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Differentiation

Section: Mathematics

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Question : 47 of 79

Marks: +1, -0

The derivative of

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

with respect to

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

x=\frac{1}{2}

[5 Sep 2020 Shift 2]

$\frac{\sqrt{3}}{12}$
$\frac{\sqrt{3}}{10}$
$\frac{2\sqrt{3}}{5}$
$\frac{2\sqrt{3}}{3}$

Solution:

Let

f=\tan^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)

Put

x=\tan \Rightarrow \theta = \tan^{-1} x

f=\tan^{-1}\left(\frac{\sec \theta - 1}{\tan \theta}\right)

f=\tan^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right) = \frac{\theta}{2}

f = \frac{\tan^{-1} x}{2} \Rightarrow \frac{df}{dx}

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

....(i)

Let

g = \tan^{-1}\left(\frac{2x\sqrt{1-x^{2}}}{1-2x^{2}}\right)

Put

x = \sin \theta \Rightarrow \theta = \sin^{-1} x

g = \tan^{-1}\left(\frac{2 \sin \theta \cos \theta}{1-2 \sin^{2} \theta}\right)

g = \tan^{-1}(\tan 2\theta) = 2\theta

g = 2 \sin^{-1} x

\frac{dg}{dx} = \frac{2}{\sqrt{1-x^{2}}}

....(ii)

\frac{df}{dg} = \frac{1}{2(1+x^{2})} \frac{\sqrt{1-x^{2}}}{2}

at

x = \frac{1}{2} \left( \frac{df}{dg} \right)_{x=\frac{1}{2}} = \frac{\sqrt{3}}{10}

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