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Find out differentiation coefficient \(\rm \tan^{-1}{2x\over 1-x^2}\) with respect to \(\rm \sin^{-1}{2x\over 1+x^2}\)
1. 1
2. 1 + x2
3. \(\rm 1\over 1+x^2\)
4. -1

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Correct Answer - Option 1 : 1

Concept:

Parametric Form:

If f(x) and g(x) are the functions in x, then 

\(\rm df(x)\over dg(x)\) = \(\rm \frac{df(x)\over dx}{dg(x)\over dx}\) 

Calculation:

Let f = \(\rm \tan^{-1}{2x\over 1-x^2}\) and g = \(\rm \sin^{-1}{2x\over 1+x^2}\)

Put x = tan θ

\(\rm {2x\over 1-x^2} = {2\tan \theta \over 1-\tan^2 \theta} = \tan 2\theta\)

\(\rm {2x\over 1+x^2} = {2\tan \theta \over 1+\tan^2 \theta} = \sin 2\theta\)

Now,

f = \(\rm \tan^{-1}{2x\over 1-x^2}\) = tan-1 tan 2θ = 2θ            (∵ tan-1 tan x = x)

g = \(\rm \sin^{-1}{2x\over 1+x^2}\) = sin-1 sin 2θ = 2θ             (∵ sin-1 sin x = x)

If x = tan θ then θ = tan-1 x

Now,

f = 2θ = 2tan-1 x

Differentiating with respect to  x, we get

\(\rm \frac {df}{dx}= {2\over1+x^2}\)

g = 2θ = 2tan-1 x

Differentiating with respect to  x, we get

\(\rm \frac {dg}{dx}= {2\over1+x^2}\)

D = \(\rm {f'\over g'} = \frac{2\over 1+x^2}{2\over 1+x^2}\)

D = 1

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