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Continuity And Differentiability

Question
CBSEENMA12034840

Find space dy over dx when space xy equals straight e to the power of straight x minus straight y end exponent.

Solution
Here space space space space space space space space space space space space space xy equals straight e to the power of straight x minus straight y end exponent
therefore space space space space space space space space space log space left parenthesis xy right parenthesis equals log space straight e to the power of straight x minus straight y end exponent
therefore space log space straight x plus log space straight y equals left parenthesis straight x minus straight y right parenthesis log space straight e
therefore space log space straight x plus log space straight y equals straight x minus straight y
Differentiating space straight w. straight r. straight t. straight x comma space we space get comma
space space space space space space 1 over straight x plus 1 over straight y dy over dx equals 1 minus dy over dx
therefore space open parentheses 1 over straight y plus 1 close parentheses dy over dx equals 1 minus 1 over straight x space space space space space space space space space space rightwards double arrow fraction numerator 1 plus straight y over denominator straight y end fraction dy over dx equals fraction numerator straight x minus 1 over denominator straight x end fraction
therefore space space space space space space space space space space space space space space space space space dy over dx equals fraction numerator straight y left parenthesis straight x minus 1 right parenthesis over denominator straight x left parenthesis 1 plus straight y right parenthesis end fraction

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