Derivative of y(x)=-2sin5x+4ctg3x

1. Differentiate $- 2 \sin{\left(5 x \right)} + 4 \cot{\left(3 x \right)}$ term by term:

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

      1. Let $u = 5 x$.

      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 $\frac{d}{d x} 5 x$:

         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: $5$

         The result of the chain rule is:

         $$5 \cos{\left(5 x \right)}$$

      So, the result is: $- 10 \cos{\left(5 x \right)}$

   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:

            $$\cot{\left(3 x \right)} = \frac{1}{\tan{\left(3 x \right)}}$$

         2. Let $u = \tan{\left(3 x \right)}$.

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

         4. Then, apply the chain rule. Multiply by $\frac{d}{d x} \tan{\left(3 x \right)}$:

            1. Rewrite the function to be differentiated:

               $$\tan{\left(3 x \right)} = \frac{\sin{\left(3 x \right)}}{\cos{\left(3 x \right)}}$$

            2. Apply the quotient rule, which is:

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

               $f{\left(x \right)} = \sin{\left(3 x \right)}$ and $g{\left(x \right)} = \cos{\left(3 x \right)}$.

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

               1. Let $u = 3 x$.

               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 $\frac{d}{d x} 3 x$:

                  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: $3$

                  The result of the chain rule is:

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

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

               1. Let $u = 3 x$.

               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 $\frac{d}{d x} 3 x$:

                  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: $3$

                  The result of the chain rule is:

                  $$- 3 \sin{\left(3 x \right)}$$

               Now plug in to the quotient rule:

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

            The result of the chain rule is:

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

         Method #2

         1. Rewrite the function to be differentiated:

            $$\cot{\left(3 x \right)} = \frac{\cos{\left(3 x \right)}}{\sin{\left(3 x \right)}}$$

         2. Apply the quotient rule, which is:

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

            $f{\left(x \right)} = \cos{\left(3 x \right)}$ and $g{\left(x \right)} = \sin{\left(3 x \right)}$.

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

            1. Let $u = 3 x$.

            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 $\frac{d}{d x} 3 x$:

               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: $3$

               The result of the chain rule is:

               $$- 3 \sin{\left(3 x \right)}$$

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

            1. Let $u = 3 x$.

            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 $\frac{d}{d x} 3 x$:

               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: $3$

               The result of the chain rule is:

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

            Now plug in to the quotient rule:

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

      So, the result is: $- \frac{4 \left(3 \sin^{2}{\left(3 x \right)} + 3 \cos^{2}{\left(3 x \right)}\right)}{\cos^{2}{\left(3 x \right)} \tan^{2}{\left(3 x \right)}}$

   The result is: $- \frac{4 \left(3 \sin^{2}{\left(3 x \right)} + 3 \cos^{2}{\left(3 x \right)}\right)}{\cos^{2}{\left(3 x \right)} \tan^{2}{\left(3 x \right)}} - 10 \cos{\left(5 x \right)}$

2. Now simplify:

   $$- 10 \cos{\left(5 x \right)} - \frac{12}{\sin^{2}{\left(3 x \right)}}$$

The answer is:

$$- 10 \cos{\left(5 x \right)} - \frac{12}{\sin^{2}{\left(3 x \right)}}$$