# Need help computing the partial derivatives of a vector function.

I need help computing the partial derivative shown below. I've never taken a course in vector analysis so I'm not if my previous attempts at solving the problem were even on the right track. If someone could show me step by step what to do, that would be a great help to me.

I need to compute

$$\frac{\partial{f}}{\partial{X_{0}}}$$

where

$$f(X_{0},X_{1},X_{2}) =\frac{W_{u}}{\mid\mid W_{u} \mid\mid}$$

and

$$W_{u} = a(X_{1} - X_{0}) + c(X_{2} - X_{0} )$$

All of the variables above denoted by capital letters are vectors in $\mathbb{R}^{3}$. I've been stuck on a couple of different aspects of this. Firstly, is there a "quotient rule" for taking this type of derivative? And does it work the same way as it does in single variable calculus? I assumed that this was the case and went along with the problem however when I did this I eded up with an equation that didn't really make sense to me (matrices, vectors, and scalars added together). If someone could please break down the process for solving this I would much appreciate it.

Thanks!

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## 2 Answers

Considering $f,X_0,X_1,X_2:{\Bbb{R}}^9\to{\Bbb{R}}^9$, instead of $f,X_0,X_1,X_2:{\Bbb{R}}^3\to{\Bbb{R}}^3$, this via the embedding, which asign -for example- $$\left(\!\begin{array}{c}\ x\\ y\\ z\\ \end{array}\!\right) \hookrightarrow \left(\!\begin{array}{c} x\\ y\\ z\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ \end{array}\!\right) = x \left(\!\begin{array}{c} 1\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ \end{array}\!\right)+y \left(\!\begin{array}{c} 0\\ 1\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ \end{array}\!\right)+\cdots$$ in one hand. And in the other being key to consider the linear map $W:{\Bbb{R}}^9\to{\Bbb{R}}^9$ as $$\left(\!\begin{array}{c} x\\ y\\ z\\ s\\ t\\ u\\ r\\ v\\ w \end{array}\!\right) \longmapsto \left(\!\begin{array}{c} -(a+b)x\\ -(a+b)y\\ -(a+b)z\\ as\\ at\\ au\\ br\\ bv\\ bw\end{array}\!\right),$$ from where one can get the $JW$, needed in $D_{X_0}W$ and intervine in $D_{X_0}f$, what the OP requires.

So, according to my computations the solution could be: $$D_{X_0}f=-\frac{(a+b)^2\|X_0\||}{\|W\|^2}f-\frac{a+b}{\|W\|}X_0.$$ I hope someone could corroborate this.

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standard ${\Bbb{R}}^9$-calculus allows to calculate $D_Xf=[Jf]X$ which is the usual matrices' multiplication. –  janmarqz Jan 10 '14 at 18:50

To differentiate with respect to a variable (i.e. $X_{i}$), you keep other variables constant and differentiate as you would in single variable calculus given that the coordinates are orthonormal (a result that is often taken for granted). You must calculate the norm of $W_{u}$ explicitly. The part I do not quite understand is what you mean by $||W_{u}||$ since $W_{u}$ is a real valued function. Anyway, I hope this helps!

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Here $f:{\Bbb{R}}^3\times{\Bbb{R}}^3\times{\Bbb{R}}^3\to{\Bbb{R}}^3$. So it makes sense to ask for $$D_{X_0}f$$. –  janmarqz Jan 10 '14 at 16:26