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

   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:

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

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

   1. Differentiate $1 - \cot{\left(2 x \right)}$ 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. 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. Let $u = 2 x$.

               2. $$\frac{d}{d u} \tan{\left(u \right)} = \frac{1}{\cos^{2}{\left(u \right)}}$$

               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:

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

               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:

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

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

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

   Now plug in to the quotient rule:

   $\frac{\frac{\left(1 - \cot{\left(2 x \right)}\right) \left(2 \sin^{2}{\left(2 x \right)} + 2 \cos^{2}{\left(2 x \right)}\right)}{\cos^{2}{\left(2 x \right)}} - \frac{2 \sin^{2}{\left(2 x \right)} + 2 \cos^{2}{\left(2 x \right)}}{\cos^{2}{\left(2 x \right)} \tan{\left(2 x \right)}}}{\left(1 - \cot{\left(2 x \right)}\right)^{2}}$

2. Now simplify:

   $$\frac{2 \left(\tan{\left(2 x \right)} - 2\right)}{\left(\cot{\left(2 x \right)} - 1\right)^{2} \cos^{2}{\left(2 x \right)} \tan{\left(2 x \right)}}$$

The answer is: