Derivative of ln(e^x+e^-x)

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

2. The derivative of $\log{\left(u \right)}$ is $\frac{1}{u}$.

3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(e^{x} + e^{- x}\right)$:

   1. Differentiate $e^{x} + e^{- x}$ term by term:

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

      2. Let $u = - x$.

      3. The derivative of $e^{u}$ is itself.

      4. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(- x\right)$:

         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: $-1$

         The result of the chain rule is:

         $$- e^{- x}$$

      The result is: $e^{x} - e^{- x}$

   The result of the chain rule is:

   $$\frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}$$

4. Now simplify:

   $$\tanh{\left(x \right)}$$

The answer is:

$$\tanh{\left(x \right)}$$