# Exact differential

by zeshkani
Tags: differential, exact
 Sci Advisor HW Helper PF Gold P: 4,771 The definition of the total differential of a function f(x,y) is the expression $$df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$$ Now with the z you're given, is the expression for dz correct? That is to say, is $$\frac{\partial z}{\partial x}=(y+0)$$ and $$\frac{\partial z}{\partial y}=(x-1)$$ ??
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,491 Yes. z=xy-y+lnx+2 so dz= (y+ 1/x)dx + (x-1)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 $$(y+ 1/x)= \frac{\partial z}{\partial x}$$ for some function z and $$x-1= \frac{\partial z}{\partial y}$$ then $$\frac{\partial(y+ 1/x)}{\partial y}= \frac{\partial^2 z}{\partial x\partial y}$$ and $$\frac{\partial(x-1)}{\partial x}= \frac{\partial^2 z}{\partial y\partial x}$$ 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 $$x-1= \frac{\partial z}{\partial y}$$ 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, $$\frac{\partial z}{\partial x}= y+ C'(x)= y+ 1/x$$ 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+ (x-1)dy.