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I have a question regarding the differential $d_{\textbf a} f$.

Suppose we have the function $f(x,y)= xy$, and the vectors $\textbf a = (1,1)$ and $\textbf u = (2,1)$. Then, if I understand this correctly, $d_{\textbf a} f(\textbf u) = \nabla f(\textbf a) \cdot \textbf u = (1,1)\cdot (2,1) = 2+1 = 3,$ where $\nabla f(\textbf a) = (\partial f/\partial x, \partial f/\partial y)$. But what if my assignment is to calculate $d_{\textbf a} f$? I don't know what it means. Do they want me to calculate $d_{\textbf a} f(x,y) = (1,1)\cdot (x,y) = x+y$, or something else?

Edit: Note that it is not the directional derivative that I'm asking about.

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    By the way: your calculations are correct and you're almost done, the calculations need only be interpreted and I'll explain it in$a$moment. Concerning your question to Jesse: Yes, $\mathbf{a}$ is$a$point in $\mathbb{R}^2$, that is to say a vector.2011-05-30

2 Answers 2

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Essentially, you have worked out everything already, but there seems to be a bit of confusion about the definitions, so let me try to set this straight.

The differential of $f$ at the point $\mathbf{a} \in \mathbb{R}^2$ is the row matrix $ d_{\mathbf{a}}f = \begin{pmatrix} \frac{\partial}{\partial x} f(\mathbf{a}) & \frac{\partial}{\partial y}f (\mathbf{a}) \end{pmatrix}.$

Now if you write $d_{\mathbf{a}}f (\mathbf{u})$ for $\mathbf{u} = \begin{pmatrix} u_1 \\\ u_2 \end{pmatrix} \in \mathbb{R}^2$ you're meaning the matrix product $d_{\mathbf{a}}f (\mathbf{u}) = \begin{pmatrix} \frac{\partial}{\partial x} f(\mathbf{a}) & \frac{\partial}{\partial y}f (\mathbf{a}) \end{pmatrix} \cdot \begin{pmatrix} u_1 \\\ u_2 \end{pmatrix} = \frac{\partial}{\partial x} f(\mathbf{a}) \cdot u_1 + \frac{\partial}{\partial y}f (\mathbf{a}) \cdot u_2 .$

On the other hand, $\nabla f (\mathbf{a})$ is the column vector $ \nabla f (\mathbf{a}) = \begin{pmatrix} \frac{\partial}{\partial x} f(\mathbf{a}) \\\ \frac{\partial}{\partial y}f (\mathbf{a}) \end{pmatrix}$ and when you're writing $\nabla f (\mathbf{a}) \cdot \mathbf{u}$ you're meaning the scalar product $\nabla f( \mathbf{a}) \cdot u = \begin{pmatrix} \frac{\partial}{\partial x} f(\mathbf{a}) \\\ \frac{\partial}{\partial y}f (\mathbf{a}) \end{pmatrix} \cdot \begin{pmatrix} u_1 \\\ u_2 \end{pmatrix} = \frac{\partial}{\partial x} f(\mathbf{a}) \cdot u_1 + \frac{\partial}{\partial y}f (\mathbf{a}) \cdot u_2 . $

So we see that for $f(x,y) = xy$ $d_{\mathbf{a}}f = \begin{pmatrix} y & x \end{pmatrix} \qquad \text{while} \qquad \nabla f (\mathbf{a}) = \begin{pmatrix} y \\\ x \end{pmatrix}.$

Now the confused reaction was due to the fact that the notation used here for the derivative of $f$ at the point $\mathbf{a}$ is often used as the directional derivative, and as you rightly pointed out in a comment, we have the relations $ D_{\mathbf{u}} f (\mathbf{a}) : = d_{\mathbf{a}} f (\mathbf{u}) = \nabla f(\mathbf{a}) \cdot \mathbf{u},$ and everything should be fine now, no?

Since you made the computations yourself already, I'll not repeat them here.

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    @Eivind: In fact, it took me quite some time to see why these rather subtle distinctions are made and why they are so important. Don't worry too much about them now, they will become much clearer soon. Try to get used to calculating the various forms of derivatives, try to get some feeling for them and as soon as you've mastered the formal busines, the theoretical distinctions will become much easier than they may seem now. That's the best piece of advice I can give you right now (and many people might disagree).2011-05-31
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In the case of functions $f\colon \mathbb{R}^n \to \mathbb{R}$, like $f(x,y) = xy$ as you have, the differential $d_af$ is the same thing as the gradient $\nabla f(a)$.

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    @Theo: Yes, you're right of course, but I was trying to keep it simple.2011-05-30