If \( y=\sqrt{\displaystyle\frac{1-x}{1+x}} \), then the value of \( (1-x^{2})\displaystyle\frac{dy}{dx}+\sqrt{\displaystyle\frac{1-x}{1+x}} \) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0
Since \( y \) is a quotient of square roots, logarithmic differentiation is the easiest route:
\( \displaystyle \log y=\frac{1}{2}\left[\log(1-x)-\log(1+x)\right] \)
Differentiating both sides with respect to \( x \):
\( \displaystyle \frac{1}{y}\frac{dy}{dx}=\frac{1}{2}\left[\frac{-1}{1-x}-\frac{1}{1+x}\right] \)
\( \displaystyle =\frac{1}{2}\cdot\frac{-(1+x)-(1-x)}{1-x^{2}} \)
\( \displaystyle =-\frac{1}{1-x^{2}} \)
This gives \( (1-x^{2})\frac{dy}{dx}=-y \).
The square root in the given expression is \( y \) itself, so the expression equals \( -y+y=0 \).