Jacobian matrix and chain rule

**Jonroberts74** · Jun 28th 2014, 11:09 AM

question gives

$x=t^2-s^2, y=ts,u=x,v=-y$

a) compute derivative matrix (jacobian matrix)

$f(u,v)=f(t^2-s^2,-(ts))$

$\vec{D}f(u,v) = \left[ \begin{array}{cc}2t&-2s\\-s&-t\end{array}\right]$

b) express (u,v) in terms of (t,s)

$(u,v) = (t^2-s^2, -(ts))$

c) calculate $\vec{D}(u,v)$

$\left[\begin{array}{cc}1&-1\end{array}\right] \left[\begin{array}{cc}2t&-2s\\s&t\end{array}\right] = \left[\begin{array}{cc}2t-s&-2s-t\end{array}\right]$

d) verify chain rule holds

$f(t^2-s^2,-(ts))$ not sure how to go about verifying this when im not sure if the rest is correct

**Jonroberts74** · Jun 29th 2014, 02:41 PM

re-did this one

a) compute derivative matrices

$\vec{D}f(x,y) = \left[\begin{array}{cc}2t&-2s\\s&t\end{array}\right]$

$\vec{D}f(u,v) = \left[\begin{array}{cc}1&0\\0&-1\end{array}\right]$

b) express (u,v) in terms of (t,s)

$f(u(x,y),v(x,y) = (t^2-s2,-(ts))$

c) evaluate $\vec{D}(u,v)$

$\vec{D}(u,v) = \left[\begin{array}{cc}1&0\\0&-1\end{array}\right] \left[\begin{array}{cc}2t&-2s\\s&t\end{array}\right]$

$= \left[\begin{array}{cc}2t&-2s\\-s&-t\end{array}\right]$

d) verify chain rule holds

this one I am still not sure how to show if it holds

**Dark Sun** · Jul 5th 2014, 01:09 AM

You have already practically shown d)

$Df(u(x,y),v(x,y))=\left[\begin{array}{cc}2t & -2s \\ -s & -t \end{array} \right]=\left[\begin{array}{cc}1 & 0 \\ 0 & -1 \end{array} \right]\cdot \left[\begin{array}{cc}2t & -2s \\ s & t \end{array} \right]=Df(u,y)\cdot Df(x,y)$

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