Calculating short partial derivatives

Question

We have $u(x,y) $ and $v(x,y)$ satisfying $x=uv$ , $ y=1/u$. I got

\begin{equation} \frac{\partial u}{\partial x} = y , \frac{\partial u}{\partial y} = x \\ \frac{\partial v}{\partial x} = 1-y , \frac{\partial v}{\partial y} = -1/y^2 \end{equation}

Are these correct?

Surely you notice that $u=1/y$, $v=xy$. – John B Dec 29 '17 at 21:55 — John B, Dec 29 '17 at 21:55
I accidentally switched up u and v, thanks! – szantamano Dec 29 '17 at 22:03 — szantamano, Dec 29 '17 at 22:03

score 0 · Accepted Answer · answered Dec 29 '17 at 21:59

0

$u=\frac1y$ and $v=xy$ thus $$u_x=0 \quad u_y=-\frac1{y^2}$$

$$v_x=y \quad v_y=x$$

answered Dec 29 '17 at 21:59

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Calculating short partial derivatives

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