If` y = (e^(x)+1)/(e^(x)-1), " then" (y^(2))/2 + (dy)/(dx) ` is equal to

Options

A. 1
B. `-1`
C. `-1/2`
D. `1/2`

Answer

To solve the problem, we need to find the expression for \( \frac{y^2}{2} + \frac{dy}{dx} \) given that \( y = \frac{e^x + 1}{e^x - 1} \).

### Step 1: Find \( y^2 \)
We start with the expression for \( y \):
\[ y = \frac{e^x + 1}{e^x - 1} \]
Now, we square \( y \):
\[ y^2 = \left(\frac{e^x + 1}{e^x - 1}\right)^2 = \frac{(e^x + 1)^2}{(e^x - 1)^2} \]

### Step 2: Calculate \( \frac{y^2}{2} \)
Now, we divide \( y^2 \) by 2:
\[ \frac{y^2}{2} = \frac{1}{2} \cdot \frac{(e^x + 1)^2}{(e^x - 1)^2} = \frac{(e^x + 1)^2}{2(e^x - 1)^2} \]

### Step 3: Find \( \frac{dy}{dx} \)
To find \( \frac{dy}{dx} \), we differentiate \( y \) using the quotient rule:
\[ \frac{dy}{dx} = \frac{(e^x - 1)(e^x) - (e^x + 1)(e^x)}{(e^x - 1)^2} \]
Simplifying the numerator:
\[ = \frac{e^{2x} - e^x - e^{2x} - e^x}{(e^x - 1)^2} = \frac{-2e^x}{(e^x - 1)^2} \]

### Step 4: Combine \( \frac{y^2}{2} \) and \( \frac{dy}{dx} \)
Now we add \( \frac{y^2}{2} \) and \( \frac{dy}{dx} \):
\[ \frac{y^2}{2} + \frac{dy}{dx} = \frac{(e^x + 1)^2}{2(e^x - 1)^2} - \frac{2e^x}{(e^x - 1)^2} \]
To combine these fractions, we need a common denominator:
\[ = \frac{(e^x + 1)^2 - 4e^x}{2(e^x - 1)^2} \]

### Step 5: Expand and simplify the numerator
Now we expand the numerator:
\[ (e^x + 1)^2 = e^{2x} + 2e^x + 1 \]
So,
\[ (e^x + 1)^2 - 4e^x = e^{2x} + 2e^x + 1 - 4e^x = e^{2x} - 2e^x + 1 \]
Thus, we have:
\[ \frac{y^2}{2} + \frac{dy}{dx} = \frac{e^{2x} - 2e^x + 1}{2(e^x - 1)^2} \]

### Step 6: Factor the numerator
Notice that the numerator can be factored:
\[ e^{2x} - 2e^x + 1 = (e^x - 1)^2 \]
Thus, we have:
\[ \frac{y^2}{2} + \frac{dy}{dx} = \frac{(e^x - 1)^2}{2(e^x - 1)^2} \]
This simplifies to:
\[ \frac{1}{2} \]

### Final Answer
Therefore, the final result is:
\[ \frac{y^2}{2} + \frac{dy}{dx} = \frac{1}{2} \]