Question

$Find space dy over dx space when space straight y equals left parenthesis sin space straight x right parenthesis to the power of straight x plus sin to the power of negative 1 end exponent square root of straight x.$

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Solution

Here space straight y equals left parenthesis sin space straight x right parenthesis to the power of straight x plus sin to the power of negative 1 end exponent square root of straight x Put space space left parenthesis sin space straight x right parenthesis to the power of straight x equals straight u comma space sin to the power of negative 1 end exponent square root of straight x 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 Now space straight u equals left parenthesis sin space straight x right parenthesis to the power of straight x space rightwards double arrow space log space straight u equals log left parenthesis sin space straight x right parenthesis to the power of straight x rightwards double arrow space log space straight u equals straight x. space log space sin space straight x. Differentiating space both space sides space straight w. straight r. straight t. straight x. 1 over straight u du over dx equals straight x. open parentheses fraction numerator 1 over denominator sin space straight x end fraction cross times cos space straight x close parentheses plus left parenthesis log space sin space straight x right parenthesis.1 rightwards double arrow space du over dx equals straight u left square bracket straight x space cot space straight x plus log space sin space straight x right square bracket rightwards double arrow space du over dx equals left parenthesis sin space straight x right parenthesis to the power of straight x left square bracket straight x space cot space straight x plus log space sin space straight x right square bracket Also space space space space straight v equals sin to the power of negative 1 end exponent square root of straight x therefore space space dv over dx equals fraction numerator 1 over denominator square root of 1 minus open parentheses square root of straight x close parentheses squared end root end fraction cross times straight d over dx left parenthesis square root of straight x right parenthesis equals fraction numerator 1 over denominator square root of 1 minus straight x end root end fraction cross times fraction numerator 1 over denominator 2 square root of straight x end fraction therefore space space dv over dx equals fraction numerator 1 over denominator 2 square root of straight x left parenthesis 1 minus straight x right parenthesis end root end fraction 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. dy over dx equals left parenthesis sin space straight x right parenthesis to the power of straight x left square bracket straight x space cot space straight x plus log space sin space straight x right square bracket plus fraction numerator 1 over denominator 2 square root of straight x left parenthesis 1 minus straight x right parenthesis end root end fraction

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Continuity And Differentiability

$Find space dy over dx space when space straight y equals left parenthesis sin space straight x right parenthesis to the power of straight x plus sin to the power of negative 1 end exponent square root of straight x.$

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