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Continuity And Differentiability

Question
CBSEENMA12034807
Wired Faculty App

If space straight y square root of straight x squared plus 1 end root equals log open parentheses straight x plus square root of straight x squared plus 1 end root close parentheses comma space show space that space left parenthesis straight x squared plus 1 right parenthesis dy over dx plus xy minus 1 equals 0

Solution
Here space straight y square root of straight x squared plus 1 end root equals log open parentheses straight x plus square root of straight x squared plus 1 end root close parentheses Differentiating space both space sides space straight w. straight r. straight t. straight x comma space we space get comma straight y. fraction numerator 2 straight x over denominator 2 square root of straight x squared plus 1 end root end fraction plus square root of straight x squared plus 1 end root dy over dx equals fraction numerator 1 over denominator straight x plus square root of straight x squared plus 1 end root end fraction cross times open square brackets 1 plus fraction numerator 2 straight x over denominator 2 square root of straight x squared plus 1 end root end fraction close square brackets 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 open square brackets because straight d over dx left square bracket space left curly bracket space straight f left parenthesis straight x right parenthesis space right curly bracket to the power of straight n space right square bracket equals straight n left curly bracket space left square bracket space straight f left parenthesis straight x right parenthesis space right square bracket space right curly bracket to the power of straight n minus 1 end exponent space straight f apostrophe left parenthesis straight x right parenthesis close square brackets therefore fraction numerator xy over denominator square root of straight x squared plus 1 end root end fraction plus square root of straight x squared plus 1 end root dy over dx equals fraction numerator 1 over denominator straight x plus square root of straight x squared plus 1 end root end fraction cross times fraction numerator square root of straight x squared plus 1 end root plus 1 over denominator square root of straight x squared plus 1 end root end fraction therefore fraction numerator xy over denominator square root of straight x squared plus 1 end root end fraction plus square root of straight x squared plus 1 end root dy over dx equals fraction numerator 1 over denominator square root of straight x squared plus 1 end root end fraction comma Multiplying space both space sides space by space square root of straight x squared plus 1 end root comma space space space space xy plus left parenthesis straight x squared plus 1 right parenthesis dy over dx equals 1 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 or space left parenthesis straight x squared plus 1 right parenthesis dy over dx plus xy minus 1 equals 0

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