IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

Xavier4

If y=sec^-1(2x / 1+x²) + sin^-1(x-1 / x+1), 0<x<1 then calculate dy/dx.

Xavier4

Someone please reply meeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

Prajju

sec-1 function is <=-1 & >=1
so let us bound it ---------
2x/(1+x^2) >=1
=> 2x > = 1 + x^2 ( since 1 + x^2 is+ve so it cud be cross multiplied)
=> 0 > = (1-x)^2 which is not possible so sec-1 { 2x / (1 + x^2 ) }will not be defined and so it wont be differentiated..........
thus,
now if we naglect sec-1 fun.......
y= { sin-1[(x - 1) / (x + 1)] }
now since x lies bet 0 to1 therefore (x - 1) / (x + 1)wud be -ve
so y = { sin-1[- (1 - x)/(1 + x )]}
and now differentiate.........
Hope you are able to understand........plz solve and tell me wether my approach is correct or not........

shinee

Prajju
can u plz tell me why in y= { sin-1[(x - 1) / (x + 1)] }, the value of x lies between 0 and 1, if we put 2, we get 1/2 which is also part of the domain

karthik2007

What are you guys doing!!! Its a straightforward sum. Put x = tan@!

Prajju

But putting x= tan@ will make the problem more complicated.......
and Shinee its given in d ques itself that 0 < x < 1 then how can we put 2.......

vnkt.swaroop

write sec invrse function as cos inverse of reciprocal of the given function.

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