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Continuity And Differentiability

Question
CBSEENMA12034833

If space straight y equals straight a to the power of straight a to the power of straight a to the power of straight a........... infinity end exponent end exponent end exponent comma prove space that space dy over dx equals fraction numerator straight y squared log space straight y over denominator straight x left parenthesis 1 minus straight y space log space straight x. log space straight y right parenthesis end fraction

Solution
Here space straight y equals straight a to the power of straight a to the power of straight a to the power of straight a........... infinity end exponent end exponent end exponent
therefore space straight y equals straight a to the power of left parenthesis straight a to the power of straight x right parenthesis end exponent
therefore space log space straight y equals log space straight a to the power of left parenthesis straight x to the power of straight x right parenthesis end exponent space space space space space space space space rightwards double arrow space log space straight y equals straight x to the power of straight y log space straight a
therefore space log left parenthesis log space straight y right parenthesis equals log left parenthesis straight x to the power of straight y log space straight a right parenthesis space rightwards double arrow space log left parenthesis log space straight y right parenthesis equals log space straight x to the power of straight y plus log left parenthesis log space straight a right parenthesis
therefore space log left parenthesis log space straight y right parenthesis equals straight y space log space straight x plus log left parenthesis log space straight a right parenthesis
differentiating space straight w. straight r. straight t. space straight x comma space we space get comma
space space space space fraction numerator 1 over denominator log space straight y end fraction.1 over straight y dy over dx equals straight y.1 over straight x plus log space straight a. dy over dx plus 0
therefore space open parentheses fraction numerator 1 over denominator straight y space log space straight y end fraction minus log space straight x close parentheses dy over dx equals straight y over straight x
therefore space open parentheses fraction numerator 1 minus straight y space log space xlog space straight y over denominator straight y space log space straight y end fraction close parentheses dy over dx equals straight y over straight x
therefore space dy over dx equals fraction numerator straight y squared log space straight y over denominator straight x left parenthesis 1 minus straight y space log space straight y space log space straight x right parenthesis end fraction

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