# Ex 5.3, 6 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at April 16, 2024 by Teachoo

Last updated at April 16, 2024 by Teachoo

Ex 5.3, 6 Find šš¦/šš„ 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.š”.š„ . š(š„3 + š„2š¦ + š„š¦2 + š¦3)/šš„ = (š (81))/šš„ š(š„3)/šš„ + š(š„2š¦)/šš„ + (š(š„š¦2))/šš„ + š(š¦3)/šš„ =0 3š„^(3ā1) + (š(š„2š¦))/šš„ + (š(š„š¦2))/šš„ + š(š¦3)/šš„ Ć šš¦/šš¦ = 0 3š„^2 + (š(š„2š¦))/šš„ + (š(š„š¦2))/šš„ + š(š¦3)/šš„ Ć šš¦/šš„ = 0 3š„^2+ (š(š„2š¦))/šš„ + (š(š„š¦2))/šš„ +3š¦^(3ā1) . šš¦/šš„ = 0 3š„2 + (š(š„2š¦))/šš„ + (š(š„š¦2))/šš„ +3š¦^2 šš¦/šš„ = 0 Using product rule in š„2š¦ & š„š¦2 (uv)ā = uāv + vāu 3š„2 + ((š(š„2))/šš„.š¦+š„2 .(š(š¦))/šš„)+((š(š„))/šš„.š¦2+ .(š(š¦2))/šš„ š„ )+ 3y2 šš¦/( šš„) = 0 3š„2 + (2š„.š¦+š„2 (š(š¦))/šš„) + (1.š¦2+š„ .(š(š¦2))/šš„ )+ 3y2 šš¦/( šš„) = 0 3š„2 + (2š„.š¦+š„2 (š(š¦))/šš„) + (š¦2+š„ .(š(š¦2))/šš„ Ć šš¦/šš¦)+ 3y2 šš¦/( šš„) = 0 3š„2 + 2š„š¦+š„2 šš¦/šš„+š¦2+š„.(š(š¦2))/šš¦ Ćšš¦/šš„+3š¦2 šš¦/( šš„) = 0 3š„2 + 2š„š¦+š„2 šš¦/šš„+š¦2 + š„. 2š¦^(2ā1) (šš¦/šš„)+3š¦2 šš¦/šš„=0 3š„2 + 2š„š¦+š¦2 + x2 šš¦/šš„+š„.2š¦ šš¦/šš„+3š¦2 šš¦/šš„=0 "(3" š„"2 "+" " 2š„š¦+š¦2")" + šš¦/šš„ (š„2+2š„š¦+3š¦2)=0 šš¦/šš„ (š„2+2š„š¦+3š¦2)=ā "(3" š„"2 "+" " 2š„š¦+š¦2")" š š/š š= (ā "(" šš"2 " +" " ššš + šš")" )/((šš + ššš + ššš) )