Derivative of cot(4*x-3)

1. There are multiple ways to do this derivative.

   Method #1

   1. Rewrite the function to be differentiated:

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

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

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

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

      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 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: $4$

               2. The derivative of the constant $\left(-1\right) 3$ is zero.

               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 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: $4$

               2. The derivative of the constant $\left(-1\right) 3$ is zero.

               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:

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

   Method #2

   1. Rewrite the function to be differentiated:

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

   2. Apply the quotient rule, which is:

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

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

            2. The derivative of the constant $\left(-1\right) 3$ is zero.

            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 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: $4$

            2. The derivative of the constant $\left(-1\right) 3$ is zero.

            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:

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

2. Now simplify:

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

The answer is:

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