- #1
richyw
- 180
- 0
Homework Statement
Let [itex](u,v)=\mathbf{f}(x,y,x)=(2x+y^3,xe^{5y-7z})[/itex]
Compute [itex]D\mathbf{f}(x,y,z),\;\partial (u,v)/\partial(x,y),\;\partial (u,v)/\partial(y,z)\text{ and }\partial (u,v)/\partial(x,z)[/itex]
Homework Equations
-chain rule
The Attempt at a Solution
well I can get [tex]D\mathbf{f}(x,y,z)=\left[\begin{matrix} \partial u / \partial x & \partial u / \partial y & \partial u / \partial z \\ \partial v / \partial x & \partial v / \partial y & \partial v/ \partial z \end{matrix}\right]=\left[\begin{matrix} 2 & 3y^2 & 0 \\ e^{5y-7z} & 5xe^{5y-7z} & -7xe^{5y-7z}\end{matrix}\right][/tex]and I can see that if I ignore the variable that is not in the partial, and take the determinant of that 2x2 matrix, then I get the partial derivative I want. My questions is what am I doing? what happens if it is a 3x2 matrix instead of a 2x3? does this "method" still work?