Derivative of 2ctg(x/2)

1. 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:

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

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

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

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

         1. Rewrite the function to be differentiated:

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

         2. Apply the quotient rule, which is:

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

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

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

            1. Let $u = \frac{x}{2}$.

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

               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: $\frac{1}{2}$

               The result of the chain rule is:

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

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

            1. Let $u = \frac{x}{2}$.

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

               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: $\frac{1}{2}$

               The result of the chain rule is:

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

            Now plug in to the quotient rule:

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

         The result of the chain rule is:

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

      Method #2

      1. Rewrite the function to be differentiated:

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

      2. Apply the quotient rule, which is:

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

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

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

         1. Let $u = \frac{x}{2}$.

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

            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: $\frac{1}{2}$

            The result of the chain rule is:

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

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

         1. Let $u = \frac{x}{2}$.

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

            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: $\frac{1}{2}$

            The result of the chain rule is:

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

         Now plug in to the quotient rule:

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

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

2. Now simplify:

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

The answer is:

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