## A Treatise on the Differential Calculus |

### From inside the book

Results 1-5 of 42

Page 21

... dz + dx dx

... dz + dx dx

**dz dx**In like manner , y being supposed variable and x constant , ( 1 ) . Du du du dz dz + · dy dy dz dy dz ( 2 ) . In these equations and are the partial differential coeffi- dx dy du cients of z with regard to x and y ... Page 22

... dz accordingly also the same in both . The equations written in the most expressive form would accordingly be Du du du dz = + dx dx

... dz accordingly also the same in both . The equations written in the most expressive form would accordingly be Du du du dz = + dx dx

**Dz dx**Du y dy = du du dz dy Dz dy ( 3 ) , ( 4 ) . Owing to the complexity of the notation in ( 3 ) and ... Page 23

... dx , Du dx2 2 = du dy , dx dy , dx du + 1 · du dy · dx dy , dx Du du • dx2 dy , Ax dy2 2 + du dys dx3 dy , dx , + ... dz = 0 . = + dx dx

... dx , Du dx2 2 = du dy , dx dy , dx du + 1 · du dy · dx dy , dx Du du • dx2 dy , Ax dy2 2 + du dys dx3 dy , dx , + ... dz = 0 . = + dx dx

**dz dx**Again , supposing y variable and a constant , we shall have also Du du du dz = + • dy dy ... Page 39

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**dx**du**dz**dy = COS 2 ,**dz**dy = COS y , = cos x :**dx**but , by Art . ( 19 ) , Cor . du ༄ །Ë = du**dz**dy**dx dz**dy**dx**hence Ex . 11. Given u = du = cos z cos y cos x .**dx**sin ( ay ) + sin ( ẞz ) + tan ̄1 ( yz ) , y = sec x , z = cosec x ... Page 41

... dx cos y ( a2 — x2 ) * * u = Xyz , = 0 , z being a function of x and y by virtue of the equation let it be proposed to find z = sin ( xy ) , du du du Du Du Du . dx ' dy dy '

... dx cos y ( a2 — x2 ) * * u = Xyz , = 0 , z being a function of x and y by virtue of the equation let it be proposed to find z = sin ( xy ) , du du du Du Du Du . dx ' dy dy '

**dz dx**' dy du Xyz By Art . ( 30 ) , = yz . dx du by Art ...### Contents

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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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