Derivative of y=(e^x-e^-x)/2x

1. Apply the quotient rule, which is:

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

   $f{\left(x \right)} = x \left(e^{2 x} - 1\right)$ and $g{\left(x \right)} = 2 e^{x}$.

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

   1. Apply the product rule:

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

      $f{\left(x \right)} = x$; to find $\frac{d}{d x} f{\left(x \right)}$:

      1. Apply the power rule: $x$ goes to $1$

      $g{\left(x \right)} = e^{2 x} - 1$; to find $\frac{d}{d x} g{\left(x \right)}$:

      1. Differentiate $e^{2 x} - 1$ term by term:

         1. The derivative of the constant $-1$ is zero.

         2. Let $u = 2 x$.

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

         4. Then, apply the chain rule. Multiply by $\frac{d}{d x} 2 x$:

            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: $2$

            The result of the chain rule is:

            $$2 e^{2 x}$$

         The result is: $2 e^{2 x}$

      The result is: $2 x e^{2 x} + e^{2 x} - 1$

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

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

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

      So, the result is: $2 e^{x}$

   Now plug in to the quotient rule:

   $\frac{\left(- 2 x \left(e^{2 x} - 1\right) e^{x} + 2 \left(2 x e^{2 x} + e^{2 x} - 1\right) e^{x}\right) e^{- 2 x}}{4}$

2. Now simplify:

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

The answer is:

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