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UPSC NDA I 2015 Mathematics Paper

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Question : 65 of 120

Marks: +1, -0

What is the derivative of

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

with respect to

\tan^{-1} x

?

0
½
1
x

Solution:

Concept:

Use substitution

x = \tan \theta

and standard trigonometric identities to simplify the given expression before differentiating.

Explanation:

Let

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

and

v = \tan^{-1} x

.

Substitute

x = \tan \theta

, so

\theta = \tan^{-1} x

and

\sqrt{1+x^2} = \sqrt{1+\tan^2 \theta} = \sec \theta

.

Then

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

.

Use identities

1 - \cos \theta = 2\sin^2\frac{\theta}{2}

and

\sin \theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}

.

Thus

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

.

Therefore

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

.

Differentiate:

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

and

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

.

Using

\frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}}

, we get

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

.

Answer:

\frac{1}{2}

(Option B).

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