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Continuity And Differentiability

Question
CBSEENMA12035431
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If space straight y equals fraction numerator sin to the power of negative 1 end exponent straight x over denominator square root of 1 minus straight x squared end root end fraction comma space prove space that space left parenthesis 1 minus straight x squared right parenthesis straight y subscript 2 minus 3 space straight x space straight y subscript 1 minus straight y equals 0.

Solution
straight y equals fraction numerator sin to the power of negative 1 end exponent straight x over denominator square root of 1 minus straight x squared end root end fraction space space space space space rightwards double arrow space square root of 1 minus straight x squared end root space straight y equals sin to the power of negative 1 end exponent straight x Diff. space straight w. straight r. straight t. straight x. space space straight y subscript 1 plus straight y. fraction numerator negative 2 space straight x over denominator 2 square root of 1 minus straight x squared end root end fraction equals fraction numerator 1 over denominator square root of 1 minus straight x squared end root end fraction or space space space square root of 1 minus straight x squared end root space straight y subscript 1 minus fraction numerator straight x space straight y over denominator square root of 1 minus straight x squared end root end fraction equals fraction numerator 1 over denominator square root of 1 minus straight x squared end root end fraction or space space space space left parenthesis 1 minus straight x squared right parenthesis straight y subscript 1 minus straight x space straight y equals 1. again space differentiating space straight w. straight r. straight t. straight x comma space left parenthesis 1 minus straight x squared right parenthesis. straight y subscript 2 plus straight y subscript 1 left parenthesis negative 2 straight x right parenthesis minus space xy subscript 1 minus straight y subscript 1 equals 0 or space left parenthesis 1 minus straight x squared right parenthesis. straight y subscript 2 minus 3 space straight x space minus straight y equals 0

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