Derivative of (-2)sin2x+5ctg6x

1. Differentiate $- 2 \sin{\left(2 x \right)} + 5 \cot{\left(6 x \right)}$ term by term:

   1. The derivative of a constant times a function is the constant times the derivative of the function.

      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:

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

      So, the result is: $- 4 \cos{\left(2 x \right)}$

   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:

            $$\cot{\left(6 x \right)} = \frac{1}{\tan{\left(6 x \right)}}$$

         2. Let $u = \tan{\left(6 x \right)}$.

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

         4. Then, apply the chain rule. Multiply by $\frac{d}{d x} \tan{\left(6 x \right)}$:

            1. Rewrite the function to be differentiated:

               $$\tan{\left(6 x \right)} = \frac{\sin{\left(6 x \right)}}{\cos{\left(6 x \right)}}$$

            2. Apply the quotient rule, which is:

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

               $f{\left(x \right)} = \sin{\left(6 x \right)}$ and $g{\left(x \right)} = \cos{\left(6 x \right)}$.

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

               1. Let $u = 6 x$.

               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 $\frac{d}{d x} 6 x$:

                  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: $6$

                  The result of the chain rule is:

                  $$6 \cos{\left(6 x \right)}$$

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

               1. Let $u = 6 x$.

               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 $\frac{d}{d x} 6 x$:

                  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: $6$

                  The result of the chain rule is:

                  $$- 6 \sin{\left(6 x \right)}$$

               Now plug in to the quotient rule:

               $\frac{6 \sin^{2}{\left(6 x \right)} + 6 \cos^{2}{\left(6 x \right)}}{\cos^{2}{\left(6 x \right)}}$

            The result of the chain rule is:

            $$- \frac{6 \sin^{2}{\left(6 x \right)} + 6 \cos^{2}{\left(6 x \right)}}{\cos^{2}{\left(6 x \right)} \tan^{2}{\left(6 x \right)}}$$

         Method #2

         1. Rewrite the function to be differentiated:

            $$\cot{\left(6 x \right)} = \frac{\cos{\left(6 x \right)}}{\sin{\left(6 x \right)}}$$

         2. Apply the quotient rule, which is:

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

            $f{\left(x \right)} = \cos{\left(6 x \right)}$ and $g{\left(x \right)} = \sin{\left(6 x \right)}$.

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

            1. Let $u = 6 x$.

            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 $\frac{d}{d x} 6 x$:

               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: $6$

               The result of the chain rule is:

               $$- 6 \sin{\left(6 x \right)}$$

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

            1. Let $u = 6 x$.

            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 $\frac{d}{d x} 6 x$:

               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: $6$

               The result of the chain rule is:

               $$6 \cos{\left(6 x \right)}$$

            Now plug in to the quotient rule:

            $\frac{- 6 \sin^{2}{\left(6 x \right)} - 6 \cos^{2}{\left(6 x \right)}}{\sin^{2}{\left(6 x \right)}}$

      So, the result is: $- \frac{5 \left(6 \sin^{2}{\left(6 x \right)} + 6 \cos^{2}{\left(6 x \right)}\right)}{\cos^{2}{\left(6 x \right)} \tan^{2}{\left(6 x \right)}}$

   The result is: $- \frac{5 \left(6 \sin^{2}{\left(6 x \right)} + 6 \cos^{2}{\left(6 x \right)}\right)}{\cos^{2}{\left(6 x \right)} \tan^{2}{\left(6 x \right)}} - 4 \cos{\left(2 x \right)}$

2. Now simplify:

   $$- \frac{- 2 \left(\cos{\left(12 x \right)} - 1\right) \cos{\left(2 x \right)} + 30 \sin^{2}{\left(6 x \right)} + 30 \cos^{2}{\left(6 x \right)}}{\cos^{2}{\left(6 x \right)} \tan^{2}{\left(6 x \right)}}$$

The answer is: