   Chapter 14, Problem 41RE

Chapter
Section
Textbook Problem

If z =f(u, v), where u = xy, v = y/x, and f has continuous second partial derivatives, show that x 2 ∂ 2 z ∂ x 2 − y 2 ∂ 2 z ∂ y 2 = − 4 u v ∂ 2 z ∂ u   ∂ v + 2 v ∂ z ∂ v

To determine

To show: The equation x22zx2y22zy2=4uv2zuv+2vzv if z=f(u,v),u=xyandv=yx also f hascontinuous second order partial derivatives.

Explanation

Proof:

Given:

The functions are, z=f(u,v),u=xyandv=yx .

Calculation:

Compute the value of zx by using Chain Rule,

zx=zuux+zvvx=zux(xy)+zvx(yx)=zuy+zv(yx2)

Compute the value of zy by using Chain Rule,

zy=zuuy+zvvy=zuy(xy)+zvy(yx)=zux+zv(1x)

Take the partial derivative of zx with respect to x ,

2zx2=x(zuy+zv(yx2))=yx(zu)+x(yx2)zv+(yx2)x(zv)=y[2zu2y+2zvu(yx2)]+2yx3zvyx2[2zuvy+2zv2(yx2)]

Simplify further,

2zx2=y[2zu2y+2zvu(yx2)]+2yx3zvyx2[2zuvy+2zv2(yx2)]=2yx3zvyx2[2zuvy+2zv2(yx2)]+y[2zu2y+2zvu(yx2)]=2yx3zvy2x22zuv+y2x42zv2+y22zu2y2x22zvu=2yx3zv2y2x22zuv+y2x42zv2+y22zu2(2z

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