If `e^(y)=(sqrt(1+x)+sqrt(1-x))/(sqrt(1+x)-sqrt(1-x))," then "(dy)/(dx)=`

Options

A. `(1)/(xsqrt(1-x^(2)))`
B. `(1)/(sqrt(1-x^(2)))`
C. `(-1)/(xsqrt(1-x^(2)))`
D. `(-1)/(sqrt(1-x^(2)))`

Answer

To solve the problem \( e^{y} = \frac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} - \sqrt{1-x}} \) and find \( \frac{dy}{dx} \), we will follow these steps:

### Step 1: Simplify the Expression
We start with the given equation:
\[ e^{y} = \frac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} - \sqrt{1-x}} \]

To simplify, we can multiply the numerator and denominator by the conjugate of the denominator:
\[ e^{y} = \frac{(\sqrt{1+x} + \sqrt{1-x})^2}{(\sqrt{1+x})^2 - (\sqrt{1-x})^2} \]

### Step 2: Expand the Numerator and Denominator
Calculating the numerator:
\[ (\sqrt{1+x} + \sqrt{1-x})^2 = (1+x) + (1-x) + 2\sqrt{(1+x)(1-x)} = 2 + 2\sqrt{1-x^2} \]

Calculating the denominator:
\[ (\sqrt{1+x})^2 - (\sqrt{1-x})^2 = (1+x) - (1-x) = 2x \]

Thus, we have:
\[ e^{y} = \frac{2 + 2\sqrt{1-x^2}}{2x} \]
This simplifies to:
\[ e^{y} = \frac{1 + \sqrt{1-x^2}}{x} \]

### Step 3: Differentiate Both Sides
Now we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(e^{y}) = \frac{d}{dx}\left(\frac{1 + \sqrt{1-x^2}}{x}\right) \]

Using the chain rule on the left side:
\[ e^{y} \frac{dy}{dx} \]

For the right side, we will use the quotient rule:
\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \]
where \( u = 1 + \sqrt{1-x^2} \) and \( v = x \).

Calculating \( \frac{du}{dx} \):
\[ \frac{du}{dx} = 0 + \frac{1}{2\sqrt{1-x^2}}(-2x) = -\frac{x}{\sqrt{1-x^2}} \]

Calculating \( \frac{dv}{dx} \):
\[ \frac{dv}{dx} = 1 \]

Applying the quotient rule:
\[ \frac{d}{dx}\left(\frac{1 + \sqrt{1-x^2}}{x}\right) = \frac{x\left(-\frac{x}{\sqrt{1-x^2}}\right) - (1 + \sqrt{1-x^2})}{x^2} \]
This simplifies to:
\[ \frac{-\frac{x^2}{\sqrt{1-x^2}} - (1 + \sqrt{1-x^2})}{x^2} \]

### Step 4: Set the Derivatives Equal
Setting the derivatives equal gives us:
\[ e^{y} \frac{dy}{dx} = \frac{-\frac{x^2}{\sqrt{1-x^2}} - (1 + \sqrt{1-x^2})}{x^2} \]

### Step 5: Solve for \( \frac{dy}{dx} \)
Now, isolate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{-\frac{x^2}{\sqrt{1-x^2}} - (1 + \sqrt{1-x^2})}{x^2 e^{y}} \]

### Step 6: Substitute \( e^{y} \)
Recall that \( e^{y} = \frac{1 + \sqrt{1-x^2}}{x} \):
\[ \frac{dy}{dx} = \frac{-\frac{x^2}{\sqrt{1-x^2}} - (1 + \sqrt{1-x^2})}{x^2 \cdot \frac{1 + \sqrt{1-x^2}}{x}} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{-\frac{x^2}{\sqrt{1-x^2}} - (1 + \sqrt{1-x^2})}{x(1 + \sqrt{1-x^2})} \]

### Final Result
Thus, the final expression for \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = -\frac{1}{x \sqrt{1-x^2}} \]