Derivative of xsin(3x+1)+2ctgx*(3x+1)

1. Differentiate $x \sin{\left(3 x + 1 \right)} + \left(3 x + 1\right) 2 \cot{\left(x \right)}$ term by term:

   1. Apply the product rule:

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

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

      1. Apply the power rule: $x$ goes to $1$

      $g{\left(x \right)} = \sin{\left(3 x + 1 \right)}$; to find $\frac{d}{d x} g{\left(x \right)}$:

      1. Let $u = 3 x + 1$.

      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 $\frac{d}{d x} \left(3 x + 1\right)$:

         1. Differentiate $3 x + 1$ term by term:

            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: $3$

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

            The result is: $3$

         The result of the chain rule is:

         $$3 \cos{\left(3 x + 1 \right)}$$

      The result is: $3 x \cos{\left(3 x + 1 \right)} + \sin{\left(3 x + 1 \right)}$

   2. Apply the product rule:

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

      $f{\left(x \right)} = 2 \cot{\left(x \right)}$; 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:

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

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

      $g{\left(x \right)} = 3 x + 1$; to find $\frac{d}{d x} g{\left(x \right)}$:

      1. Differentiate $3 x + 1$ term by term:

         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: $3$

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

         The result is: $3$

      The result is: $- \frac{2 \left(3 x + 1\right) \left(\sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}\right)}{\cos^{2}{\left(x \right)} \tan^{2}{\left(x \right)}} + 6 \cot{\left(x \right)}$

   The result is: $3 x \cos{\left(3 x + 1 \right)} - \frac{2 \left(3 x + 1\right) \left(\sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}\right)}{\cos^{2}{\left(x \right)} \tan^{2}{\left(x \right)}} + \sin{\left(3 x + 1 \right)} + 6 \cot{\left(x \right)}$

2. Now simplify:

   $$3 x \cos{\left(3 x + 1 \right)} - \frac{6 x}{\sin^{2}{\left(x \right)}} + \sin{\left(3 x + 1 \right)} + \frac{6}{\tan{\left(x \right)}} - \frac{2}{\sin^{2}{\left(x \right)}}$$

The answer is:

$$3 x \cos{\left(3 x + 1 \right)} - \frac{6 x}{\sin^{2}{\left(x \right)}} + \sin{\left(3 x + 1 \right)} + \frac{6}{\tan{\left(x \right)}} - \frac{2}{\sin^{2}{\left(x \right)}}$$