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: Use logarithmic differentiation:
Take logs of $y = \left(\dfrac{1-x}{1+x}\right)^{1/2}$: $\ln y = \dfrac{1}{2}[\ln(1-x) - \ln(1+x)]$.

Step 2: Differentiate both sides:
\[ \frac{1}{y}\frac{dy}{dx} = \frac{1}{2}\left[\frac{-1}{1-x} - \frac{1}{1+x}\right] = \frac{1}{2}\cdot\frac{-(1+x)-(1-x)}{1-x^2} = \frac{-1}{1-x^2} \]

Step 3: Solve for dy/dx:
So $\dfrac{dy}{dx} = \dfrac{-y}{1-x^2}$.

Step 4: Substitute in the expression:
$(1 - x^2)\dfrac{dy}{dx} = -y$, so $(1-x^2)\dfrac{dy}{dx} + y = -y + y = 0$.

Step 5: Match option:
The value 0 is option 3.

Final Answer:
The expression equals 0. \[ \boxed{0} \]