Derivative of y=tgx+cosx

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

   1. Rewrite the function to be differentiated:

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

   2. Apply the quotient rule, which is:

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

      $f{\left(x \right)} = \sin{\left(x \right)}$ 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:

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

   3. The derivative of cosine is negative sine:

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

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

2. Now simplify:

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

The answer is:

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