Derivative of ctg(9^x)

1. There are multiple ways to do this derivative.

   Method #1

   1. Rewrite the function to be differentiated:

      $$\cot{\left(9^{x} \right)} = \frac{1}{\tan{\left(9^{x} \right)}}$$

   2. Let $u = \tan{\left(9^{x} \right)}$.

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

   4. Then, apply the chain rule. Multiply by $\frac{d}{d x} \tan{\left(9^{x} \right)}$:

      1. Rewrite the function to be differentiated:

         $$\tan{\left(9^{x} \right)} = \frac{\sin{\left(9^{x} \right)}}{\cos{\left(9^{x} \right)}}$$

      2. Apply the quotient rule, which is:

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

         $f{\left(x \right)} = \sin{\left(9^{x} \right)}$ and $g{\left(x \right)} = \cos{\left(9^{x} \right)}$.

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

         1. Let $u = 9^{x}$.

         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 $\frac{d}{d x} 9^{x}$:

            1. $\frac{d}{d x} 9^{x} = 9^{x} \log{\left(9 \right)}$

            The result of the chain rule is:

            $$9^{x} \log{\left(9 \right)} \cos{\left(9^{x} \right)}$$

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

         1. Let $u = 9^{x}$.

         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 $\frac{d}{d x} 9^{x}$:

            1. $\frac{d}{d x} 9^{x} = 9^{x} \log{\left(9 \right)}$

            The result of the chain rule is:

            $$- 9^{x} \log{\left(9 \right)} \sin{\left(9^{x} \right)}$$

         Now plug in to the quotient rule:

         $\frac{9^{x} \log{\left(9 \right)} \sin^{2}{\left(9^{x} \right)} + 9^{x} \log{\left(9 \right)} \cos^{2}{\left(9^{x} \right)}}{\cos^{2}{\left(9^{x} \right)}}$

      The result of the chain rule is:

      $$- \frac{9^{x} \log{\left(9 \right)} \sin^{2}{\left(9^{x} \right)} + 9^{x} \log{\left(9 \right)} \cos^{2}{\left(9^{x} \right)}}{\cos^{2}{\left(9^{x} \right)} \tan^{2}{\left(9^{x} \right)}}$$

   Method #2

   1. Rewrite the function to be differentiated:

      $$\cot{\left(9^{x} \right)} = \frac{\cos{\left(9^{x} \right)}}{\sin{\left(9^{x} \right)}}$$

   2. Apply the quotient rule, which is:

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

      $f{\left(x \right)} = \cos{\left(9^{x} \right)}$ and $g{\left(x \right)} = \sin{\left(9^{x} \right)}$.

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

      1. Let $u = 9^{x}$.

      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 $\frac{d}{d x} 9^{x}$:

         1. $\frac{d}{d x} 9^{x} = 9^{x} \log{\left(9 \right)}$

         The result of the chain rule is:

         $$- 9^{x} \log{\left(9 \right)} \sin{\left(9^{x} \right)}$$

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

      1. Let $u = 9^{x}$.

      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 $\frac{d}{d x} 9^{x}$:

         1. $\frac{d}{d x} 9^{x} = 9^{x} \log{\left(9 \right)}$

         The result of the chain rule is:

         $$9^{x} \log{\left(9 \right)} \cos{\left(9^{x} \right)}$$

      Now plug in to the quotient rule:

      $\frac{- 9^{x} \log{\left(9 \right)} \sin^{2}{\left(9^{x} \right)} - 9^{x} \log{\left(9 \right)} \cos^{2}{\left(9^{x} \right)}}{\sin^{2}{\left(9^{x} \right)}}$

2. Now simplify:

   $$- \frac{\log{\left(9^{9^{x}} \right)}}{\sin^{2}{\left(9^{x} \right)}}$$

The answer is:

$$- \frac{\log{\left(9^{9^{x}} \right)}}{\sin^{2}{\left(9^{x} \right)}}$$