Derivative of y=sin5x+4ctg(3-4x)

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

   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:

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

   4. 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:

            $$\cot{\left(3 - 4 x \right)} = - \frac{1}{\tan{\left(4 x - 3 \right)}}$$

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

            1. Let $u = \tan{\left(4 x - 3 \right)}$.

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

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

               1. Rewrite the function to be differentiated:

                  $$\tan{\left(4 x - 3 \right)} = \frac{\sin{\left(4 x - 3 \right)}}{\cos{\left(4 x - 3 \right)}}$$

               2. Apply the quotient rule, which is:

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

                  $f{\left(x \right)} = \sin{\left(4 x - 3 \right)}$ and $g{\left(x \right)} = \cos{\left(4 x - 3 \right)}$.

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

                  1. Let $u = 4 x - 3$.

                  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 $\frac{d}{d x} \left(4 x - 3\right)$:

                     1. Differentiate $4 x - 3$ term by term:

                        1. The derivative of the constant $-3$ is zero.

                        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: $4$

                        The result is: $4$

                     The result of the chain rule is:

                     $$4 \cos{\left(4 x - 3 \right)}$$

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

                  1. Let $u = 4 x - 3$.

                  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 $\frac{d}{d x} \left(4 x - 3\right)$:

                     1. Differentiate $4 x - 3$ term by term:

                        1. The derivative of the constant $-3$ is zero.

                        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: $4$

                        The result is: $4$

                     The result of the chain rule is:

                     $$- 4 \sin{\left(4 x - 3 \right)}$$

                  Now plug in to the quotient rule:

                  $\frac{4 \sin^{2}{\left(4 x - 3 \right)} + 4 \cos^{2}{\left(4 x - 3 \right)}}{\cos^{2}{\left(4 x - 3 \right)}}$

               The result of the chain rule is:

               $$- \frac{4 \sin^{2}{\left(4 x - 3 \right)} + 4 \cos^{2}{\left(4 x - 3 \right)}}{\cos^{2}{\left(4 x - 3 \right)} \tan^{2}{\left(4 x - 3 \right)}}$$

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

         Method #2

         1. Rewrite the function to be differentiated:

            $$\cot{\left(3 - 4 x \right)} = - \frac{\cos{\left(4 x - 3 \right)}}{\sin{\left(4 x - 3 \right)}}$$

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

            1. Apply the quotient rule, which is:

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

               $f{\left(x \right)} = \cos{\left(4 x - 3 \right)}$ and $g{\left(x \right)} = \sin{\left(4 x - 3 \right)}$.

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

               1. Let $u = 4 x - 3$.

               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 $\frac{d}{d x} \left(4 x - 3\right)$:

                  1. Differentiate $4 x - 3$ term by term:

                     1. The derivative of the constant $-3$ is zero.

                     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: $4$

                     The result is: $4$

                  The result of the chain rule is:

                  $$- 4 \sin{\left(4 x - 3 \right)}$$

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

               1. Let $u = 4 x - 3$.

               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 $\frac{d}{d x} \left(4 x - 3\right)$:

                  1. Differentiate $4 x - 3$ term by term:

                     1. The derivative of the constant $-3$ is zero.

                     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: $4$

                     The result is: $4$

                  The result of the chain rule is:

                  $$4 \cos{\left(4 x - 3 \right)}$$

               Now plug in to the quotient rule:

               $\frac{- 4 \sin^{2}{\left(4 x - 3 \right)} - 4 \cos^{2}{\left(4 x - 3 \right)}}{\sin^{2}{\left(4 x - 3 \right)}}$

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

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

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

2. Now simplify:

   $$5 \cos{\left(5 x \right)} + 16 + \frac{16}{\tan^{2}{\left(4 x - 3 \right)}}$$

The answer is:

$$5 \cos{\left(5 x \right)} + 16 + \frac{16}{\tan^{2}{\left(4 x - 3 \right)}}$$