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Continuity And Differentiability

Question
CBSEENMA12035459

If y = ea cos-1 x , show that
left parenthesis 1 minus straight x squared right parenthesis fraction numerator straight d squared straight y over denominator dx squared end fraction minus straight x dy over dx minus straight a squared straight y equals 0.

Solution
space space straight y equals straight e to the power of straight m space sin to the power of negative 1 end exponent straight x end exponent 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 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 parenthesis 1 right parenthesis
therefore space straight y subscript 1 equals space space straight y equals straight e to the power of straight m space sin to the power of negative 1 end exponent straight x end exponent. fraction numerator straight m over denominator square root of 1 minus straight x squared end root end fraction space space rightwards double arrow space space square root of 1 minus straight x squared end root. straight y subscript 1 equals straight m space straight e to the power of straight m space sin to the power of negative 1 end exponent straight x end exponent
rightwards double arrow space square root of 1 minus straight x squared end root. straight y subscript 1 equals straight m space 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 space space space space space space space left square bracket because space of space left parenthesis 1 right parenthesis right square bracket
rightwards double arrow space left parenthesis 1 minus straight x squared right parenthesis straight y subscript 1 squared equals straight m squared straight y squared

Differentiating both sides w.r.t. x, we get

(1-x2)·2 y1 y2+y12 (-2 x)=a2·2 y y1

Dividing both sides by 2 y1 , we get,

left parenthesis 1 minus straight x squared right parenthesis straight y subscript 2 minus straight x space straight y subscript 1 equals straight a squared straight y space space or space space left parenthesis 1 minus straight x squared right parenthesis fraction numerator straight d squared straight y over denominator dx squared end fraction minus straight x dy over dx minus straight a squared straight y equals 0.

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