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Ex 5.3, 6 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

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Ex 5.3, 6 - Find dy/dx in x3 + x2y + xy2 + y3 = 81 - CBSE

Ex 5.3, 6 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.3, 6 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

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Ex 5.3, 6 Find š‘‘š‘¦/š‘‘š‘„ in, š‘„3 + š‘„2š‘¦ + š‘„š‘¦2 + š‘¦3 = 81 š‘„3 + š‘„2š‘¦ + š‘„š‘¦2 + š‘¦3 = 81 Differentiating both sides š‘¤.š‘Ÿ.š‘”.š‘„ . š‘‘(š‘„3 + š‘„2š‘¦ + š‘„š‘¦2 + š‘¦3)/š‘‘š‘„ = (š‘‘ (81))/š‘‘š‘„ š‘‘(š‘„3)/š‘‘š‘„ + š‘‘(š‘„2š‘¦)/š‘‘š‘„ + (š‘‘(š‘„š‘¦2))/š‘‘š‘„ + š‘‘(š‘¦3)/š‘‘š‘„ =0 3š‘„^(3āˆ’1) + (š‘‘(š‘„2š‘¦))/š‘‘š‘„ + (š‘‘(š‘„š‘¦2))/š‘‘š‘„ + š‘‘(š‘¦3)/š‘‘š‘„ Ɨ š‘‘š‘¦/š‘‘š‘¦ = 0 3š‘„^2 + (š‘‘(š‘„2š‘¦))/š‘‘š‘„ + (š‘‘(š‘„š‘¦2))/š‘‘š‘„ + š‘‘(š‘¦3)/š‘‘š‘„ Ɨ š‘‘š‘¦/š‘‘š‘„ = 0 3š‘„^2+ (š‘‘(š‘„2š‘¦))/š‘‘š‘„ + (š‘‘(š‘„š‘¦2))/š‘‘š‘„ +3š‘¦^(3āˆ’1) . š‘‘š‘¦/š‘‘š‘„ = 0 3š‘„2 + (š‘‘(š‘„2š‘¦))/š‘‘š‘„ + (š‘‘(š‘„š‘¦2))/š‘‘š‘„ +3š‘¦^2 š‘‘š‘¦/š‘‘š‘„ = 0 Using product rule in š‘„2š‘¦ & š‘„š‘¦2 (uv)ā€™ = uā€™v + vā€™u 3š‘„2 + ((š‘‘(š‘„2))/š‘‘š‘„.š‘¦+š‘„2 .(š‘‘(š‘¦))/š‘‘š‘„)+((š‘‘(š‘„))/š‘‘š‘„.š‘¦2+ .(š‘‘(š‘¦2))/š‘‘š‘„ š‘„ )+ 3y2 š‘‘š‘¦/( š‘‘š‘„) = 0 3š‘„2 + (2š‘„.š‘¦+š‘„2 (š‘‘(š‘¦))/š‘‘š‘„) + (1.š‘¦2+š‘„ .(š‘‘(š‘¦2))/š‘‘š‘„ )+ 3y2 š‘‘š‘¦/( š‘‘š‘„) = 0 3š‘„2 + (2š‘„.š‘¦+š‘„2 (š‘‘(š‘¦))/š‘‘š‘„) + (š‘¦2+š‘„ .(š‘‘(š‘¦2))/š‘‘š‘„ Ɨ š‘‘š‘¦/š‘‘š‘¦)+ 3y2 š‘‘š‘¦/( š‘‘š‘„) = 0 3š‘„2 + 2š‘„š‘¦+š‘„2 š‘‘š‘¦/š‘‘š‘„+š‘¦2+š‘„.(š‘‘(š‘¦2))/š‘‘š‘¦ Ɨš‘‘š‘¦/š‘‘š‘„+3š‘¦2 š‘‘š‘¦/( š‘‘š‘„) = 0 3š‘„2 + 2š‘„š‘¦+š‘„2 š‘‘š‘¦/š‘‘š‘„+š‘¦2 + š‘„. 2š‘¦^(2āˆ’1) (š‘‘š‘¦/š‘‘š‘„)+3š‘¦2 š‘‘š‘¦/š‘‘š‘„=0 3š‘„2 + 2š‘„š‘¦+š‘¦2 + x2 š‘‘š‘¦/š‘‘š‘„+š‘„.2š‘¦ š‘‘š‘¦/š‘‘š‘„+3š‘¦2 š‘‘š‘¦/š‘‘š‘„=0 "(3" š‘„"2 "+" " 2š‘„š‘¦+š‘¦2")" + š‘‘š‘¦/š‘‘š‘„ (š‘„2+2š‘„š‘¦+3š‘¦2)=0 š‘‘š‘¦/š‘‘š‘„ (š‘„2+2š‘„š‘¦+3š‘¦2)=āˆ’ "(3" š‘„"2 "+" " 2š‘„š‘¦+š‘¦2")" š’…š’š/š’…š’™= (āˆ’ "(" šŸ‘š’™"2 " +" " šŸš’™š’š + š’ššŸ")" )/((š’™šŸ + šŸš’™š’š + šŸ‘š’ššŸ) )

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