Derivative of x^9+tan(x)+e^x

1. Differentiate $e^{x} + \left(x^{9} + \tan{\left(x \right)}\right)$ term by term:

   1. Differentiate $x^{9} + \tan{\left(x \right)}$ term by term:

      1. Apply the power rule: $x^{9}$ goes to $9 x^{8}$

      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: $9 x^{8} + \frac{\sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}}{\cos^{2}{\left(x \right)}}$

   2. The derivative of $e^{x}$ is itself.

   The result is: $9 x^{8} + \frac{\sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}}{\cos^{2}{\left(x \right)}} + e^{x}$

2. Now simplify:

   $$9 x^{8} + e^{x} + \frac{1}{\cos^{2}{\left(x \right)}}$$

The answer is:

$$9 x^{8} + e^{x} + \frac{1}{\cos^{2}{\left(x \right)}}$$