Define z as a function of x and y

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Homework Statement


The equations x=uv, y=u+v and z=u^2-v^2 define z as a function of x and y. Find \frac{\partial u}{\partial x}<br />

Homework Equations



Chain rule

The Attempt at a Solution


Once I get z as a function of x and y the solution seems pretty straight forward, but how exactly is that done. Would it be along the lines of z(x(u,v),y(u,v))? That doesn't seem right though. Thanks for any help in advance.
 
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Did you mean to say you're trying to find \partial u/\partial x and not \partial z/\partial x?
 
No, I meant what I stated. I realized you solve for u and v and then plug into z=u^2-v^2. You get u=x/v and substitute that into y=u+v and then multiply both sides by v to get a quadratic equation. You do that for both u and v and plug them into z and its smooth sailing from there
 
Frankly, I don't see why z has anything to do with the problem then.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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