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Consider the next transformations $$ S:\left\{\begin{array}{ll} x=u^{2}-v^{2}\\y=3uv & and \quad T:\left\{\begin{array}{l} u=t^{2}-s+s t \\ v=s^{2}-\frac{2 s}{t}-3 \end{array}\right. \end{array}\right. $$ Calculate the Jacobian matrix $\frac{\partial(x, y)}{\partial(s, t)}(-1,2)$

What should I do?

do $x=(t^2-s+st)^{2}-(s^2-\frac{2s}{t}-2)^2$ and

$y=3(t^2-s+st)(s^2-\frac{2s}{t}-2)^2$

and then build the matrix $2\times2$ with the partials respect $s$ and $t$ of $x=(t^2-s+st)^{2}-(s^2-\frac{2s}{t}-2)^2$ and $y=3(t^2-s+st)(s^2-\frac{2s}{t}-2)^2$ ?

what I mean

$\begin{equation} \begin{pmatrix} \frac{\partial x}{\partial s} & \frac{\partial y}{\partial s}\\ \frac{\partial x}{\partial t} & \frac{\partial y}{\partial t} \end{pmatrix} \end{equation}$ in $(-1,2)=$

$\begin{equation} \begin{pmatrix} 0 & 17\\ -30 & \frac{-27}{2} \end{pmatrix} \end{equation}$

is right?

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1 Answer 1

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That is definitely a correct way to go about this problem. Note that what you did is apply the chain rule for a scalar function multiple times. This is equivalent to using the chain rule for Jacobian-style derivatives as

$$\frac{\partial (x,y)}{\partial (s,t)} = \frac{\partial (x,y)}{\partial (u,v)}\frac{\partial (u,v)}{\partial (s,t)} = \begin{bmatrix} \frac{\partial x}{\partial u} \frac{\partial u}{\partial s} + \frac{\partial x}{\partial v} \frac{\partial v}{\partial s} & \frac{\partial x}{\partial u} \frac{\partial u}{\partial t} + \frac{\partial x}{\partial v} \frac{\partial v}{\partial t} \\ \frac{\partial y}{\partial u} \frac{\partial u}{\partial s} + \frac{\partial y}{\partial v} \frac{\partial v}{\partial s} & \frac{\partial y}{\partial u} \frac{\partial u}{\partial t} + \frac{\partial y}{\partial v} \frac{\partial v}{\partial t}\end{bmatrix}\; .$$

Once again, you are completely correct in your approach but I want to provide a slightly different way to compute it that you may find more elegant.

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