Derivative of y=ctg^4(7x+3)

1. Let $u = \cot{\left(7 x + 3 \right)}$.

2. Apply the power rule: $u^{4}$ goes to $4 u^{3}$

3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \cot{\left(7 x + 3 \right)}$:

   1. There are multiple ways to do this derivative.

      Method #1

      1. Rewrite the function to be differentiated:

         $$\cot{\left(7 x + 3 \right)} = \frac{1}{\tan{\left(7 x + 3 \right)}}$$

      2. Let $u = \tan{\left(7 x + 3 \right)}$.

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

      4. Then, apply the chain rule. Multiply by $\frac{d}{d x} \tan{\left(7 x + 3 \right)}$:

         1. Rewrite the function to be differentiated:

            $$\tan{\left(7 x + 3 \right)} = \frac{\sin{\left(7 x + 3 \right)}}{\cos{\left(7 x + 3 \right)}}$$

         2. Apply the quotient rule, which is:

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

            $f{\left(x \right)} = \sin{\left(7 x + 3 \right)}$ and $g{\left(x \right)} = \cos{\left(7 x + 3 \right)}$.

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

            1. Let $u = 7 x + 3$.

            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 $\frac{d}{d x} \left(7 x + 3\right)$:

               1. Differentiate $7 x + 3$ 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: $7$

                  2. The derivative of the constant $3$ is zero.

                  The result is: $7$

               The result of the chain rule is:

               $$7 \cos{\left(7 x + 3 \right)}$$

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

            1. Let $u = 7 x + 3$.

            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 $\frac{d}{d x} \left(7 x + 3\right)$:

               1. Differentiate $7 x + 3$ 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: $7$

                  2. The derivative of the constant $3$ is zero.

                  The result is: $7$

               The result of the chain rule is:

               $$- 7 \sin{\left(7 x + 3 \right)}$$

            Now plug in to the quotient rule:

            $\frac{7 \sin^{2}{\left(7 x + 3 \right)} + 7 \cos^{2}{\left(7 x + 3 \right)}}{\cos^{2}{\left(7 x + 3 \right)}}$

         The result of the chain rule is:

         $$- \frac{7 \sin^{2}{\left(7 x + 3 \right)} + 7 \cos^{2}{\left(7 x + 3 \right)}}{\cos^{2}{\left(7 x + 3 \right)} \tan^{2}{\left(7 x + 3 \right)}}$$

      Method #2

      1. Rewrite the function to be differentiated:

         $$\cot{\left(7 x + 3 \right)} = \frac{\cos{\left(7 x + 3 \right)}}{\sin{\left(7 x + 3 \right)}}$$

      2. Apply the quotient rule, which is:

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

         $f{\left(x \right)} = \cos{\left(7 x + 3 \right)}$ and $g{\left(x \right)} = \sin{\left(7 x + 3 \right)}$.

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

         1. Let $u = 7 x + 3$.

         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 $\frac{d}{d x} \left(7 x + 3\right)$:

            1. Differentiate $7 x + 3$ 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: $7$

               2. The derivative of the constant $3$ is zero.

               The result is: $7$

            The result of the chain rule is:

            $$- 7 \sin{\left(7 x + 3 \right)}$$

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

         1. Let $u = 7 x + 3$.

         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 $\frac{d}{d x} \left(7 x + 3\right)$:

            1. Differentiate $7 x + 3$ 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: $7$

               2. The derivative of the constant $3$ is zero.

               The result is: $7$

            The result of the chain rule is:

            $$7 \cos{\left(7 x + 3 \right)}$$

         Now plug in to the quotient rule:

         $\frac{- 7 \sin^{2}{\left(7 x + 3 \right)} - 7 \cos^{2}{\left(7 x + 3 \right)}}{\sin^{2}{\left(7 x + 3 \right)}}$

   The result of the chain rule is:

   $$- \frac{4 \left(7 \sin^{2}{\left(7 x + 3 \right)} + 7 \cos^{2}{\left(7 x + 3 \right)}\right) \cot^{3}{\left(7 x + 3 \right)}}{\cos^{2}{\left(7 x + 3 \right)} \tan^{2}{\left(7 x + 3 \right)}}$$

4. Now simplify:

   $$- \frac{28 \cos^{3}{\left(7 x + 3 \right)}}{\sin^{5}{\left(7 x + 3 \right)}}$$

The answer is:

$$- \frac{28 \cos^{3}{\left(7 x + 3 \right)}}{\sin^{5}{\left(7 x + 3 \right)}}$$