Derivative of tan(x-pi/4)

1. Rewrite the function to be differentiated:

   $$\tan{\left(x - \frac{\pi}{4} \right)} = \frac{\sin{\left(x - \frac{\pi}{4} \right)}}{\cos{\left(x - \frac{\pi}{4} \right)}}$$

2. Apply the quotient rule, which is:

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

   $f{\left(x \right)} = \sin{\left(x - \frac{\pi}{4} \right)}$ and $g{\left(x \right)} = \cos{\left(x - \frac{\pi}{4} \right)}$.

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

   1. Let $u = x - \frac{\pi}{4}$.

   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 $\frac{d}{d x} \left(x - \frac{\pi}{4}\right)$:

      1. Differentiate $x - \frac{\pi}{4}$ term by term:

         1. Apply the power rule: $x$ goes to $1$

         2. The derivative of the constant $- \frac{\pi}{4}$ is zero.

         The result is: $1$

      The result of the chain rule is:

      $$\cos{\left(x - \frac{\pi}{4} \right)}$$

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

   1. Let $u = x - \frac{\pi}{4}$.

   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 $\frac{d}{d x} \left(x - \frac{\pi}{4}\right)$:

      1. Differentiate $x - \frac{\pi}{4}$ term by term:

         1. Apply the power rule: $x$ goes to $1$

         2. The derivative of the constant $- \frac{\pi}{4}$ is zero.

         The result is: $1$

      The result of the chain rule is:

      $$- \sin{\left(x - \frac{\pi}{4} \right)}$$

   Now plug in to the quotient rule:

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

3. Now simplify:

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

The answer is: