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Continuity And Differentiability

Question
CBSEENMA12035446
Wired Faculty App

If space straight y equals left parenthesis tan to the power of negative 1 end exponent space straight x right parenthesis squared comma space show space that space left parenthesis straight x squared space plus space 1 right parenthesis 2 space straight y subscript 2 plus 2 space straight x left parenthesis straight x squared plus 1 right parenthesis space straight y to the power of 1 equals 2.

Solution
space space space space space space straight y equals left parenthesis tan to the power of negative 1 end exponent space straight x right parenthesis 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 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 space space space space space dy over dx equals 2 space tan to the power of negative 1 end exponent space straight x. fraction numerator 1 over denominator 1 plus straight x squared end fraction rightwards double arrow space space space space space space space open parentheses 1 plus straight x squared close parentheses dy over dx equals 2 space tan to the power of negative 1 end exponent straight x rightwards double arrow space space space space space space space open parentheses 1 plus straight x squared close parentheses open parentheses dy over dx close parentheses squared equals 4 open parentheses tan to the power of negative 1 end exponent straight x close parentheses squared rightwards double arrow space space space space space space space open parentheses 1 plus straight x squared close parentheses squared open parentheses dy over dx close parentheses squared equals 4 space 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 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 space space space space space space space space space space space open parentheses 1 plus straight x squared close parentheses squared.2 dy over dx fraction numerator straight d squared straight y over denominator dx squared end fraction plus open parentheses dy over dx close parentheses squared.2 left parenthesis 1 plus straight x squared right parenthesis.2 straight x equals 4 dy over dx Differentiating space both space sides space by space 2 dy over dx comma space we space get comma space space space space space space space space space space space space open parentheses 1 plus straight x squared close parentheses squared. fraction numerator straight d squared straight y over denominator dx squared end fraction plus 2 straight x left parenthesis 1 plus straight x squared right parenthesis dy over dx equals 2 Or space space space space space space space space left parenthesis straight x squared plus 1 right parenthesis straight y subscript 2 plus 2 straight x left parenthesis straight x squared plus 1 right parenthesis straight y subscript 1

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