`(e^x+e^-x)/(e^x-e^(-x))`

Answer

To differentiate the function \( y = \frac{e^x + e^{-x}}{e^x - e^{-x}} \), we will use the quotient rule. The quotient rule states that if you have a function in the form \( \frac{u}{v} \), then its derivative is given by:

\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

### Step 1: Identify \( u \) and \( v \)

Let:
- \( u = e^x + e^{-x} \)
- \( v = e^x - e^{-x} \)

### Step 2: Differentiate \( u \) and \( v \)

Now we need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \).

1. **Differentiate \( u \)**:
\[ \frac{du}{dx} = \frac{d}{dx}(e^x + e^{-x}) = e^x - e^{-x} \]

2. **Differentiate \( v \)**:
\[ \frac{dv}{dx} = \frac{d}{dx}(e^x - e^{-x}) = e^x + e^{-x} \]

### Step 3: Apply the Quotient Rule

Now, we can apply the quotient rule:

\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

Substituting \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \):

\[ \frac{dy}{dx} = \frac{(e^x - e^{-x})(e^x - e^{-x}) - (e^x + e^{-x})(e^x + e^{-x})}{(e^x - e^{-x})^2} \]

### Step 4: Simplify the Numerator

Let's simplify the numerator:

1. **Expand the first term**:
\[ (e^x - e^{-x})^2 = e^{2x} - 2 + e^{-2x} \]

2. **Expand the second term**:
\[ (e^x + e^{-x})^2 = e^{2x} + 2 + e^{-2x} \]

Now substituting back into the numerator:

\[ e^{2x} - 2 + e^{-2x} - (e^{2x} + 2 + e^{-2x}) = e^{2x} - 2 + e^{-2x} - e^{2x} - 2 - e^{-2x} \]

This simplifies to:

\[ -4 \]

### Step 5: Write the Final Derivative

Thus, we have:

\[ \frac{dy}{dx} = \frac{-4}{(e^x - e^{-x})^2} \]

### Final Answer

The derivative of \( y = \frac{e^x + e^{-x}}{e^x - e^{-x}} \) is:

\[ \frac{dy}{dx} = \frac{-4}{(e^x - e^{-x})^2} \]