Question

If \(y = \sqrt{\dfrac{1-x}{1+x}}\), then the value of \((1 - x^2)\dfrac{dy}{dx} + y\) is

• A. \(\dfrac{1}{1+x}\)
• B. \(\dfrac{-1}{(1+x)^2}\)
• C. 0
• D. 1

Hint:

Take \(\ln y = \tfrac12[\ln(1-x) - \ln(1+x)]\) and differentiate. You will find \((1-x^2)y' = -y\).

Answer 1

The Correct Option is C

Step 1: Rewrite y:
\(y = \sqrt{\dfrac{1-x}{1+x}} = \dfrac{1-x}{\sqrt{1-x^2}}\), obtained by multiplying top and bottom by \(\sqrt{1-x}\). This form is easier to differentiate.

Step 2: Differentiate with the quotient rule:
Let \(u = 1 - x\) and \(v = (1-x^2)^{1/2}\). Then \(u' = -1\) and \(v' = \dfrac{-x}{\sqrt{1-x^2}}\).
\[ \frac{dy}{dx} = \frac{u'v - uv'}{v^2} = \frac{-\sqrt{1-x^2} + \dfrac{x(1-x)}{\sqrt{1-x^2}}}{1-x^2} \] Put the top over one denominator:
\[ \frac{dy}{dx} = \frac{-(1-x^2) + x(1-x)}{(1-x^2)^{3/2}} = \frac{-1 + x^2 + x - x^2}{(1-x^2)^{3/2}} = \frac{x - 1}{(1-x^2)^{3/2}} \]

Step 3: Multiply by \((1 - x^2)\):
\[ (1-x^2)\frac{dy}{dx} = \frac{x-1}{\sqrt{1-x^2}} = -\frac{1-x}{\sqrt{1-x^2}} = -y \]

Step 4: Add y:
\[ (1-x^2)\frac{dy}{dx} + y = -y + y = 0 \]

Step 5: Check the options:
The result is 0, not 1, and not an expression in \(x\). So option 3 is correct.

Final Answer:
The value is 0, option 3. \[ \boxed{0} \]