Examples

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

### Transcript

Example 38 (Method 1) If y = ćš ššć^(ā1) š„, show that (1 ā š„2) š2š¦/šš„2 ā š„ šš¦/šš„ = 0 . We have š¦ = ćš ššć^(ā1) š„ Differentiating š¤.š.š”.š„ šš¦/šš„ = š(ćš ššć^(ā1) š„)/šš„ šš¦/šš„ = 1/ā(ć1 ā š„ć^2 ) ā((šāš^š ) ) š^ā² = š Squaring both sides ("As " š(ćš ššć^(ā1) š„)/šš„ " = " 1/ā(ć1 ā š„ć^2 )) (ā((1āš„^2 ) ) š¦^ā² )^2 = 1^2 (1āš„^2 )(š¦^ā² )^2 = 1 Again Differentiating š¤.š.š”.š„ š/šš„ ((1āš„^2 )(š¦^ā² )^2 ) = (š(1))/šš„ d(1 ā x^2 )/šš„ (š¦^ā² )^2+(1āš„^2 ) š((š¦^ā² )^2 )/šš„ = 0 ā2š„(š¦^ā² )^2+(1āš„^2 ) 2š¦^ā² Ć š¦^ā²ā² = 0 ć2yć^ā² [āšš^ā²+(šāš^š ) š^ā²ā² ] = 0 āš„š¦^ā²+(1āš„^2 ) š¦^ā²ā²=0 (ćšāšć^š ) (š^š š)/ćššć^š ā š . šš/šš = 0 Example 38 (Method 2) If y = ćš ššć^(ā1) š„, show that (1 ā š„2) š2š¦/šš„2 ā š„ šš¦/šš„ = 0 . We have š¦ = ćš ššć^(ā1) š„ Differentiating š¤.š.š”.š„ šš¦/šš„ = š(ćš ššć^(ā1) š„)/šš„ šš¦/šš„ = 1/ā(ć1 ā š„ć^2 ) šš/šš = (ćšāšć^š )^((āš)/( š)) ("As " š(ćš ššć^(ā1) š„)/šš„ " = " 1/ā(ć1 ā š„ć^2 )) Again Differentiating š¤.š.š”.š„ š/šš„ (šš¦/šš„) = (š(ć1 ā š„ć^2 )^((ā1)/( 2)))/šš„ (š^2 š¦)/ćšš„ć^2 = (ā1)/( 2) (ć1āš„ć^2 )^((ā1)/( 2) ā1) . š(ć1 ā š„ć^2 )/šš„ (š^2 š¦)/ćšš„ć^2 = (ā1)/( 2) (ć1āš„ć^2 )^((ā3)/2 ). (0ā2š„) (š^2 š¦)/ćšš„ć^2 = (ā1)/( 2) (ć1āš„ć^2 )^((ā3)/2 ). (ā2š„) (š^š š)/ćššć^š = š(ćšāšć^š )^((āš)/š ) Now, We need to prove (ć1āš„ć^2 ) (š^2 š¦)/ćšš„ć^2 ā š„ . šš¦/šš„ = 0 Solving LHS (ć1āš„ć^2 ) (š^2 š¦)/ćšš„ć^2 ā š„ . šš¦/šš„ = (ć1āš„ć^2 ) . (š„ć (ć1āš„ć^2 )ć^((ā3)/2 ) ) ā š„ (ć1āš„ć^2 )^((ā1)/( 2)) = š„ć (ć1āš„ć^2 )ć^(š + ((āš)/š) )āš„ (ć1āš„ć^2 )^((ā1)/( 2)) = š„ć (ć1āš„ć^2 )ć^((ā1)/( 2))āš„ (ć1āš„ć^2 )^((ā1)/( 2)) = 0 = RHS Hence proved