# Ex 5.7, 12 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at Sept. 17, 2019 by Teachoo

Last updated at Sept. 17, 2019 by Teachoo

Transcript

Ex 5.7, 12 If y= γπππ γ^(β1) π₯ , Find π2π¦/ππ₯2 in terms of π¦ alone. Let y = γπππ γ^(β1) π₯ Differentiating π€.π.π‘.π₯ ππ¦/ππ₯ = (π(γπππ γ^(β1) π₯))/ππ₯ ππ¦/ππ₯ = (β1)/β(1 β π₯^2 ) ("As" π(γπππ γ^(β1) π₯)/ππ₯=(β1)/β(1 β π₯^2 )) Again Differentiating π€.π.π‘.π₯ π/ππ₯ (ππ¦/ππ₯) = π/ππ₯ ((β1)/β(1 β π₯^2 )) (π^2 π¦)/(ππ₯^2 ) = π/ππ₯ ((β1)/β(1 β π₯^2 )) using Quotient Rule As, (π’/π£)^β²= (π’βπ£ β π£βπ’)/π£^2 where u = 1 & v = β(1βπ₯^2 ) (π^2 π¦)/(ππ₯^2 ) = β [ ((π(1) )/ππ₯ . β(1 β π₯^2 )β (π (β(1 β π₯^2 ) ))/ππ₯ . 1)/(β(1 β π₯^2 ) )^2 ] (π^2 π¦)/(ππ₯^2 ) = β[ (0 . β(1 β π₯^2 )β 1/(2 β(1 β π₯^2 )) . π/ππ₯ (1 β π₯^2) )/(β(1 β π₯^2 ) )^2 ] = β[ ( 1/(2 β(1 β π₯^2 )) . (0β2π₯))/((1 β π₯^2 ) ) ] = β (β1 (β 2π₯))/(2 (β(1 β π₯^2 ) ) (1 β π₯^2 ) ) = ( β2π₯)/(2 (β(1 β π₯^2 ) ) (1 β π₯^2 ) ) = (β π₯)/( (1 β π₯^2 )^(1/2) (1 β π₯^2 ) ) = (β π₯)/( (1 β π₯^2 )^(1/2 + 1) ) (π ^π π)/(π π^π ) = (β π)/( (π β π^π )^(π/π ) ) But we need to calculate (π^2 π¦)/(ππ₯^2 ) in terms of y . y = γπππ γ^(β1) π₯ cosβ‘π¦ = π₯ π₯ = cosβ‘π¦ Putting π₯ = cosβ‘π¦ in equation (π^2 π¦)/(ππ₯^2 ) = (β π₯)/( (1 β π₯^2 )^(3/2 ) ) (π^2 π¦)/(ππ₯^2 ) = (βcosβ‘π¦ " " )/( (1 β γ(cosβ‘π¦)γ^2 )^(3/2 ) ) = (β cosβ‘π¦ " " )/( γ(sin2β‘γπ¦) γγ^(3/2 ) ) = (β cosβ‘π¦ " " )/( γ(sinβ‘γπ¦) γγ^(2 Γ 3/2 ) ) = (β cosβ‘π¦ " " )/( γ(sinβ‘γπ¦) γγ^(3 ) ) = (βcosβ‘π¦ " " )/( sin π¦ ) Γ 1/sin2β‘π¦ (π ^π π)/(π π^π ) = βπππβ‘π. ππππππ π

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.