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Continuity And Differentiability

Question
CBSEENMA12034813
Wired Faculty App

If space straight x square root of 1 plus straight y end root plus straight y square root of 1 plus straight x end root equals 0 comma space show space that space dy over dx equals negative left parenthesis 1 plus straight x right parenthesis to the power of negative 2 end exponent

Solution
because space straight x square root of 1 plus straight y end root plus straight y square root of 1 plus straight x end root equals 0 space space space space rightwards double arrow space space straight x square root of 1 plus straight y end root equals negative straight y square root of 1 plus straight x end root rightwards double arrow space straight x squared left parenthesis 1 plus straight y right parenthesis equals straight y squared left parenthesis 1 plus straight x right parenthesis space space space space space space space space space space rightwards double arrow space space straight x squared plus straight x squared straight y equals straight y squared plus straight y squared straight x rightwards double arrow space left parenthesis straight x squared minus straight y squared right parenthesis plus left parenthesis straight x squared straight y minus xy squared right parenthesis equals 0 space rightwards double arrow space left parenthesis straight x minus straight y right parenthesis left parenthesis straight x plus straight y right parenthesis plus xy left parenthesis straight x minus straight y right parenthesis equals 0 rightwards double arrow space straight x plus straight y plus xy equals 0 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 straight y not equal to straight x right square bracket rightwards double arrow space straight y plus xy equals negative straight x space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space straight y left parenthesis 1 plus straight x right parenthesis equals negative straight x rightwards double arrow space straight y equals negative fraction numerator straight x over denominator 1 plus straight x end fraction therefore space dy over dx equals negative open square brackets fraction numerator left parenthesis 1 plus straight x right parenthesis 1 minus straight x.1 over denominator left parenthesis 1 plus straight x right parenthesis squared end fraction close square brackets equals negative fraction numerator 1 over denominator left parenthesis 1 plus straight x right parenthesis squared end fraction equals negative left parenthesis 1 plus straight x right parenthesis to the power of negative 2 end exponent

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