Derivative of y=ln(tg((2x+1)/4))

1. Let $u = \tan{\left(\frac{2 x + 1}{4} \right)}$.

2. The derivative of $\log{\left(u \right)}$ is $\frac{1}{u}$.

3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \tan{\left(\frac{2 x + 1}{4} \right)}$:

   1. Rewrite the function to be differentiated:

      $$\tan{\left(\frac{2 x + 1}{4} \right)} = \frac{\sin{\left(\frac{2 x + 1}{4} \right)}}{\cos{\left(\frac{2 x + 1}{4} \right)}}$$

   2. Apply the quotient rule, which is:

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

      $f{\left(x \right)} = \sin{\left(\frac{2 x + 1}{4} \right)}$ and $g{\left(x \right)} = \cos{\left(\frac{2 x + 1}{4} \right)}$.

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

      1. Let $u = \frac{2 x + 1}{4}$.

      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 $\frac{d}{d x} \frac{2 x + 1}{4}$:

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

            1. Differentiate $2 x + 1$ 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: $2$

               2. The derivative of the constant $1$ is zero.

               The result is: $2$

            So, the result is: $\frac{1}{2}$

         The result of the chain rule is:

         $$\frac{\cos{\left(\frac{2 x + 1}{4} \right)}}{2}$$

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

      1. Let $u = \frac{2 x + 1}{4}$.

      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 $\frac{d}{d x} \frac{2 x + 1}{4}$:

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

            1. Differentiate $2 x + 1$ 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: $2$

               2. The derivative of the constant $1$ is zero.

               The result is: $2$

            So, the result is: $\frac{1}{2}$

         The result of the chain rule is:

         $$- \frac{\sin{\left(\frac{2 x + 1}{4} \right)}}{2}$$

      Now plug in to the quotient rule:

      $\frac{\frac{\sin^{2}{\left(\frac{2 x + 1}{4} \right)}}{2} + \frac{\cos^{2}{\left(\frac{2 x + 1}{4} \right)}}{2}}{\cos^{2}{\left(\frac{2 x + 1}{4} \right)}}$

   The result of the chain rule is:

   $$\frac{\frac{\sin^{2}{\left(\frac{2 x + 1}{4} \right)}}{2} + \frac{\cos^{2}{\left(\frac{2 x + 1}{4} \right)}}{2}}{\cos^{2}{\left(\frac{2 x + 1}{4} \right)} \tan{\left(\frac{2 x + 1}{4} \right)}}$$

4. Now simplify:

   $$\frac{1}{\left(\cos{\left(x + \frac{1}{2} \right)} + 1\right) \tan{\left(\frac{x}{2} + \frac{1}{4} \right)}}$$

The answer is:

$$\frac{1}{\left(\cos{\left(x + \frac{1}{2} \right)} + 1\right) \tan{\left(\frac{x}{2} + \frac{1}{4} \right)}}$$