Derivative of ctg^2(x)+1

1. Differentiate $\cot^{2}{\left(x \right)} + 1$ term by term:

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

   2. Apply the power rule: $u^{2}$ goes to $2 u$

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

      1. There are multiple ways to do this derivative.

         Method #1

         1. Rewrite the function to be differentiated:

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

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

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

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

            1. Rewrite the function to be differentiated:

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

            2. Apply the quotient rule, which is:

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

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

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

               1. The derivative of sine is cosine:

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

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

               1. The derivative of cosine is negative sine:

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

               Now plug in to the quotient rule:

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

            The result of the chain rule is:

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

         Method #2

         1. Rewrite the function to be differentiated:

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

         2. Apply the quotient rule, which is:

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

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

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

            1. The derivative of cosine is negative sine:

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

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

            1. The derivative of sine is cosine:

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

            Now plug in to the quotient rule:

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

      The result of the chain rule is:

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

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

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

2. Now simplify:

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

The answer is:

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