I have to find the derivative $dw$ if $w=F(u,v,z)$ where $u=x^2+y^2, v=x^2-y^2$ and $z=2xy$. So $dw=\frac{\partial w}{\partial z}dz + \frac{\partial w}{\partial u}du + \frac{\partial w}{\partial v}dv$. How to find $dw$ now?
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1HINT Apply the chain rule. – Fredrik Meyer Apr 23 '13 at 09:52
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$dw$ is called a differential. It measures the tendency for the dependent variable $w$ to change as the independent variables $x$ and $y$ change. To compute $dw$, we use your formula, along with the differentials $du$, $dv$, and $dz$ (computed the same way you suggest). $$du=2xdx+2ydy\\dv=2xdx-2ydy\\dz=2ydx+2xdy$$
Next, using your formula, we see that $$\begin{align}dw&=\frac{\partial w}{\partial u}(2xdx+2ydy)+\frac{\partial w}{\partial v}(2xdx-2ydy)+\frac{\partial w}{\partial z}(2ydx+2xdy)\\&=(2x\frac{\partial w}{\partial u}+2x\frac{\partial w}{\partial v}+2y\frac{\partial w}{\partial z})dx+(2y\frac{\partial w}{\partial u}-2y\frac{\partial w}{\partial v}+2x\frac{\partial w}{\partial z})dy\end{align}$$
Without knowing an explicitly formula for $w=F(u,v,z)$, this is as explicit as we can be.
Jared
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