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Sponsor Area

Continuity And Differentiability

Question
CBSEENMA12035450

If space space straight y equals open square brackets log open parentheses straight x plus square root of straight x squared plus 1 end root close parentheses close square brackets squared comma space prove space that space left parenthesis 1 space plus space straight x squared right parenthesis straight y subscript 2 space plus space straight x space straight y subscript 1 space equals space 2.

Solution
space straight y equals open square brackets log open parentheses straight x plus square root of straight x squared plus 1 end root close parentheses close square brackets squared space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
therefore space straight y subscript 1 equals 2 space log open parentheses straight x plus square root of straight x squared plus 1 end root close parentheses. fraction numerator 1 over denominator straight x plus square root of straight x squared plus 1 end root end fraction. open parentheses 1 plus fraction numerator 2 straight x over denominator 2 square root of straight x squared plus 1 end root end fraction close parentheses
space space space space space space space space space equals 2 space log open parentheses straight x plus square root of straight x squared plus 1 end root close parentheses. fraction numerator 1 over denominator straight x plus square root of straight x squared plus 1 end root end fraction. fraction numerator straight x plus square root of straight x squared plus 1 end root over denominator square root of straight x squared plus 1 end root end fraction
therefore space straight y subscript 1 equals fraction numerator 2 over denominator square root of straight x squared plus 1 end root end fraction log open parentheses straight x plus square root of straight x squared plus 1 end root close parentheses space space rightwards double arrow space space square root of 1 plus straight x squared end root. straight y subscript 1 equals 2 space log open parentheses straight x plus square root of straight x squared plus 1 end root close parentheses
rightwards double arrow space open parentheses square root of 1 plus straight x squared end root. close parentheses straight y subscript 1 squared equals 4 open square brackets log open parentheses straight x plus square root of straight x squared plus 1 end root close parentheses close square brackets squared
rightwards double arrow space open parentheses square root of 1 plus straight x squared end root. close parentheses straight y subscript 1 squared equals 4 straight y space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket because space of space left parenthesis 1 right parenthesis right square bracket
Differentiating space both space sides space straight w. straight r. straight t. straight x comma space we space get comma space
left parenthesis 1 plus straight x squared right parenthesis. space 2 space straight y subscript 1 space straight y subscript 2 plus straight y subscript 1 squared space. space 2 space straight x space equals space 4 space straight y subscript 1 space
space Dividing space both space sides space by space 2 space straight y 1 comma space we space get comma
left parenthesis 1 plus straight x squared right parenthesis space straight y subscript 2 plus straight x space straight y subscript 1 equals 2

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