If y=cos−1(x−x−1x+x−1) then dydx=

y=cos−1(x−1xx+1x) ⇒ y=cos−1(x2−1x2+1) ⇒ y=cos−1(−(1−x2)1+x2) ( cos−1(−x)=π−cos−1(x)) ⇒y=π−cos−1(1−x21+x2) 2tan−1x=cos−1(1−x21+x2) ⇒y=π−2tan−1x ⇒dydx=−21+x2

If y=cos−1(x−x−1x+x−1) then dydx=

y=cos−1(x−1xx+1x) ⇒ y=cos−1(x2−1x2+1) ⇒ y=cos−1(−(1−x2)1+x2) ( cos−1(−x)=π−cos−1(x)) ⇒y=π−cos−1(1−x21+x2) 2tan−1x=cos−1(1−x21+x2) ⇒y=π−2tan−1x ⇒dydx=−21+x2

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If $y = \cos^{- 1} (\frac{x - x^{- 1}}{x + x^{- 1}})$ then $\frac{d y}{d x} =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

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

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

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

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

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$y = \cos^{- 1} (\frac{x - \frac{1}{x}}{x + \frac{1}{x}}) \Rightarrow y = \cos^{- 1} (\frac{x^{2} - 1}{x^{2} + 1}) \Rightarrow y = \cos^{- 1} (\frac{- (1 - x^{2})}{1 + x^{2}}) (\cos^{- 1} (- x) = π - \cos^{- 1} (x)) \Rightarrow y = π - \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) (2 \tan^{- 1} x = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})) \Rightarrow y = π - 2 \tan^{- 1} x \Rightarrow \frac{d y}{d x} = \frac{- 2}{1 + x^{2}}$