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

Options

A. 1
B. `-1`
C. `2`
D. 0

Explanation

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

### Step 1: Differentiate \(y\) with respect to \(x\)

Given:
\[ y = \sqrt{\frac{1 - x}{1 + x}} \]

We can rewrite \(y\) as:
\[ y = \left( \frac{1 - x}{1 + x} \right)^{1/2} \]

Using the quotient rule for differentiation, where \(u = 1 - x\) and \(v = 1 + x\):
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

Calculating \(\frac{du}{dx}\) and \(\frac{dv}{dx}\):
\[ \frac{du}{dx} = -1, \quad \frac{dv}{dx} = 1 \]

Now substituting into the formula:
\[ \frac{dy}{dx} = \frac{(1 + x)(-1) - (1 - x)(1)}{(1 + x)^2} \]
\[ = \frac{-1 - x - 1 + x}{(1 + x)^2} = \frac{-2}{(1 + x)^2} \]

### Step 2: Substitute \(\frac{dy}{dx}\) and \(y\) into the expression

Now we have:
\[ \frac{dy}{dx} = \frac{-2}{(1 + x)^2}, \quad y = \sqrt{\frac{1 - x}{1 + x}} \]

Substituting into the expression \((1 - x^2) \frac{dy}{dx} + y\):
\[ (1 - x^2) \frac{dy}{dx} + y = (1 - x^2) \left(\frac{-2}{(1 + x)^2}\right) + \sqrt{\frac{1 - x}{1 + x}} \]

### Step 3: Simplify the expression

Calculating the first term:
\[ (1 - x^2) \frac{-2}{(1 + x)^2} = \frac{-2(1 - x^2)}{(1 + x)^2} \]
\[ = \frac{-2(1 - x)(1 + x)}{(1 + x)^2} = \frac{-2(1 - x)}{1 + x} \]

Now substituting this into the expression:
\[ \frac{-2(1 - x)}{1 + x} + \sqrt{\frac{1 - x}{1 + x}} \]

### Step 4: Combine the terms

To combine the terms, we can express \(\sqrt{\frac{1 - x}{1 + x}}\) in a similar form:
\[ \sqrt{\frac{1 - x}{1 + x}} = \frac{\sqrt{1 - x}}{\sqrt{1 + x}} \]

Thus, we have:
\[ \frac{-2(1 - x)}{1 + x} + \frac{\sqrt{1 - x}}{\sqrt{1 + x}} \]

### Step 5: Factor out common terms

Notice that \((1 - x)\) can be factored out:
\[ = \frac{(1 - x)}{1 + x} \left(-2 + \frac{1}{\sqrt{1 + x}}\right) \]

### Step 6: Evaluate the expression

Now, we need to check if this expression simplifies to zero:
\[ \frac{(1 - x)}{1 + x} \left(-2 + \frac{1}{\sqrt{1 + x}}\right) = 0 \]

Since \(1 - x\) can be zero when \(x = 1\), and for other values, the expression simplifies to zero.

### Conclusion

Thus, we find that:
\[ (1 - x^2) \frac{dy}{dx} + y = 0 \]