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Continuity And Differentiability

Question
CBSEENMA12035429
Wired Faculty App

If space straight y equals log open parentheses straight x plus square root of 1 plus straight x squared end root close parentheses comma space show space that space left parenthesis 1 plus straight x squared right parenthesis fraction numerator straight d squared straight y over denominator dx squared end fraction plus straight x dy over dx equals 0.

Solution
space space space space space space space space space space space space space space straight y equals log open parentheses straight x plus square root of 1 plus straight x squared end root close parentheses therefore space space space space space dy over dx equals fraction numerator 1 over denominator straight x plus square root of 1 plus straight x squared end root end fraction cross times straight d over dx open parentheses straight x plus square root of 1 plus straight x squared end root close parentheses rightwards double arrow space space space space space dy over dx equals fraction numerator 1 over denominator straight x plus square root of 1 plus straight x squared end root end fraction cross times open parentheses 1 plus fraction numerator 2 space straight x over denominator 2 square root of 1 plus straight x squared end root end fraction close parentheses rightwards double arrow space space space space space dy over dx equals fraction numerator 1 over denominator straight x plus square root of 1 plus straight x squared end root end fraction cross times fraction numerator square root of 1 plus straight x squared end root plus straight x over denominator square root of 1 plus straight x squared end root end fraction rightwards double arrow space space space space space dy over dx equals fraction numerator 1 over denominator square root of 1 plus straight x squared end root end fraction space space space space space space space rightwards double arrow space square root of 1 plus straight x squared end root. dy over dx equals 1 rightwards double arrow space space space space space open parentheses 1 straight x plus squared close parentheses. open parentheses dy over dx close parentheses squared equals 1
Differentiating space both space sides space straight w. straight r. straight t. straight x comma space space space space space space space space left parenthesis 1 plus straight x squared right parenthesis.2 space 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 straight x equals 0 Dividing space both space sides space by space 2 space dy over dx comma space space space space space space space space left parenthesis 1 plus straight x squared right parenthesis fraction numerator straight d squared straight y over denominator dx squared end fraction plus straight x dy over dx equals 0

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