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I am trying to understand the following definiton.

$f:\mathbb{R}^n \rightarrow \mathbb{R}^m$ . The total derivative of $f$ in point $a$ is the unique linear map $Df|_a$ such that $$\lim_{h \rightarrow 0}\dfrac{f(a+h)-f(a)-Df|_a(h)}{||h||} = 0$$

Could someone explain why this definition works?

-Why should we divide by $||h||$?

-Why is $Df|_a$ linear?

-How should I interpret this linear map $Df|_a$, what is the meaning of the total derivative?

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4  
If you don't divide by $||h||$ then it will be true for all linear maps and it will just express the fact that $f$ is continuous at $a$. – Captain Lama Mar 12 at 13:10
up vote 6 down vote accepted

This is one of the most fundamental definitions in all of analysis.

It says that the increment $\Delta f:=f(a+h)-f(h)$ of the function value should in first approximation be a linear function of the increment $h$ attached at the point $a$. In other terms: We want $$f(a+h)-f(a)=Lh +r(h)\qquad(|h|\ll1)\ ,\tag{1}$$ whereby the error $r(h)$ should be smaller by magnitudes than the linear term $Lh$ when $h$ is small. Now in general $|Lh|$ will be of order $|h|$ for certain $h$. This means that we should require that $$\lim_{h\to0}{|r(h)|\over |h|}=0$$ in order to impart any real content to $(1)$. It turns out that this condition determines $L$ uniquely. If it can be satisfied then $f$ is called differentiable at $a$, and one denotes the resulting $L$ by $Df\bigr|_a$, or similar.

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I think it should be noted that in the OP's definition the existence of the differential is given for granted, which should not be the case. – Massimo Ortolano Mar 12 at 19:11

I think it's much easier to understand if you write the definition as "$Df_{|a}$ is the unique linear map $L$ (if it exists) satisfying $f(x) = f(a) + L(x-a) + o(||x-a||)$ when $x\rightarrow a$".

Maybe it's even clearer if you write "$Df_{|a}$ is the linear part of the unique affine map $A$ (if it exists) satisfying $f(x) = A(x) + o(||x-a||)$ when $x\rightarrow a$".

So you can see that $A$ is the best possible affine approximation of $f$ near $a$ (because the error you make by replacing $f$ with $A$ is negligible compared to any affine map), and $Df_{|a}$ is the linear part of this affine approximation.

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The division by $\|h\|$ here is exactly analogous with the division by $h$ in the definition of the (standard) derivative of a real-valued function of a real variable:

$\dfrac{df}{dx}=\underset{h\rightarrow 0}{\lim}\dfrac{f(x+h)-f(x)}{h}$

To answer the second question, $Df|_a$ is linear because it satisfies the linearity property, that is it commutes with addition and scalar multiplication. This is a consequence of how it is defined and is not actually the hard to prove (try using the definition to show $Df|_a+Dg|_a=D(f+g)|_a$ and $D(\alpha f)|_a=\alpha Df|_a$ directly; hint, use some linear algebra)

To answer the final question, it is the natural analog of the familiar derivative of a function $f:\mathbb{R} \rightarrow \mathbb{R}$ for the case of a function from $f:\mathbb{R}^n \rightarrow \mathbb{R}^m$.

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You are correct that the linearity of $Df\mid_a$ can be proved easily from the definition. This is the proof: "The definition requires $Df\mid_a$ to be linear. Ergo, if it exists, it is linear". It is possible to pick non-linear functions with this same property. But the very purpose of the derivative is to be a linear approximation to the original function. So in the definition is intentionally restricted to linear functions. – Paul Sinclair Mar 12 at 17:34
    
I was specifically referring to how $Df|_a$ is defined (by the determinantal formula, which only works if $f$ is differentiable), it is a consequence of that definition that it is linear. – Justin Benfield Mar 12 at 17:38
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Not sure what you even mean by the "determinantal formula" (the matrix of partial derivatives, maybe?), but the definition is as given in the OP. Everything else is a theorem, not a definition. – Paul Sinclair Mar 13 at 0:27
    
Yes, I was referring to the matrix of partial derivatives (actually the $\det$ of that matrix, which is the formula). It can just as well be taken as the definition of the total derivative in this case (I think, maybe some other assumptions are needed, but the given definition should be able to derived from it). Looking at it from that formula helps motivate the more abstract definition given to the OP. – Justin Benfield Mar 13 at 16:12
    
The jacobian is not the same thing as the derivative. It is useful for change of coordinates in integration and a few other applications, but the purpose of the derivative is to act as a linearized version of the original function, and the jacobian is just not up to that task. The matrix itself (as a linear operator) instead of its determinant does serve as a derivative, but is defined in terms of a specific coordinate system. Thus the version given in the OP is preferred, as it does not depend on a coordinate system. – Paul Sinclair Mar 13 at 17:57

Just to add a bit to the previous answers, you can check that this definition meets the "usual" requirement for differentiability if $f:\mathbb R\to \mathbb R.$

Take $x=a$. Then, according to the "new" definition, to find the derivative of $f$ at $a$, we seek an

$L(a):\mathcal L(\mathbb R,\mathbb R)\to \mathcal L(\mathbb R,\mathbb R)$ such that

$\lim _{h\to 0}\frac{f(a+h)-f(a)-L(a)h}{h}=0$.

Now, the linear tranformations from $\mathbb R\to \mathbb R$ are of the form $fh=bh$ for some $b\in \mathbb R$, so we have now, with $f=L(a)$

$\lim _{h\to 0}\frac{f(a+h)-f(a)-bh}{h}=0$, which simplifies to

$\lim _{h\to 0}\left ( \frac{f(a+h)-f(a)}{h}-b \right )=0$ so that $b=f'(a)$, that is, $L(a)h=f'(a)h$

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