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Continuity And Differentiability

Question
CBSEENMA12034843

Here space If space straight y equals square root of fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction end root space
therefore space log space straight y equals log open parentheses fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction close parentheses to the power of 1 half end exponent space space space space space space space rightwards double arrow space log space straight y equals 1 half log space open parentheses fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction close parentheses
rightwards double arrow space log space straight y equals 1 half left square bracket log left parenthesis 1 minus straight x right parenthesis minus log left parenthesis 1 plus straight x right parenthesis right square bracket
Differentiating space both space sides space straight w. straight r. straight t. straight x comma space we space get comma
space space space space space 1 over straight y dy over dx equals 1 half open square brackets fraction numerator 1 over denominator 1 minus straight x end fraction left parenthesis negative 1 right parenthesis minus fraction numerator 1 over denominator 1 plus straight x end fraction close square brackets
therefore space 1 over straight y dy over dx equals negative 1 half open square brackets fraction numerator 1 over denominator 1 minus straight x end fraction plus fraction numerator 1 over denominator 1 plus straight x end fraction close square brackets
rightwards double arrow space 1 over straight y dy over dx equals negative 1 half open square brackets fraction numerator 1 plus straight x plus negative straight x over denominator left parenthesis 1 minus straight x right parenthesis left parenthesis 1 plus straight x right parenthesis end fraction close square brackets
rightwards double arrow space 1 over straight y dy over dx equals negative 1 half open square brackets fraction numerator 2 over denominator 1 minus straight x squared end fraction close square brackets space space space space rightwards double arrow space 1 over straight y dy over dx equals negative fraction numerator 1 over denominator 1 minus straight x squared end fraction
therefore left parenthesis 1 minus straight x squared right parenthesis dy over dx equals negative straight y
therefore left parenthesis 1 minus straight x squared right parenthesis dy over dx plus straight y equals 0

Solution

Some More Questions From Continuity and Differentiability Chapter