Find d^2y/dx^2 for 5x^2-4y^2=20?

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Answer 1 · 2018-04-19T12:22:46

#(d^2y)/(dx^2)=-25/(4y^3)#

#5x^2-4y^2=20...to(A)#

Diff.w.r.t. #x#

#10x-8y(dy)/(dx)=0#

#=>8y(dy)/(dx)=10x#

#=>4y(dy)/(dx)=5x..........to(dy)/(dx)=(5x)/(4y)to(1)#

Again diff.w.r.t. #x#

#4y(d^2y)/(dx^2)+(dy)/(dx)*4(dy)/(dx)=5#

#=>4y(d^2y)/(dx^2)+4((dy)/(dx))^2=5...toApply (1)#

#=>4y(d^2y)/(dx^2)+4((5x)/(4y))^2=5#

#=>4y(d^2y)/(dx^2)=5-4((25x^2)/(16y^2))#

#=>4y(d^2y)/(dx^2)=5-((25x^2)/(4y^2))#

#=>4y(d^2y)/(dx^2)=(20y^2-25x^2)/(4y^2)#

#=>4y(d^2y)/(dx^2)=-5((5x^2-4y^2)/(4y^2))#

#=>4y(d^2y)/(dx^2)=-5(20/(4y^2))...toFrom(A)#

#=>4y(d^2y)/(dx^2)=-25/y^2#

#=>(d^2y)/(dx^2)=-25/(4y^3)#