Editing Cramer's rule (section)

====Computing derivatives implicitly====
Consider the two equations <math>F(x, y, u, v) = 0</math> and <math>G(x, y, u, v) = 0</math>. When ''u'' and ''v'' are independent variables, we can define <math>x = X(u, v)</math> and <math>y = Y(u, v).</math>

An equation for <math>\dfrac{\partial x}{\partial u}</math> can be found by applying Cramer's rule.

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First, calculate the first derivatives of ''F'', ''G'', ''x'', and ''y'':

:<math>\begin{align}
dF &= \frac{\partial F}{\partial x} dx + \frac{\partial F}{\partial y} dy +\frac{\partial F}{\partial u} du +\frac{\partial F}{\partial v} dv = 0 \\[6pt]
dG &= \frac{\partial G}{\partial x} dx + \frac{\partial G}{\partial y} dy +\frac{\partial G}{\partial u} du +\frac{\partial G}{\partial v} dv = 0 \\[6pt]
dx &= \frac{\partial X}{\partial u} du + \frac{\partial X}{\partial v} dv \\[6pt]
dy &= \frac{\partial Y}{\partial u} du + \frac{\partial Y}{\partial v} dv.
\end{align}</math>

Substituting ''dx'', ''dy'' into ''dF'' and ''dG'', we have:

:<math>\begin{align}
dF &= \left(\frac{\partial F}{\partial x} \frac{\partial x}{\partial u} +\frac{\partial F}{\partial y} \frac{\partial y}{\partial u} + \frac{\partial F}{\partial u} \right) du + \left(\frac{\partial F}{\partial x} \frac{\partial x}{\partial v} +\frac{\partial F}{\partial y} \frac{\partial y}{\partial v} +\frac{\partial F}{\partial v} \right) dv = 0 \\ [6pt]
dG &= \left(\frac{\partial G}{\partial x} \frac{\partial x}{\partial u} +\frac{\partial G}{\partial y} \frac{\partial y}{\partial u} +\frac{\partial G}{\partial u} \right) du + \left(\frac{\partial G}{\partial x} \frac{\partial x}{\partial v} +\frac{\partial G}{\partial y} \frac{\partial y}{\partial v} +\frac{\partial G}{\partial v} \right) dv = 0.
\end{align}</math>

Since ''u'', ''v'' are both independent, the coefficients of ''du'', ''dv'' must be zero.  So we can write out equations for the coefficients:

:<math>\begin{align}
\frac{\partial F}{\partial x} \frac{\partial x}{\partial u} +\frac{\partial F}{\partial y} \frac{\partial y}{\partial u} & = -\frac{\partial F}{\partial u} \\[6pt]
\frac{\partial G}{\partial x} \frac{\partial x}{\partial u} +\frac{\partial G}{\partial y} \frac{\partial y}{\partial u} & = -\frac{\partial G}{\partial u} \\[6pt]
\frac{\partial F}{\partial x} \frac{\partial x}{\partial v} +\frac{\partial F}{\partial y} \frac{\partial y}{\partial v} & = -\frac{\partial F}{\partial v} \\[6pt]
\frac{\partial G}{\partial x} \frac{\partial x}{\partial v} +\frac{\partial G}{\partial y} \frac{\partial y}{\partial v} & = -\frac{\partial G}{\partial v}.
\end{align}</math>

Now, by Cramer's rule, we see that:

:<math>\frac{\partial x}{\partial u} = \frac{\begin{vmatrix} -\frac{\partial F}{\partial u} & \frac{\partial F}{\partial y} \\ -\frac{\partial G}{\partial u} & \frac{\partial G}{\partial y}\end{vmatrix}}{\begin{vmatrix}\frac{\partial F}{\partial x} & \frac{\partial F}{\partial y} \\ \frac{\partial G}{\partial x} & \frac{\partial G}{\partial y}\end{vmatrix}}.</math>

This is now a formula in terms of two [[Jacobian matrix and determinant|Jacobian]]s:

:<math>\frac{\partial x}{\partial u} = -\frac{\left(\frac{\partial (F, G)}{\partial (u, y)}\right)}{\left(\frac{\partial (F, G)}{\partial(x, y)}\right)}.</math>

Similar formulas can be derived for <math>\frac{\partial x}{\partial v}, \frac{\partial y}{\partial u}, \frac{\partial y}{\partial v}.</math>
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