Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Stuck here, help me understand: - Limit , continuity and differentiability - JEE Main-2

If   y= \sec \left ( \tan ^{-1}x \right ),then\: \: \frac{dy}{dx}\: \: at\: x= 1is equal to:

  • Option 1)

    \sqrt{2}

  • Option 2)

    \frac{1}{\sqrt{2}}

  • Option 3)

    \frac{1}{2}

  • Option 4)

    1

 

Answers (1)

best_answer

As we learnt in

Trigonometric functions -

\frac{d}{dx}(sinx)=cosx

\frac{d}{dx}(secx)=secx\:tanx

\frac{d}{dx}(cosec\:f(x))=-cosec\:\left \{ f(x) \right \}\:cot\:\left \{ f(x) \right \}\;f'(x)

-

 

 y=\sec (\tan ^{-1}x)

\frac{dy}{dx}= \sec (\tan^{-1}x)\tan\: (\tan ^{-1})\times \frac{1}{1+x^{2}}

Put x=1

  sec\left (\frac{\pi}{4}\right )\tan \left ( \frac{\pi}{4} \right )\times \frac{1}{2}

\therefore \sqrt 2 \times 1 \times \frac{1}{2}=\frac{1}{\sqrt 2}


Option 1)

\sqrt{2}

This option is incorrect.

Option 2)

\frac{1}{\sqrt{2}}

This option is correct.

Option 3)

\frac{1}{2}

This option is incorrect.

Option 4)

1

This option is incorrect.

Posted by

Plabita

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Advanced 2024: Chemical engineering cut-offs at IITs; know opening, closing ranks
anam-khan

IIT JEE Mains Result 2024 Session 2: How is overall merit list prepared?
anam-khan

JEE Mains Session 2 result 2024 soon; List of top NITs in India

Latest Question