JEE Main 2022 · Mathematics · Differentiation and Application of Derivatives · medium

If y=tan ^-1 ((1+x)/(1-x)), then dy/dx at x=2 is

Single correct option

y=\tan^{-1}\!\left(\dfrac{1+x}{1-x}\right)

, then

\dfrac{dy}{dx}

x=2

Show answer and solution

Answer: (C) $\frac15$

Since

\tan^{-1}\frac{1+x}{1-x}=\tan^{-1}1+\tan^{-1}x

up to a constant (valid on each branch),

\frac{dy}{dx}=\frac{1}{1+x^2}

. At

x=2

this is

\frac15

. Direct differentiation with the chain rule gives the same value after simplifying

\frac{2}{(1-x)^2}\big/\big(1+\frac{(1+x)^2}{(1-x)^2}\big)=\frac{2}{2+2x^2}

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