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Continuity And Differentiability

Question
CBSEENMA12035417
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If y=xx, prove that fraction numerator straight d squared straight y over denominator dx squared end fraction minus 1 over straight y open parentheses dy over dx close parentheses squared minus straight y over straight x equals 0.

Solution

Here y=xx
∴ log y=log xx ⇒log y=x.log x
Differentiating w.r.t.x, we get,
space 1 over straight y dy over dx equals straight x.1 over straight x plus log space straight x.1 comma space or space space dy over dx equals straight y left parenthesis 1 plus log space straight x right parenthesis space space space space space space space space space 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 fraction numerator straight d squared straight y over denominator dx squared end fraction equals straight y. dy over dx left parenthesis 1 plus log space straight x right parenthesis plus left parenthesis 1 plus log space straight x right parenthesis. dy over dx therefore space fraction numerator straight d squared straight y over denominator dx squared end fraction equals straight y.1 over straight x plus left parenthesis 1 plus log space straight x right parenthesis. straight y left parenthesis 1 plus log space straight x right parenthesis space space space space space space space space space space space space space space 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 therefore space fraction numerator straight d squared straight y over denominator dx squared end fraction equals straight y over straight x plus straight y left parenthesis 1 plus log space straight x right parenthesis squared therefore space fraction numerator straight d squared straight y over denominator dx squared end fraction equals straight y over straight x plus straight y. open parentheses 1 over straight y dy over dx close parentheses 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 left square bracket because space of space left parenthesis 1 right parenthesis right square bracket therefore space fraction numerator straight d squared straight y over denominator dx squared end fraction equals straight y over straight x plus 1 over straight y open parentheses dy over dx close parentheses squared therefore space fraction numerator straight d squared straight y over denominator dx squared end fraction equals 1 over straight y open parentheses dy over dx close parentheses squared minus straight y over straight x equals 0

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