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Continuity And Differentiability

Question
CBSEENMA12034869
Wired Faculty App

Differentiate the following w.r.t. x 
   xx + (1 + x)log x

Solution
Let space straight y equals straight x to the power of straight x plus left parenthesis 1 space plus space straight x right parenthesis to the power of log space straight x end exponent Put space straight x to the power of straight x equals straight u comma space left parenthesis 1 space plus space straight x right parenthesis to the power of log space straight x end exponent equals straight v therefore space straight y equals straight u plus straight v space therefore space dy over dx equals du over dx plus dv over dx 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 Now space straight u equals straight x to the power of straight x therefore space log space straight u equals log space straight x to the power of straight x therefore space 1 over straight u du over dx equals straight x.1 over straight x plus log space straight x.1 therefore space du over dx equals straight x to the power of straight x left square bracket 1 plus log space straight x right square bracket 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 2 right parenthesis Also space straight v equals left parenthesis 1 space plus space straight x right parenthesis to the power of log space straight x end exponent therefore space log space straight v equals log left parenthesis 1 space plus space straight x right parenthesis to the power of log space straight x end exponent equals log left parenthesis 1 plus straight x right parenthesis therefore space 1 over straight v dv over dx equals left parenthesis log space straight x right parenthesis cross times fraction numerator 1 over denominator 1 plus straight x end fraction plus log left parenthesis 1 plus straight x right parenthesis therefore space dv over dx equals left parenthesis 1 space plus space straight x right parenthesis to the power of log space straight x end exponent open square brackets fraction numerator log space straight x over denominator 1 plus straight x end fraction plus fraction numerator log left parenthesis 1 plus straight x right parenthesis over denominator straight x 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... left parenthesis 3 right parenthesis From space left parenthesis 1 right parenthesis comma space left parenthesis 2 right parenthesis space and space left parenthesis 3 right parenthesis comma space we space get comma space space space space space dy over dx equals straight x to the power of straight x left parenthesis 1 plus log space straight x right parenthesis plus left parenthesis 1 plus straight x right parenthesis to the power of log space straight x end exponent open square brackets fraction numerator log space straight x over denominator 1 plus straight x end fraction plus fraction numerator log left parenthesis 1 plus straight x right parenthesis over denominator straight x end fraction close square brackets

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