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Continuity And Differentiability

Question

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$Find space dy over dx space when space sin left parenthesis straight x space straight y right parenthesis plus straight x over straight y equals straight x squared minus straight y.$

Solution

space space space space space sin left parenthesis straight x space straight y right parenthesis plus straight x over straight y equals straight x squared minus straight y. Differentiating space both space sides space straight w. straight r. straight t. space straight x comma cos left parenthesis straight x space straight y right parenthesis. straight d over dx left parenthesis straight x space straight y right parenthesis plus fraction numerator straight y begin display style straight d over dx end style left parenthesis straight x right parenthesis minus straight x begin display style straight d over dx end style left parenthesis straight y right parenthesis over denominator straight y squared end fraction equals 2 straight x minus dy over dx therefore space cos left parenthesis straight x space straight y right parenthesis cross times open square brackets straight x dy over dx plus straight y.1 close square brackets plus fraction numerator straight y.1 minus straight x begin display style dy over dx end style over denominator straight y squared end fraction equals 2 straight x minus dy over dx therefore space straight x space straight y squared cos left parenthesis straight x space straight y right parenthesis dy over dx plus straight y cubed cos left parenthesis straight x space straight y right parenthesis plus straight y minus straight x dy over dx equals 2 straight x space straight y squared minus straight y squared dy over dx therefore space left square bracket straight x space straight y squared cos left parenthesis straight x space straight y right parenthesis minus straight x plus straight y squared right square bracket dy over dx equals 2 space straight x space straight y squared minus straight y cubed space cos left parenthesis straight x space straight y right parenthesis minus straight y therefore space dy over dx equals fraction numerator straight y left square bracket 2 space straight x space straight y minus straight y squared space cos left parenthesis straight x space straight y right parenthesis minus 1 right square bracket over denominator straight x space straight y squared cos left parenthesis straight x space straight y right parenthesis minus straight x plus straight y squared end fraction.

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