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Continuity And Differentiability

Question
CBSEENMA12034832

If space straight y equals straight e to the power of straight x plus straight e to the power of straight x plus straight e to the power of straight x plus........ infinity end exponent end exponent comma space show space that space dy over dx equals fraction numerator straight y over denominator 1 minus straight y end fraction.

Solution
because space straight y equals straight e to the power of straight x plus straight e to the power of straight x plus straight e to the power of straight x plus........ infinity end exponent end exponent
therefore space straight y equals straight e to the power of straight x plus straight y end exponent
rightwards double arrow space log space straight y equals log space straight e to the power of straight x plus straight y end exponent space space space space space space space space space space space rightwards double arrow space log space straight y equals left parenthesis straight x plus straight y right parenthesis log space straight e
rightwards double arrow space log space straight y equals straight x plus straight y 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 space space space space space space space space space space space space space space space space left square bracket because space log space straight e equals 1 right square bracket
Differentiating space both space sides space straight w. straight r. straight t. space straight x comma
space 1 over straight y dy over dx equals 1 plus dy over dx space rightwards double arrow space open parentheses 1 over straight y minus 1 close parentheses dy over dx equals 1 space space space rightwards double arrow open parentheses fraction numerator 1 minus straight y over denominator straight y end fraction close parentheses dy over dx equals 1
rightwards double arrow space space dy over dx equals fraction numerator straight y over denominator 1 minus straight y end fraction

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