I need help in finding the partial derivative in a specific point.

Let $x$, $y$, $u$ and $v$ be variables for which this relationship is true:

$$ \left\{ \begin{array}{c} x^2 + xy - y^2=u \\ 2xy + y^2=v \\ \end{array} \right. $$ How do I find $(\frac{\partial x}{\partial u})_{v}$ for $x = 2$ and $y = -1$ ? As far as I know you need an expression for $x = x(u,v)$ in terms of only $v$ and $u$, right ?

  • $\begingroup$ What does the notation $(\frac{\partial x}{\partial u})_{v}$ stand for? (that $v$ there) And by the way, have you tried implicit derivation on each equation? $\endgroup$
    – Javi
    Jun 10, 2018 at 20:14

1 Answer 1


We have that

$$du = \frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy=(2x+y)dx+(x-2y)dy=3\,dx+4\,dy$$

$$dv = \frac{\partial v}{\partial x}dx+\frac{\partial v}{\partial y}dy=2ydx+(2x+2y)dy=-2\,dx+2\,dy$$

from here we can find

$$dx = \frac{\partial x}{\partial u}du+\frac{\partial x}{\partial v}dv$$

$$dy = \frac{\partial y}{\partial u}du+\frac{\partial y}{\partial v}dv$$


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