I am currently reading R.G Bartle's "Elements of Real Analysis" for a one semester course in Advanced Real Analysis. In the chapter on Differentiation in $\mathbb{R}^p$, I am confused regarding the terminologies and the notations used. I list the basic doubts :

$1$. What do you mean when you say derivative of $f:A\subseteq\mathbb{R}^p\to\mathbb{R}^q$ is a linear map L from $\mathbb{R}^p\to \mathbb{R}^q $ ?

$2.$ What does the notation $Df(c)(u)$ mean ?

$3$. What do you mean when you say partial derivative of f with respect to an arbitratry vector $u\in \mathbb{R}^p $ ?

$4.$ Given $f(x) = \begin{cases} \dfrac{xy^2}{x^2+y^2}, &\text{if }(x,y)\ne(0,0) \\ 0, &\text{if }(x,y)=(0,0). \end{cases}$

How is $D_{(a,b)}f(0,0)=\dfrac{ab^2}{a^2+b^2},(a,b)\ne(0,0 )?$

I need help in understanding these. Thanks

  • $\begingroup$ There are lots of questions here! Looking at the first one. The crux of several variables, is that it is no longer helpful to think of the derivative as a number (even a vector). You have to start thinking of it as a linear approximation. In other words a map. Hence tangent spaces, fibres etc. But I am the wrong person to answer this - I know nothing about Bartle's book. $\endgroup$ – almagest Sep 13 '14 at 6:40
  • $\begingroup$ But it is certainly a strange introductory book if he does not explain basic notation ... :) $\endgroup$ – almagest Sep 13 '14 at 6:53
  • $\begingroup$ If these things are all obscure to you, you might want to get a different book to complement the Bartle. Hubbard & Hubbard, Vector Calculus is the best book on the subject of multivariable calculus. $\endgroup$ – Kevin Arlin Sep 13 '14 at 7:06
  • $\begingroup$ Are you reading the first or second edition?? My knowledge on $\Bbb R^p$ is minimal. But I have the book. And the notations seem to be well-defined in pages 347 - 349 in the second edition. $\endgroup$ – Ishfaaq Sep 13 '14 at 7:09
  • $\begingroup$ @Ishfaaq, 2nd edition $\endgroup$ – creative Sep 13 '14 at 9:00

Disclaimer: I don't know the book you are reading, but I know something about differentiation on $\Bbb{R}^n$ (or more generally on Banach spaces).

  1. Here, it is important to say that the derivative of $f$ at some fixed point $x \in A$ is a linear map, written $Df(x) : \Bbb{R}^p \to \Bbb{R}^q$, so that the value of the linear map $Df(x)$ at $v \in \Bbb{R}^p$ is $Df(x)(v) \in \Bbb{R}^q$. This also answers part (2) of your question. Personally, I prefer to write this as $Df(x)\langle v \rangle$.

    Here, I have written $\Vert \cdot \Vert$ and $|\cdot|$ to distinguish the different norms (one on $\Bbb{R}^q$, one on $\Bbb{R}^p$).

    For the exact definition of this map, $Df(x)$ is the unique linear map so that $$ \lim_{y \to x} \frac{\Vert f(y) - f(x) - Df(x) \langle y-x \rangle \Vert}{|x-y|} = 0, $$ if such a map exists (then it is unique). This means that the map $y \mapsto f(y) + Df(x)\langle y -x \rangle$ is the best (affine-) linear approximation to $f$ at $x$.

  2. See (1)
  3. This is normally called the directional derivative and not the partial derivative w.r.t. an arbitrary vector. The directional derivative $D_v f (x)$ of $f$ in direction $v$ at $x$ is usually defined as $$ D_v f(x) := \lim_{t \to 0} \frac{f(x + tv) - f(x)}{t}. $$ You should check that $D_v f(x) = Df(x)\langle v \rangle$ holds, if $Df(x)$ exists, i.e. if $f$ is differentiable at $x$.
  4. Let us plug in the definition and calculate. \begin{eqnarray*} D_{\left(a,b\right)}f\left(0,0\right) & = & \lim_{t\to0}\frac{f\left(\left(0,0\right)+t\left(a,b\right)\right)-f\left(\left(0,0\right)\right)}{t}\\ & = & \lim_{t\to0}\frac{\left(ta\right)\left(tb\right)^{2}}{\left(ta\right)^{2}+\left(tb\right)^{2}}\\ & = & \lim_{t\to0}\frac{t^{3}}{t^{2}}\frac{ab^{2}}{a^{2}+b^{2}}=0\neq\frac{ab^{2}}{a^{2}+b^{2}} \end{eqnarray*} in general, so I think that you (or the book) miswrote the definition of $f$ (check it).

