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Exact differential 
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#1
Sep207, 03:27 PM

P: 29

how would i show that this is the exact differential
z=xyy+lnx+2 dz= (y+ 0)dx + (x1)dy (i hope the total differential is right) so how would i show that dz is the exact differential ??? 


#2
Sep207, 03:36 PM

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PF Gold
P: 4,771

The definition of the total differential of a function f(x,y) is the expression
[tex]df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy[/tex] Now with the z you're given, is the expression for dz correct? That is to say, is [tex]\frac{\partial z}{\partial x}=(y+0)[/tex] and [tex]\frac{\partial z}{\partial y}=(x1)[/tex] ?? 


#3
Sep207, 03:39 PM

P: 29

i believe dz is correct because i used partial differentials to solve it, and i hope its correct



#4
Sep207, 04:21 PM

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PF Gold
P: 4,771

Exact differential
what is the derivative of ln(x)?



#5
Sep307, 05:55 AM

Math
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PF Gold
P: 39,363

Once you fixed the derivative of ln x thing, the fact that dz is the total derivative of z pretty much means it is an "exact" differential. That's the definition of "exact" differential!



#6
Sep307, 01:35 PM

P: 29

thx and the derivative of lnx is just 1/x



#7
Sep307, 04:39 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,363

Yes.
z=xyy+lnx+2 so dz= (y+ 1/x)dx + (x1)dy And that is an "exact" differential precisely because it is the differential of z. Now suppose you were given the differential form (y+ 1/x)dx+ (x 1)dy without having been given z. How would you determine whether it was an exact differential? If [tex](y+ 1/x)= \frac{\partial z}{\partial x}[/tex] for some function z and [tex]x1= \frac{\partial z}{\partial y}[/tex] then [tex]\frac{\partial(y+ 1/x)}{\partial y}= \frac{\partial^2 z}{\partial x\partial y}[/tex] and [tex]\frac{\partial(x1)}{\partial x}= \frac{\partial^2 z}{\partial y\partial x}[/tex] and so must be equal. Fortunately, it is easy to see that each second derivative is 1 and so they are in fact equal. Okay, now, how would we find z? Knowing that [tex]x1= \frac{\partial z}{\partial y}[/tex] integrating with respect to y (treating x as a constant) we get z= xy y+ C, except that, since we are treating x as a constant, that "constant of integration", C, may depend on x: z= xy y+ C(x). Differentiating that with respect to x, [tex]\frac{\partial z}{\partial x}= y+ C'(x)= y+ 1/x[/tex] Notice that the "y" terms cancel (it was the "cross condition" above that guarenteed that) and so we have C'(x)= 1/x. Then C(x)= ln(x)+ C where "C" now really is a constant. Any z(x,y)= xy y+ ln(x)+ C satisfies dz= (y+ 1/x)dx+ (x1)dy. 


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