Advertisements
Advertisements
प्रश्न
If y = `cos^-1 ((x + 1)/(x - 1)) + "cosec"^-1 ((x - 1)/(x + 1))`, then `("d"y)/("d"x)` is ______.
पर्याय
1
`(x - 1)/(x + 1)`
`(x + 1)/(x - 1)`
0
MCQ
रिकाम्या जागा भरा
Advertisements
उत्तर
If y = `cos^-1 ((x + 1)/(x - 1)) + "cosec"^-1 ((x - 1)/(x + 1))`, then `("d"y)/("d"x)` is 0.
Explanation:
y = `cos^-1 ((x + 1)/(x - 1)) + "cosec"^-1 ((x - 1)/(x + 1))`
= `cos^-1 ((x + 1)/(x - 1)) + sin^-1 ((x + 1)/(x- 1))`
∴ y = `pi/2` ......`[because sin^-1x + cos^-1x = pi/2]`
∴ `("d"y)/("d"x)` = 0
shaalaa.com
Derivative of Inverse Functions
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
