Derivative of y=ctg5x^2+2x-7

1. Differentiate $2 x + \cot^{2}{\left(5 x \right)} - 7$ term by term:

   1. Let $u = \cot{\left(5 x \right)}$.

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

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

      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 of the chain rule is:

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

   4. 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: $2$

   5. The derivative of the constant $\left(-1\right) 7$ is zero.

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

2. Now simplify:

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

The answer is:

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