    The point of an example like this (replace e.g. the $b^2$ in the nominator by $b$ and repeat the calculation) is to show that it can happen that all partial derivatives exist without the function being differentiable at $x$ (at $0$ in this case). You can even deduce that $f$ is not differentiable, because otherwise, part (3) would show that the directional-derivative map $$ v \mapsto D_v f(x) = Df(x)\langle v \rangle $$ is linear, which is not the case in your example.

EDIT: Maybe the following can also help you. In the setting of $\Bbb{R}^p$ and $\Bbb{R}^q$, any linear map $T : \Bbb{R}^p \to \Bbb{R}^q$ can always be identified with its matrix (w.r.t. the standard basis), i.e. we identify the matrix $A \in \Bbb{R}^{q \times p}$ with the linear map

$$ x \mapsto Ax. $$

With this notation, the linear map $Df(x)$ is identified with the Jacobian matrix containing the partial derivatives, i.e.

$$ Df(x) \leftrightarrow \left(\frac{\partial f_i (x)}{\partial x_j} \right)_{i,j}, $$ where I used the standard notation

$$ \frac{\partial f_i (x)}{\partial x_j} := D_{e_j} f_i (x), $$ where $e_j$ is the $j$-th vector of the standard basis and $f_i$ is the $i$-th component of $f$.

In fact, sometimes the derivative in this setting is defined as a matrix instead of a linear map, so that $\widetilde{D} f(x)$ (I just invented that notation) is defined as the unique matrix such that

$$ \lim_{y \to x} \frac{\Vert f(y) - f(x) - \widetilde{D}f(x) \cdot (y-x) \Vert}{|y - x|} = 0. $$

So, why do we bother with the (seemingly) more complicated definition using linear maps instead of plain old matrices? The answer is that this becomes much more convenient in the setting of spaces that are finite-dimensional, but do not look like $\Bbb{R}^n$ in a natural way. For example, consider the determinant map

$$ \det : \Bbb{R}^{n\times n} \to \Bbb{R}. $$

You could of course identify $\Bbb{R}^{n \times n}$ with $\Bbb{R}^{n^2}$ by introducing some arbitrary enumeration of the coordinates (or even some other basis), but this will make it complicated (not painless) to compute the matrix-derivative.

Instead, it is not so hard to compute the linear-map version of the derivative (at least at the identity matrix $I$). First, note that the determinant is certainly differentiable (because it is a polynomial map), so that it suffices to calculate the directional derivative

$$ D_A \det(I) = \lim_{t \to 0} \frac{\det (I + tA) - \det(I)}{t} = \rm{tr}(A), $$

where the last equality comes from the fact that the Leibniz formula yields

$$ \det(I + tA) = \sum_\sigma \rm{sgn}(\sigma) \cdot \prod_i (I + tA)_{i, \sigma(i)} = 1 + t \cdot \rm{trace}(A) + t^2 \cdot \ast, $$

where $\ast$ is a polynomial in $t$, as you can check.

One can also calculate the derivative at other points than at the identity, but I willl skip that here.

  • $\begingroup$ Thanks a lot !! I am able to get some picture now. I have one doubt. Can you give me a non trivial example of $f:\mathbb{R}^p \to \mathbb{R}^q$ and show me that the derivative is a linear map indeed ? $\endgroup$ – creative Sep 19 '14 at 7:25
  • $\begingroup$ I don't completely understand the question. The derivative is a linear map by definition if it exists. Or do you want to have an explicit proof of $\Vert f(x + h) - f(x) - Df(x)\langle h \rangle \Vert/|h| \to 0$ for some non trivial, but explicit, $f$? $\endgroup$ – PhoemueX Sep 19 '14 at 8:39
  • $\begingroup$ I just mean that take an example and show me, if it's possible $\endgroup$ – creative Sep 19 '14 at 8:50

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