Derivative of y=4ctg(2x)*sinx

1. Apply the product rule:

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

   $f{\left(x \right)} = 4 \cot{\left(2 x \right)}$; to find $\frac{d}{d x} f{\left(x \right)}$:

   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:

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

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

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

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

            1. Rewrite the function to be differentiated:

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

            2. Apply the quotient rule, which is:

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

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

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

               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:

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

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

               1. Let $u = 2 x$.

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

                  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:

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

               Now plug in to the quotient rule:

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

            The result of the chain rule is:

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

         Method #2

         1. Rewrite the function to be differentiated:

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

         2. Apply the quotient rule, which is:

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

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

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

            1. Let $u = 2 x$.

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

               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:

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

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

            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:

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

            Now plug in to the quotient rule:

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

      So, the result is: $- \frac{4 \left(2 \sin^{2}{\left(2 x \right)} + 2 \cos^{2}{\left(2 x \right)}\right)}{\cos^{2}{\left(2 x \right)} \tan^{2}{\left(2 x \right)}}$

   $g{\left(x \right)} = \sin{\left(x \right)}$; to find $\frac{d}{d x} g{\left(x \right)}$:

   1. The derivative of sine is cosine:

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

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

2. Now simplify:

   $$- 4 \left(1 + \frac{1}{2 \cos^{2}{\left(x \right)}}\right) \sin{\left(x \right)}$$

The answer is:

$$- 4 \left(1 + \frac{1}{2 \cos^{2}{\left(x \right)}}\right) \sin{\left(x \right)}$$