If {{y}} = { ^{ - 1}}≤ft( {{{2{{x}}}}{{1 + {{{x}}²}}} ), then {{{{dy}}}}{{{{dx}}}} is equal to

Correct Answer is : −21+x2- {2}{{1 + {{{x}}²}}−1+x22 for all | x |

- {2}{{1 + {{{x}}²}} for all | x |

- {2}{{1 + {{{x}}²}} for all | x | > 1

{2}{{1 + {{{x}}²}} for all | x |

Correct Answer is : −21+x2- {2}{{1 + {{{x}}²}}−1+x22 for all | x |

Test Index

UPSC NDA 2 2017 Mathematics Solved Paper

Question : 82 of 120

Marks: +1, -0

{\rm{y}} = {\cos ^{ - 1}}\left( {\frac{{2{\rm{x}}}}{{1 + {{\rm{x}}^2}}}} \right)

\frac{{{\rm{dy}}}}{{{\rm{dx}}}}

is equal to

Solution:

Concept:

Use a standard inverse trigonometric identity to simplify the given function before differentiating.

Explanation:

Given:

y = \cos^{-1}\left(\frac{2x}{1+x^2}\right)

For

|x| < 1

, use the identity

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

Thus,

y = \frac{\pi}{2} - 2\tan^{-1}x

Differentiate with respect to

x

\frac{dy}{dx} = 0 - 2\cdot\frac{1}{1+x^2} = -\frac{2}{1+x^2}

This derivative is valid only for

|x| < 1

, where the identity holds.

Answer:

Option A:

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

for all

|x| < 1

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