# Jacobian and chain rule

Let $u$, $v$ be differentiable functions of $x$, $y$ and $x$, $y$ be differentiable functions of $s$, $t$. Then the connection

$\frac{\partial (u,v)}{\partial (s,t)}}={\displaystyle \frac{\partial (u,v)}{\partial (x,y)}}\cdot {\displaystyle \frac{\partial (x,y)}{\partial (s,t)}$ | (1) |

between the Jacobian determinants is in .

*Proof.* Starting from the right hand side of (1), where one can multiply the determinants^{} (http://planetmath.org/Determinant2) similarly as the corresponding matrices (http://planetmath.org/MatrixMultiplication), we have

$$\left|\begin{array}{cc}\hfill {u}_{x}\hfill & \hfill {u}_{y}\hfill \\ \hfill {v}_{x}\hfill & \hfill {v}_{y}\hfill \end{array}\right|\cdot \left|\begin{array}{cc}\hfill {x}_{s}\hfill & \hfill {x}_{t}\hfill \\ \hfill {y}_{s}\hfill & \hfill {y}_{t}\hfill \end{array}\right|=\left|\begin{array}{cc}\hfill {u}_{x}{x}_{s}+{u}_{y}{y}_{s}\hfill & \hfill {u}_{x}{x}_{t}+{u}_{y}{y}_{t}\hfill \\ \hfill {v}_{x}{x}_{s}+{v}_{y}{y}_{s}\hfill & \hfill {v}_{x}{x}_{t}+{v}_{y}{y}_{t}\hfill \end{array}\right|=\left|\begin{array}{cc}\hfill {u}_{s}\hfill & \hfill {u}_{t}\hfill \\ \hfill {v}_{s}\hfill & \hfill {v}_{t}\hfill \end{array}\right|.$$ |

Here, the last stage has been written according to the general chain rule^{} (http://planetmath.org/ChainRuleSeveralVariables). But thus we have arrived at the left hand side of the equation (1), which hereby has been proved.

Remark. The rule (1) is only a visualisation of the more general one concerning the case of functions of $n$ variables.

Title | Jacobian and chain rule |
---|---|

Canonical name | JacobianAndChainRule |

Date of creation | 2013-03-22 18:59:45 |

Last modified on | 2013-03-22 18:59:45 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 4 |

Author | pahio (2872) |

Entry type | Example |

Classification | msc 15-00 |

Classification | msc 26B05 |

Classification | msc 26B10 |