If \( y=\cot^{-1}\left(\displaystyle\frac{1-x}{1+x}\right) \) then \( \displaystyle\frac{dy}{dx} \) equals
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{1}{1+x^2} \)
By the chain rule, with \( \frac{d}{du}\cot^{-1}u=-\frac{1}{1+u^2} \) and \( u=\frac{1-x}{1+x} \):
\( \displaystyle \frac{du}{dx}=\frac{(1+x)(-1)-(1-x)(1)}{(1+x)^2} \)
\( \displaystyle =\frac{-2}{(1+x)^2} \)
\( \displaystyle 1+u^2=\frac{(1+x)^2+(1-x)^2}{(1+x)^2} \)
\( \displaystyle =\frac{2(1+x^2)}{(1+x)^2} \)
\( \displaystyle \frac{dy}{dx}=-\frac{1}{1+u^2}\cdot\frac{du}{dx} \)
\( \displaystyle =-\frac{(1+x)^2}{2(1+x^2)}\cdot\frac{-2}{(1+x)^2} \)
\( \displaystyle =\frac{1}{1+x^2} \)
Shortcut: Put \( x=\tan\theta \). Then
\( \displaystyle \frac{1-\tan\theta}{1+\tan\theta}=\tan\left(\frac{\pi}{4}-\theta\right) \)
\( \displaystyle =\cot\left(\frac{\pi}{4}+\theta\right) \)
\( \displaystyle y=\frac{\pi}{4}+\theta \)
\( \displaystyle =\frac{\pi}{4}+\tan^{-1}x \)
\( \displaystyle \frac{dy}{dx}=\frac{1}{1+x^2} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{-1}{1+x^2} \) — Incorrect. The derivative of \( \cot^{-1}u \) carries a minus sign, but \( \frac{du}{dx} \) is also negative; the two signs cancel, so the answer is positive.
Option (3) \( \displaystyle\frac{1}{(1+x^2)^{3/2}} \) — Incorrect. No power 3/2 can arise: \( 1+u^2 \) simplifies to \( \frac{2(1+x^2)}{(1+x)^2} \), with no square root.
Option (4) \( \displaystyle\frac{-1}{(1+x^2)^{3/2}} \) — Incorrect. The power 3/2 is wrong, and so is the sign.