Derivative of (6x^8)*cot(5x)

1. Apply the product rule:

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

   $f{\left(x \right)} = 6 x^{8}$; 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. Apply the power rule: $x^{8}$ goes to $8 x^{7}$

      So, the result is: $48 x^{7}$

   $g{\left(x \right)} = \cot{\left(5 x \right)}$; to find $\frac{d}{d x} g{\left(x \right)}$:

   1. There are multiple ways to do this derivative.

      Method #1

      1. Rewrite the function to be differentiated:

         $$\cot{\left(5 x \right)} = \frac{1}{\tan{\left(5 x \right)}}$$

      2. Let $u = \tan{\left(5 x \right)}$.

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

      4. Then, apply the chain rule. Multiply by $\frac{d}{d x} \tan{\left(5 x \right)}$:

         1. Rewrite the function to be differentiated:

            $$\tan{\left(5 x \right)} = \frac{\sin{\left(5 x \right)}}{\cos{\left(5 x \right)}}$$

         2. Apply the quotient rule, which is:

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

            $f{\left(x \right)} = \sin{\left(5 x \right)}$ and $g{\left(x \right)} = \cos{\left(5 x \right)}$.

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

            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:

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

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

            1. Let $u = 5 x$.

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

               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:

               $$- 5 \sin{\left(5 x \right)}$$

            Now plug in to the quotient rule:

            $\frac{5 \sin^{2}{\left(5 x \right)} + 5 \cos^{2}{\left(5 x \right)}}{\cos^{2}{\left(5 x \right)}}$

         The result of the chain rule is:

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

      Method #2

      1. Rewrite the function to be differentiated:

         $$\cot{\left(5 x \right)} = \frac{\cos{\left(5 x \right)}}{\sin{\left(5 x \right)}}$$

      2. Apply the quotient rule, which is:

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

         $f{\left(x \right)} = \cos{\left(5 x \right)}$ and $g{\left(x \right)} = \sin{\left(5 x \right)}$.

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

         1. Let $u = 5 x$.

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

            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:

            $$- 5 \sin{\left(5 x \right)}$$

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

         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:

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

         Now plug in to the quotient rule:

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

   The result is: $- \frac{6 x^{8} \left(5 \sin^{2}{\left(5 x \right)} + 5 \cos^{2}{\left(5 x \right)}\right)}{\cos^{2}{\left(5 x \right)} \tan^{2}{\left(5 x \right)}} + 48 x^{7} \cot{\left(5 x \right)}$

2. Now simplify:

   $$x^{7} \left(- \frac{30 x}{\sin^{2}{\left(5 x \right)}} + \frac{48}{\tan{\left(5 x \right)}}\right)$$

The answer is:

$$x^{7} \left(- \frac{30 x}{\sin^{2}{\left(5 x \right)}} + \frac{48}{\tan{\left(5 x \right)}}\right)$$