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Continuity And Differentiability

Question
CBSEENMA12034866

Differentiate the following w.r.t. x :
     xx + x1/x

Solution
Let space space space straight y equals straight x to the power of straight x space plus space straight x to the power of 1 divided by straight x end exponent
Put space straight x to the power of straight x equals straight u comma space space straight x to the power of 1 divided by straight x end exponent equals straight v
therefore space straight y equals straight u plus straight v
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... 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 equals straight x. log space straight x
therefore space 1 over straight u equals 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... left parenthesis 2 right parenthesis
Also space straight v equals straight x to the power of 1 over straight x end exponent
therefore space log space straight v equals log space straight x to the power of 1 over straight x end exponent equals 1 over straight x log space straight x equals straight x to the power of negative 1 end exponent. log space straight x
therefore space 1 over straight v dv over dx equals straight x to the power of negative 1 end exponent.1 over straight x plus left parenthesis log space straight x right parenthesis. left parenthesis negative straight x to the power of negative 2 end exponent right parenthesis
therefore space dv over dx equals straight x to the power of 1 over straight x end exponent open square brackets 1 over straight x squared minus fraction numerator log space straight x over denominator straight x squared 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... 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 dy 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 plus straight x to the power of 1 over straight x end exponent open square brackets 1 over straight x squared minus fraction numerator log space straight x over denominator straight x squared end fraction close square brackets

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