`(d)/(dx) "" tan^(-1) ((2e^(x))/(1 - e^(2x)))= `

Options

A. `(2e^(x))/(1 - e^(2x))`
B. `(e^(x))/(1 - e^(2x))`
C. `(2e^(x))/(1 + e^(2x))`
D. `(e^(x))/(1 + e^(2x))`

Explanation

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

### Step 1: Simplify the expression inside the arctangent
We can use the identity for the tangent of a double angle:
\[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \]
Let \( e^x = \tan(\theta) \). Then, we have:
\[ \tan(\theta) = e^x \quad \text{and} \quad \tan^2(\theta) = e^{2x} \]
Thus, we can rewrite the expression:
\[ \tan^{-1} \left( \frac{2e^x}{1 - e^{2x}} \right) = \tan^{-1}(\tan(2\theta)) = 2\theta \]

### Step 2: Substitute back for \(\theta\)
Since \( \theta = \tan^{-1}(e^x) \), we have:
\[ y = 2\tan^{-1}(e^x) \]

### Step 3: Differentiate the function
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 2 \cdot \frac{d}{dx} \left( \tan^{-1}(e^x) \right) \]
Using the derivative of \( \tan^{-1}(u) \), which is \( \frac{1}{1 + u^2} \cdot \frac{du}{dx} \), we get:
\[ \frac{d}{dx} \left( \tan^{-1}(e^x) \right) = \frac{1}{1 + (e^x)^2} \cdot \frac{d}{dx}(e^x) \]
Calculating \( \frac{d}{dx}(e^x) \):
\[ \frac{d}{dx}(e^x) = e^x \]
Thus, we have:
\[ \frac{d}{dx} \left( \tan^{-1}(e^x) \right) = \frac{e^x}{1 + e^{2x}} \]

### Step 4: Combine the results
Now substituting back into our derivative:
\[ \frac{dy}{dx} = 2 \cdot \frac{e^x}{1 + e^{2x}} = \frac{2e^x}{1 + e^{2x}} \]

### Final Answer
Therefore, the derivative of \( y = \tan^{-1} \left( \frac{2e^x}{1 - e^{2x}} \right) \) is:
\[ \frac{dy}{dx} = \frac{2e^x}{1 + e^{2x}} \]

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