d2x / d2y = - ( d2y / dx2) * (dy/dx)^-3 please explain this with an example .
i tried with x = 3y^2 but i cant get this!!!!
i tried with x = 3y^2 but i cant get this!!!!
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We have : dx/dy = 1 / ( dy/dx ).
∴ d²x/dy²
= d/dy ( dx/dy )
= d/dy [ 1 / ( dy/dx ) ]
= d/dy ( 1/ u ), ...................... where : u = dy/dx
= d/dx ( 1/ u ) • dx/dy ........... by Chain Rule
= [ -1/ u² ] • du/dx • dx/dy
= [ -1/ ( dy/dx )² ] • d/dx ( dy/dx ) • dx/dy
= [ -1/ ( dy/dx )² ] • d²y/dx² • [ 1 / ( dy/dx ) ]
= [ -1/ ( dy/dx )³ ] • d²y/dx²
= - ( d²y/dx² ) • ( dy/dx )ֿ³ ................................... Q.E.D.
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Happy To Help !
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∴ d²x/dy²
= d/dy ( dx/dy )
= d/dy [ 1 / ( dy/dx ) ]
= d/dy ( 1/ u ), ...................... where : u = dy/dx
= d/dx ( 1/ u ) • dx/dy ........... by Chain Rule
= [ -1/ u² ] • du/dx • dx/dy
= [ -1/ ( dy/dx )² ] • d/dx ( dy/dx ) • dx/dy
= [ -1/ ( dy/dx )² ] • d²y/dx² • [ 1 / ( dy/dx ) ]
= [ -1/ ( dy/dx )³ ] • d²y/dx²
= - ( d²y/dx² ) • ( dy/dx )ֿ³ ................................... Q.E.D.
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Happy To Help !
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You are Welcome, RP !
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