Derivative of tan(e^x)

1. Rewrite the function to be differentiated:

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

2. Apply the quotient rule, which is:

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

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

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

   1. Let $u = e^{x}$.

   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 $\frac{d}{d x} e^{x}$:

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

      The result of the chain rule is:

      $$e^{x} \cos{\left(e^{x} \right)}$$

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

   1. Let $u = e^{x}$.

   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 $\frac{d}{d x} e^{x}$:

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

      The result of the chain rule is:

      $$- e^{x} \sin{\left(e^{x} \right)}$$

   Now plug in to the quotient rule:

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

3. Now simplify:

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

The answer is:

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