Differentiate the following w.r.t. x :
`tan^(-1)((e^(2x)+1)/(e^(2x)-1))`

Answer

To differentiate the function \( y = \tan^{-1}\left(\frac{e^{2x}+1}{e^{2x}-1}\right) \) with respect to \( x \), we will follow these steps:

### Step 1: Differentiate using the chain rule
We start by applying the chain rule for differentiation. The derivative of \( \tan^{-1}(u) \) is given by:
\[ \frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
where \( u = \frac{e^{2x} + 1}{e^{2x} - 1} \).

### Step 2: Differentiate \( u \)
Next, we need to find \( \frac{du}{dx} \). To differentiate \( u \), we will use the quotient rule:
\[ u = \frac{f(x)}{g(x)} \quad \text{where } f(x) = e^{2x} + 1 \text{ and } g(x) = e^{2x} - 1 \]
The quotient rule states:
\[ \frac{du}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \]
Calculating \( f'(x) \) and \( g'(x) \):
\[ f'(x) = 2e^{2x}, \quad g'(x) = 2e^{2x} \]
Now substituting into the quotient rule:
\[ \frac{du}{dx} = \frac{(2e^{2x})(e^{2x} - 1) - (e^{2x} + 1)(2e^{2x})}{(e^{2x} - 1)^2} \]
Simplifying the numerator:
\[ = \frac{2e^{4x} - 2e^{2x} - 2e^{4x} - 2e^{2x}}{(e^{2x} - 1)^2} = \frac{-4e^{2x}}{(e^{2x} - 1)^2} \]

### Step 3: Substitute back into the derivative
Now we substitute \( u \) and \( \frac{du}{dx} \) back into the derivative:
\[ \frac{dy}{dx} = \frac{1}{1 + \left(\frac{e^{2x} + 1}{e^{2x} - 1}\right)^2} \cdot \frac{-4e^{2x}}{(e^{2x} - 1)^2} \]

### Step 4: Simplify the expression
Now we need to simplify \( 1 + \left(\frac{e^{2x} + 1}{e^{2x} - 1}\right)^2 \):
\[ 1 + \left(\frac{e^{2x} + 1}{e^{2x} - 1}\right)^2 = \frac{(e^{2x} - 1)^2 + (e^{2x} + 1)^2}{(e^{2x} - 1)^2} \]
Calculating the numerator:
\[ (e^{2x} - 1)^2 + (e^{2x} + 1)^2 = (e^{4x} - 2e^{2x} + 1) + (e^{4x} + 2e^{2x} + 1) = 2e^{4x} + 2 = 2(e^{4x} + 1) \]
Thus,
\[ 1 + \left(\frac{e^{2x} + 1}{e^{2x} - 1}\right)^2 = \frac{2(e^{4x} + 1)}{(e^{2x} - 1)^2} \]

### Step 5: Final expression for the derivative
Substituting this back into our derivative:
\[ \frac{dy}{dx} = \frac{-4e^{2x}}{\frac{2(e^{4x} + 1)}{(e^{2x} - 1)^2}} = \frac{-4e^{2x}(e^{2x} - 1)^2}{2(e^{4x} + 1)} = \frac{-2e^{2x}(e^{2x} - 1)^2}{e^{4x} + 1} \]

### Final Result
Thus, the derivative of \( y \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{-2e^{2x}(e^{2x} - 1)^2}{e^{4x} + 1} \]

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