Derivative of y=-ctgx-9x

1. Differentiate $- 9 x - \cot{\left(x \right)}$ term by term:

   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:

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

         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:

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

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

   2. 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: $-9$

   The result is: $\frac{\sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}}{\cos^{2}{\left(x \right)} \tan^{2}{\left(x \right)}} - 9$

2. Now simplify:

   $$-8 + \frac{1}{\tan^{2}{\left(x \right)}}$$

The answer is:

$$-8 + \frac{1}{\tan^{2}{\left(x \right)}}$$