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Continuity And Differentiability

Question
CBSEENMA12035291

Differentiate the following w.r.t.x: straight x space to the power of sin to the power of negative 1 end exponent straight x end exponent

Solution
Let space space space space space space space space space space straight y equals straight x space to the power of sin to the power of negative 1 end exponent straight x end exponent comma space therefore space log space straight y equals log space straight x space to the power of sin to the power of negative 1 end exponent straight x end exponent space equals sin to the power of negative 1 end exponent straight x. log space straight x
therefore space 1 over straight y dy over dx equals left parenthesis sin to the power of negative 1 end exponent straight x right parenthesis.1 over straight x plus fraction numerator 1 over denominator square root of 1 minus straight x squared end root end fraction log space straight x
rightwards double arrow space space space space space space dy over dx equals straight x space to the power of sin to the power of negative 1 end exponent straight x end exponent open square brackets fraction numerator sin to the power of negative 1 end exponent straight x over denominator straight x end fraction plus fraction numerator log space straight x over denominator square root of 1 minus straight x squared end root end fraction close square brackets

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