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Continuity And Differentiability

Question

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Differentiate the following functions w.r.t. x : $straight x to the power of straight x minus 2 to the power of sin space straight x end exponent$

Solution

Let space straight y equals straight x to the power of straight x minus 2 to the power of sin space straight x end exponent Put space straight x to the power of straight x equals straight u comma space 2 to the power of sin space straight x end exponent equals straight v therefore space straight y equals straight u minus straight v therefore space dy over dx equals du over dx minus 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... left parenthesis 1 right parenthesis Now space straight u equals straight x to the power of straight x space rightwards double arrow space log space straight u equals log space straight x to the power of straight x space rightwards double arrow space log space straight u equals straight x space log space straight x therefore space 1 over straight u du over dx equals straight x.1 over straight x plus log space straight x.1 space rightwards double arrow space du over dx equals straight u left parenthesis 1 plus log space straight x right parenthesis therefore space du over dx equals straight x to the power of straight x left parenthesis 1 plus log space straight x right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 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 2 to the power of sin space straight x end exponent therefore space dv over dx equals 2 to the power of sin space straight x end exponent. log space 2. straight d over dx left parenthesis sin space straight x right parenthesis therefore space dv over dx equals 2 to the power of sin space straight x end exponent. log space 2. cos space straight x space space space space space space space space space space space space space space space space space space space space space space 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 comma space left parenthesis 3 right parenthesis comma space we space get comma dy over dx equals straight x to the power of straight x left parenthesis 1 plus log space straight x right parenthesis plus 2 to the power of sin space straight x end exponent. log space 2. space cos space straight x

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