# Homework Help: The (asserted) equivalence of first partial derivatives

1. Apr 11, 2015

### Larry Cosner

In the solution to a differential-equation problem -- proof of the existence of an integrating factor -- the following statements are made regarding a general function "u(xy)" [that is, a function of two variable that depends exclusively on the single factor "x*y"]:

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"Let xy = t

Therefore ux = uy = ut"

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I'm having a hard time seeing this as innately true. The first two functions are the partial derivatives of the function u, with respect to the two variables. If one takes a simple function like u(xy) = 3xy, then the first partial derivatives are:

ux = 3y and uy = 3x; and similarly, if one addresses the substituted variable, ut = 3.

And these are not equal.

Now I certainly see, given t = xy, that dt/dx = y and dt/dy = x. This leads to dx/dt = 1/y and dy/dt = 1/x.

Using this, one can show ut = (du/dx)(dx/dt) = (3y)(1/y) = 3. Similarly for the equivalency of ut = (du/dy)(dy/dt).

But I'm not seeing how this translates into ux = uy = ut?

And it is this exact equivalence that is used further down in the proof (which is why I care).

Thanks in advance for helping me see what I'm missing.

2. Apr 11, 2015

### wabbit

Seems weird. Assuming $u_x$ means $\frac{\partial}{\partial x}u(x,y)$ etc. then as you find, $u_x=y u_t$ and $u_y=x u_t$.

3. Apr 12, 2015

### Brian T

Perhaps you could post the entire problem or differential equation? From what you've given so far they're not equal, but perhaps we're missing some other condition or assumption