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

1. Apply the product rule:

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

   $f{\left(x \right)} = \log{\left(x \right)}$; to find $\frac{d}{d x} f{\left(x \right)}$:

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

   $g{\left(x \right)} = \tan{\left(2 x + \frac{1}{4} \right)}$; to find $\frac{d}{d x} g{\left(x \right)}$:

   1. Rewrite the function to be differentiated:

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

   2. Apply the quotient rule, which is:

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

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

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

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

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

         1. Differentiate $2 x + \frac{1}{4}$ 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 $\frac{1}{4}$ is zero.

            The result is: $2$

         The result of the chain rule is:

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

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

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

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

         1. Differentiate $2 x + \frac{1}{4}$ 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 $\frac{1}{4}$ is zero.

            The result is: $2$

         The result of the chain rule is:

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

      Now plug in to the quotient rule:

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

   The result is: $\frac{\left(2 \sin^{2}{\left(2 x + \frac{1}{4} \right)} + 2 \cos^{2}{\left(2 x + \frac{1}{4} \right)}\right) \log{\left(x \right)}}{\cos^{2}{\left(2 x + \frac{1}{4} \right)}} + \frac{\tan{\left(2 x + \frac{1}{4} \right)}}{x}$

2. Now simplify:

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

The answer is: