Derivative of y=tgx+(6^x+2x^7+8x^2/3)

1. Differentiate $6^{x} + 2 x^{7} + \frac{8 x^{2}}{3} + \tan{\left(x \right)}$ 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. $\frac{d}{d x} 6^{x} = 6^{x} \log{\left(6 \right)}$

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

      1. Apply the power rule: $x^{7}$ goes to $7 x^{6}$

      So, the result is: $14 x^{6}$

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

      1. Apply the power rule: $x^{2}$ goes to $2 x$

      So, the result is: $\frac{16 x}{3}$

   The result is: $6^{x} \log{\left(6 \right)} + 14 x^{6} + \frac{16 x}{3} + \frac{\sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}}{\cos^{2}{\left(x \right)}}$

2. Now simplify:

   $$6^{x} \log{\left(6 \right)} + 14 x^{6} + \frac{16 x}{3} + \frac{1}{\cos^{2}{\left(x \right)}}$$

The answer is:

$$6^{x} \log{\left(6 \right)} + 14 x^{6} + \frac{16 x}{3} + \frac{1}{\cos^{2}{\left(x \right)}}$$