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........
Moderation Team
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