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[tex]x=t^2-s^2, y=ts,u=x,v=-y[/tex]

a) compute derivative matrices

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

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

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

[tex]f(u(x,y),v(x,y) = (t^2-s^2,-(ts))[/tex]

c) Evaluate [tex]\vec{D}(u,v)[/tex]

[tex]\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][/tex]

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

d) verify if chain rule holds

need help with this last part, also need to know if I even did the rest correctly

a) compute derivative matrices

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

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

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

[tex]f(u(x,y),v(x,y) = (t^2-s^2,-(ts))[/tex]

c) Evaluate [tex]\vec{D}(u,v)[/tex]

[tex]\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][/tex]

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

d) verify if chain rule holds

need help with this last part, also need to know if I even did the rest correctly

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