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Example 41 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)
Last updated at March 16, 2023 by Teachoo
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Transcript
Example 41 (Method 1) If y = γπ ππγ^(β1) π₯, show that (1 β π₯2) π2π¦/ππ₯2 β π₯ ππ¦/ππ₯ = 0 . We have π¦ = γπ ππγ^(β1) π₯ Differentiating π€.π.π‘.π₯ ππ¦/ππ₯ = π(γπ ππγ^(β1) π₯)/ππ₯ ππ¦/ππ₯ = 1/β(γ1 β π₯γ^2 ) β((πβπ^π ) ) π^β² = π Squaring both sides ("As " π(γπ ππγ^(β1) π₯)/ππ₯ " = " 1/β(γ1 β π₯γ^2 )) (β((1βπ₯^2 ) ) π¦^β² )^2 = 1^2 (1βπ₯^2 )(π¦^β² )^2 = 1 Again Differentiating π€.π.π‘.π₯ π/ππ₯ ((1βπ₯^2 )(π¦^β² )^2 ) = (π(1))/ππ₯ d(1 β x^2 )/ππ₯ (π¦^β² )^2+(1βπ₯^2 ) π((π¦^β² )^2 )/ππ₯ = 0 β2π₯(π¦^β² )^2+(1βπ₯^2 ) 2π¦^β² Γ π¦^β²β² = 0 γ2yγ^β² [βππ^β²+(πβπ^π ) π^β²β² ] = 0 βπ₯π¦^β²+(1βπ₯^2 ) π¦^β²β²=0 (γπβπγ^π ) (π ^π π)/γπ πγ^π β π . π π/π π = 0 (π^2 π¦)/γππ₯γ^2 = (β1)/( 2) (γ1βπ₯γ^2 )^((β3)/2 ). (0β2π₯) (π^2 π¦)/γππ₯γ^2 = (β1)/( 2) (γ1βπ₯γ^2 )^((β3)/2 ). (β2π₯) (π ^π π)/γπ πγ^π = π(γπβπγ^π )^((βπ)/π ) Example 41 (Method 2) If y = γπ ππγ^(β1) π₯, show that (1 β π₯2) π2π¦/ππ₯2 β π₯ ππ¦/ππ₯ = 0 . We have π¦ = γπ ππγ^(β1) π₯ Differentiating π€.π.π‘.π₯ ππ¦/ππ₯ = π(γπ ππγ^(β1) π₯)/ππ₯ ππ¦/ππ₯ = 1/β(γ1 β π₯γ^2 ) π π/π π = (γπβπγ^π )^((βπ)/( π)) ("As " π(γπ ππγ^(β1) π₯)/ππ₯ " = " 1/β(γ1 β π₯γ^2 )) Again Differentiating π€.π.π‘.π₯ π/ππ₯ (ππ¦/ππ₯) = (π(γ1 β π₯γ^2 )^((β1)/( 2)))/ππ₯ (π^2 π¦)/γππ₯γ^2 = (β1)/( 2) (γ1βπ₯γ^2 )^((β1)/( 2) β1) . π(γ1 β π₯γ^2 )/ππ₯ (π^2 π¦)/γππ₯γ^2 = (β1)/( 2) (γ1βπ₯γ^2 )^((β3)/2 ). (0β2π₯) (π^2 π¦)/γππ₯γ^2 = (β1)/( 2) (γ1βπ₯γ^2 )^((β3)/2 ). (β2π₯) (π ^π π)/γπ πγ^π = π(γπβπγ^π )^((βπ)/π ) Now, We need to prove (γ1βπ₯γ^2 ) (π^2 π¦)/γππ₯γ^2 β π₯ . ππ¦/ππ₯ = 0 Solving LHS (γ1βπ₯γ^2 ) (π^2 π¦)/γππ₯γ^2 β π₯ . ππ¦/ππ₯ = (γ1βπ₯γ^2 ) . (π₯γ (γ1βπ₯γ^2 )γ^((β3)/2 ) ) β π₯ (γ1βπ₯γ^2 )^((β1)/( 2)) = π₯γ (γ1βπ₯γ^2 )γ^(π + ((βπ)/π) )βπ₯ (γ1βπ₯γ^2 )^((β1)/( 2)) = π₯γ (γ1βπ₯γ^2 )γ^((β1)/( 2))βπ₯ (γ1βπ₯γ^2 )^((β1)/( 2)) = 0 = RHS Hence proved
