The derivation of $\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$ with respect to $\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)$ at $x=\frac{1}{2}$ is:

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Differentiation

The derivation of

tan^{- 1} (\frac{1 + x ^{2} - 1}{x})

with respect to

tan^{- 1} (\frac{2 x 1 - x ^{2}}{1 - 2 x ^{2}})

x = \frac{1}{2}

is:

Answer: (B)

Put

x = tan θ

in the first function:

tan^{- 1} \frac{s e c θ - 1}{t a n θ} = \frac{θ}{2} = \frac{1}{2} tan^{- 1} x

, whose derivative is

\frac{1}{2 ( 1 + x ^{2} )}

. Put

x = sin ϕ

in the second:

tan^{- 1} \frac{2 x 1 - x ^{2}}{1 - 2 x ^{2}}

becomes

2 ϕ = 2 sin^{- 1} x

, whose derivative is

\frac{2}{1 - x ^{2}}

. Dividing, the derivative of the first with respect to the second is

\frac{1 - x ^{2}}{4 ( 1 + x ^{2} )}

. At

x = \frac{1}{2}

this is

\frac{3 /2}{4 \times 5/4} = \frac{3}{10}

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Single correct option