Derivative of y=-2x+tgx+2

1. Differentiate $\left(- 2 x + \tan{\left(x \right)}\right) + 2$ term by term:

   1. Differentiate $- 2 x + \tan{\left(x \right)}$ 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. Rewrite the function to be differentiated:

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

      3. Apply the quotient rule, which is:

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

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

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

         1. The derivative of sine is cosine:

            $$\frac{d}{d x} \sin{\left(x \right)} = \cos{\left(x \right)}$$

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

         1. The derivative of cosine is negative sine:

            $$\frac{d}{d x} \cos{\left(x \right)} = - \sin{\left(x \right)}$$

         Now plug in to the quotient rule:

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

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

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

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

2. Now simplify:

   $$\tan^{2}{\left(x \right)} - 1$$

The answer is:

$$\tan^{2}{\left(x \right)} - 1$$