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Continuity And Differentiability

Question
CBSEENMA12034865
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Differentiate the following w.r.t. x :(log x)log x + (1 + x)2x

Solution
Let space space straight y equals left parenthesis log space straight x right parenthesis to the power of log space straight x end exponent plus left parenthesis 1 plus straight x right parenthesis to the power of 2 straight x end exponent Put space left parenthesis log space straight x right parenthesis to the power of log space straight x end exponent equals straight u comma space left parenthesis 1 plus straight x right parenthesis to the power of 2 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 space space space space space space space space space space space... left parenthesis 1 right parenthesis Now space straight u equals left parenthesis log space straight x right parenthesis to the power of log space straight x end exponent therefore space log space straight u equals log left parenthesis log space straight x right parenthesis to the power of log space straight x end exponent equals log space straight x. log left parenthesis log space straight x right parenthesis therefore space 1 over straight u du over dx equals log space straight x. fraction numerator 1 over denominator log space straight x end fraction.1 over straight x plus log left parenthesis log space straight x right parenthesis.1 over straight x therefore space du over dx equals straight u open square brackets 1 over straight x plus fraction numerator log left parenthesis log space straight x right parenthesis over denominator straight x end fraction close square brackets therefore space du over dx equals left parenthesis log space straight x right parenthesis to the power of log space straight x end exponent open square brackets 1 over straight x plus fraction numerator log left parenthesis log space 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 space space space space space space space space... left parenthesis 2 right parenthesis also space straight v equals left parenthesis 1 plus straight x right parenthesis to the power of 2 straight x end exponent therefore space log space straight v equals log left parenthesis 1 plus straight x right parenthesis to the power of 2 straight x end exponent equals 2 left square bracket straight x. log left parenthesis 1 plus straight x right parenthesis right square bracket therefore space 1 over straight u du over dx equals 2 open square brackets straight x. fraction numerator 1 over denominator 1 plus straight x end fraction plus log left parenthesis 1 plus straight x right parenthesis.1 close square brackets therefore space du over dx equals 2 left parenthesis 1 plus straight x right parenthesis to the power of 2 straight x end exponent open square brackets fraction numerator straight x over denominator 1 plus straight x end fraction plus log left parenthesis 1 plus straight x right parenthesis 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... left parenthesis 3 right parenthesis

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