If \( y=\sqrt{\displaystyle\frac{1-x}{1+x}} \), then the value of \( (1-x^2)\displaystyle\frac{dy}{dx}+y \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0
Taking logarithms turns the square root and the quotient into a simple difference:
\( \displaystyle \log y=\frac{1}{2}\left[\log(1-x)-\log(1+x)\right] \)
Differentiating with respect to x,
\( \displaystyle \frac{1}{y}\frac{dy}{dx}=\frac{1}{2}\left[\frac{-1}{1-x}-\frac{1}{1+x}\right] \)
\( \displaystyle =\frac{1}{2}\cdot\frac{-(1+x)-(1-x)}{1-x^2} \)
\( \displaystyle =\frac{1}{2}\cdot\frac{-2}{1-x^2} \)
\( \displaystyle =\frac{-1}{1-x^2} \)
So
\( \displaystyle (1-x^2)\frac{dy}{dx}=-y \)
\( \displaystyle (1-x^2)\frac{dy}{dx}+y=0 \)