A Treatise on the Differential Calculus |
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Page 56
... dxdy dx d / dv dx dy d2v + + d ( do dy dv dy dy dx d'v dy dxdy dy dx Ddy dzy = Also dx dx dx2 Dv d2v dεv dy d'v dy dv d'y Hence + 2 dx2 dx2 + dxdy dx dy dx + ( 2 ) , dy dx2 d'v d2v d'v dv or D2v = dx2 + 2 • dx2 dxdy deady + dy2 + · d'y ...
... dxdy dx d / dv dx dy d2v + + d ( do dy dv dy dy dx d'v dy dxdy dy dx Ddy dzy = Also dx dx dx2 Dv d2v dεv dy d'v dy dv d'y Hence + 2 dx2 dx2 + dxdy dx dy dx + ( 2 ) , dy dx2 d'v d2v d'v dv or D2v = dx2 + 2 • dx2 dxdy deady + dy2 + · d'y ...
Page 59
... dxdy dx dy dx dz = + But , from ( 1 ) , putting Ddu dx dy du du dy ' dz = d'u + • dy dz dz dx dy " successively for u , d2u dz + dxdy dzdy dx ' D ( du d2u d2u dz + : hence D'u d'u = + = dx dz dxdz dz2 dx d'u dz d2u dz d'u dz + + dxdy dxdy ...
... dxdy dx dy dx dz = + But , from ( 1 ) , putting Ddu dx dy du du dy ' dz = d'u + • dy dz dz dx dy " successively for u , d2u dz + dxdy dzdy dx ' D ( du d2u d2u dz + : hence D'u d'u = + = dx dz dxdz dz2 dx d'u dz d2u dz d'u dz + + dxdy dxdy ...
Page 60
... dxdy dxdy dydz dx From these five equations we can determine d'u dz d'u dz dz du dz + + dxdz dy dz2 dx dy dz dxdy = 0 . dz dz d2z d2z d2z dx ' dy ' dx ' dxdy ' dy ' the partial differential coefficients of the implicit function z , in ...
... dxdy dxdy dydz dx From these five equations we can determine d'u dz d'u dz dz du dz + + dxdz dy dz2 dx dy dz dxdy = 0 . dz dz d2z d2z d2z dx ' dy ' dx ' dxdy ' dy ' the partial differential coefficients of the implicit function z , in ...
Page 74
... dxdy D3u 0 , = 0 , dx dy dy = 0 , = 0 , D3u dxdy D2u dy2 = 0 , = 0 , D3u dy3 = · 0 . Dru Dru Dru Dru Dru = 0 , = 0 , = 0 , ... = 0 , dx dx2dy " -2 ddy n - 1 dx - 1dy dx - 2dy2 the number of which is 1+ 2+ 3+ = + ( n + 1 ) = { ( n + 1 ) ...
... dxdy D3u 0 , = 0 , dx dy dy = 0 , = 0 , D3u dxdy D2u dy2 = 0 , = 0 , D3u dy3 = · 0 . Dru Dru Dru Dru Dru = 0 , = 0 , = 0 , ... = 0 , dx dx2dy " -2 ddy n - 1 dx - 1dy dx - 2dy2 the number of which is 1+ 2+ 3+ = + ( n + 1 ) = { ( n + 1 ) ...
Page 76
... dx dy dx dxdy dx dz dz dy 2 d2z dz dz d2z dz · f ' ( x ) . ( d ) 3 dy · dxdy dy dx dy ' eliminating f ' ( z ) , we have 2 dz d2z 2 dy dx2 = dz dz d2z dz 2 d2z + dx dy dxdy dx dy = 0 . Thus we see that , instead of two final equations of ...
... dx dy dx dxdy dx dz dz dy 2 d2z dz dz d2z dz · f ' ( x ) . ( d ) 3 dy · dxdy dy dx dy ' eliminating f ' ( z ) , we have 2 dz d2z 2 dy dx2 = dz dz d2z dz 2 d2z + dx dy dxdy dx dy = 0 . Thus we see that , instead of two final equations of ...
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Common terms and phrases
algebraical arbitrary functions asymptote axis Cambridge change sign constant cosec curve d'u dy d²r d²u d²x d²z d2z d2z d³u d³x d³y d³z denote df df df dx DIFFERENTIAL CALCULUS differential equations du du dy du² dv₁ dx dx dx dy dx dx dz dx² dx³ dxdy dy dF dy dx dy dy dy dy dz dy₁ dy₂ dy³ dz dx dz dy dz dz eliminate expression f(y₁ find the Differential formula ƒ Y₁ Hence implicit function increment independent variables indeterminate limit maxima and minima maximum or minimum minimum value multiplying negative partial differential coefficients points of inflection positive quantity putting regard shews Suppose tangent Taylor's Theorem total differential Trinity College University of Cambridge whence y+dy Y₁ Y₂ zero бу бх
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