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Sponsor Area

Continuity And Differentiability

Question
CBSEENMA12034949
Wired Faculty App

If space straight y equals fraction numerator sin space straight x over denominator 1 plus begin display style fraction numerator cos space straight x over denominator 1 plus begin display style fraction numerator sin space straight x over denominator 1 plus begin display style fraction numerator cos space straight x over denominator 1 plus.... infinity end fraction end style end fraction end style end fraction end style end fraction comma space prov e space that space dy over dx equals fraction numerator left parenthesis 1 plus straight y right parenthesis cos space straight x plus straight y space sin space straight x over denominator 1 plus 2 straight y plus cos space straight x minus sin space straight x end fraction

Solution
Here space straight y equals fraction numerator sin space straight x over denominator 1 plus begin display style fraction numerator cos space straight x over denominator 1 plus begin display style fraction numerator sin space straight x over denominator 1 plus begin display style fraction numerator cos space straight x over denominator 1 plus.... infinity end fraction end style end fraction end style end fraction end style end fraction therefore space space space space space space straight y equals fraction numerator sin space straight x over denominator 1 plus begin display style fraction numerator cos space straight x over denominator 1 plus straight y end fraction end style end fraction space rightwards double arrow space straight y equals fraction numerator left parenthesis 1 plus straight y right parenthesis sin space straight x over denominator 1 plus straight y plus cos space straight x end fraction rightwards double arrow space straight y plus straight y squared plus straight y space cos space straight x equals left parenthesis 1 plus straight y right parenthesis sin space straight x Differntiating space both space sides space straight w. straight r. straight t. straight x comma space we space get comma space space space space space dy over dx plus 2 straight y dy over dx plus straight y left parenthesis negative sin space straight x right parenthesis plus cos space straight x dy over dx equals left parenthesis 1 plus straight y right parenthesis cos space straight x plus sin space straight x. dy over dx therefore space left parenthesis 1 plus 2 straight y plus cos space straight x minus sin space straight x right parenthesis dy over dx equals left parenthesis 1 plus straight y right parenthesis cos space straight x plus straight y space sin space straight x therefore space dy over dx equals fraction numerator left parenthesis 1 plus straight y right parenthesis cos space straight x plus straight y space sin space straight x over denominator 1 plus 2 straight y plus cos space straight x minus sin space straight x end fraction

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