Chapter 5 Class 12 Continuity and Differentiability

Class 12
Important Questions for exams Class 12

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### Transcript

Example 38 (Method 1) If y = γπ ππγ^(β1) π₯, show that (1 β π₯2) π2π¦/ππ₯2 β π₯ ππ¦/ππ₯ = 0 . We have π¦ = γπ ππγ^(β1) π₯ Differentiating π€.π.π‘.π₯ ππ¦/ππ₯ = π(γπ ππγ^(β1) π₯)/ππ₯ ππ¦/ππ₯ = 1/β(γ1 β π₯γ^2 ) β((πβπ^π ) ) π^β² = π Squaring both sides ("As " π(γπ ππγ^(β1) π₯)/ππ₯ " = " 1/β(γ1 β π₯γ^2 )) (β((1βπ₯^2 ) ) π¦^β² )^2 = 1^2 (1βπ₯^2 )(π¦^β² )^2 = 1 Again Differentiating π€.π.π‘.π₯ π/ππ₯ ((1βπ₯^2 )(π¦^β² )^2 ) = (π(1))/ππ₯ d(1 β x^2 )/ππ₯ (π¦^β² )^2+(1βπ₯^2 ) π((π¦^β² )^2 )/ππ₯ = 0 β2π₯(π¦^β² )^2+(1βπ₯^2 ) 2π¦^β² Γ π¦^β²β² = 0 γ2yγ^β² [βππ^β²+(πβπ^π ) π^β²β² ] = 0 βπ₯π¦^β²+(1βπ₯^2 ) π¦^β²β²=0 (γπβπγ^π ) (π^π π)/γππγ^π β π . ππ/ππ = 0 Example 38 (Method 2) If y = γπ ππγ^(β1) π₯, show that (1 β π₯2) π2π¦/ππ₯2 β π₯ ππ¦/ππ₯ = 0 . We have π¦ = γπ ππγ^(β1) π₯ Differentiating π€.π.π‘.π₯ ππ¦/ππ₯ = π(γπ ππγ^(β1) π₯)/ππ₯ ππ¦/ππ₯ = 1/β(γ1 β π₯γ^2 ) ππ/ππ = (γπβπγ^π )^((βπ)/( π)) ("As " π(γπ ππγ^(β1) π₯)/ππ₯ " = " 1/β(γ1 β π₯γ^2 )) Again Differentiating π€.π.π‘.π₯ π/ππ₯ (ππ¦/ππ₯) = (π(γ1 β π₯γ^2 )^((β1)/( 2)))/ππ₯ (π^2 π¦)/γππ₯γ^2 = (β1)/( 2) (γ1βπ₯γ^2 )^((β1)/( 2) β1) . π(γ1 β π₯γ^2 )/ππ₯ (π^2 π¦)/γππ₯γ^2 = (β1)/( 2) (γ1βπ₯γ^2 )^((β3)/2 ). (0β2π₯) (π^2 π¦)/γππ₯γ^2 = (β1)/( 2) (γ1βπ₯γ^2 )^((β3)/2 ). (β2π₯) (π^π π)/γππγ^π = π(γπβπγ^π )^((βπ)/π ) Now, We need to prove (γ1βπ₯γ^2 ) (π^2 π¦)/γππ₯γ^2 β π₯ . ππ¦/ππ₯ = 0 Solving LHS (γ1βπ₯γ^2 ) (π^2 π¦)/γππ₯γ^2 β π₯ . ππ¦/ππ₯ = (γ1βπ₯γ^2 ) . (π₯γ (γ1βπ₯γ^2 )γ^((β3)/2 ) ) β π₯ (γ1βπ₯γ^2 )^((β1)/( 2)) = π₯γ (γ1βπ₯γ^2 )γ^(π + ((βπ)/π) )βπ₯ (γ1βπ₯γ^2 )^((β1)/( 2)) = π₯γ (γ1βπ₯γ^2 )γ^((β1)/( 2))βπ₯ (γ1βπ₯γ^2 )^((β1)/( 2)) = 0 = RHS Hence proved

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.