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In what sense the derivative is linear?. In Bartle's book The elements of real analysis takes as an observation the case f:R$\rightarrow$ R and says f is diferentiable at c iff the derivative f'(c) of f exists at c. In this case the derivative of f at c is linear function on R to R wich sends the real number u into the real number f'(c)u.

Then I took f(x)=$x^3$ , f'(c)=3$c^2$ at c$\neq$0 how do I define the derivative to be linear?. Obviously f'(a+b)$\neq$f'(a)+f'(b) for a, b $\neq$ 0 in general, so this isn't the sense, then I took $\phi$(x)=f'(c)(x) and yes this is linear for any f on R to R with derivative at c, because f'(c) is a number and $\phi$(x) is just a linear function in the sense $\phi$(x)=ax , where a is constant

So my question is this last sense the one to take to understand that a derivative at a point is a linear function? thanks beforehand.

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The confusion here is that there are (at least) two things that get called the derivative - the linear function $f'(c)(x - c) + f(c)$ of $x$ which approximates $f(x)$ at $c$, and the probably nonlinear function $f'(c)$ of $c$ which describes the family of linear approximations to $f$. – Qiaochu Yuan Jun 9 '11 at 18:27
The linearity of the derivative means that , locally, the change of the function can be approximated to any degree of accuracy--in a precise del-eps. sense--by a linear function. So, if f is , say $x^3$, then, near any point xo in the domain, the change $f(xo+h)-f(xo)$ is as close to the change $3xo^2(x+h)-3xo^2x$ as you want it to be, i.e., by shrinking the value of h, you can make the approximation as good as you want it to be. – gary Jun 9 '11 at 18:34
@Qiaochu Yuan but what about the fact that the derivative of $f$ at c is linear function witch sends the real number u into the real number $f'(c)u$ the other definition of derivative that you point doesn't satisfy this but the transformation $h\rightarrow f'(c)h$ does – Ivan3.14 Jun 9 '11 at 18:39
up vote 2 down vote accepted

As Chandru noted, the derivative, when viewed in generality, is simply a linear approximation of a map at a specified point that satisfies certain conditions. Intuitively, these conditions imply that the derivative is the best linear approximation at that point. To consider your specific example, as you noted, the deriviative of $f$ is just $f'(x) = 3x^2$. At any point $a$ in its domain the derivative has the value $3a^2$. This determines then a linear map $$ h \mapsto (3a^2)h $$ If you will replace this map into the definition of the derivative that Chandru provided you will see that the definition is satisfied. It can also be shown then that this is the only map that will satisfy it, i.e., it is unique.

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i'm cofused here because h$\rightarrow$3$a^2$h is not the derivative is another function let say associated function to the point a and is this one that can be pointed as linear at the point a. I can't see that the derivative at some point is linear function in the case f from R to R as Bartle says – Ivan3.14 Jun 9 '11 at 18:17
The derivative of a function $f$ from $R$ to $R$, when considered as a function $f':R\rightarrow R$, is not linear in general. The derivative is only linear once evaluated at a specific point. In the case of a real valued function of a real variable this is trivial because when evaluated at a point the derivative is just a scalar, and so it determines a linear function on reals. (edit : see Qiaochu's comment to the question) – Vhailor Jun 9 '11 at 18:31
@missing314 The map $A(h) = 3a^2h$ is, by definition, the derivative of $f$ at the point $x = a$. At some other point b, the derivative of $f$ is given by $B(h) = 3b^2h$. The key here is that the derivative, at a specified fixed point, is a linear map. I realize this point of view may be confusing since you're probably not accustomed to viewing the derivative this way. – ItsNotObvious Jun 9 '11 at 18:36
Oh so to conclude this. This $A(h)=3a^2h$ and $f'(x)=3x^2$ are called derivatives where A(h) is derivative at the point a. Ok thanks everyone – Ivan3.14 Jun 9 '11 at 18:51

Given a value $c$, you have $f(c)=c^3$. Now for a point near $c$, call it $c+\delta$, you have $f(c+\delta)=(c+\delta)^3=c^3+3c^2\delta+3c\delta^2+\delta^3$. If $\delta$ is small, the terms in $\delta^2$ and $\delta^3$ are smaller yet, so we can ignore them and get $f(c+\delta)=(c+\delta)^3\approx c^3+3c^2\delta$, which is a linear function of $\delta$. This is the linear function that best approximates $f$ near $c$ and Taylor's theorem tells you about the error committed.

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To elaborate the point a bit, the idea of the derivative is this: We are given a - maybe complicated - function $$ f : \mathbb{R} \to \mathbb{R} $$ If we fix a point p, is there then a linear function $$ a_p : \mathbb{R} \to \mathbb{R} $$ that approximates f (strictly speaking it is linear with respect to a new basepoint, namely (x, f(x)), but you'll know what I mean), so that we can use this linear function to learn something about f around the point p?

The main point is that we fixed a point p beforehand, and then asked if we can get a linear function. Therefore, if we vary p, we get different linear functions for every single p. If we denote the space of linear functions by $Hom(\mathbb{R}, \mathbb{R})$, then we actually have for differentiable $f$, that the differential is a mapping $$ p \mapsto a_p \in Hom(\mathbb{R}, \mathbb{R}) $$ For $\mathbb{R}$, with respect to a basis (consisting of exactly one base vector), every linear function can be identified with a 1-1-matrix, therefore we can write for the differential $$ p \mapsto [a_p] $$ where $[a_p]$ is the linear transformation represented as 1-1-matrix. So, in one dimension, incidentally we can abbreviate the whole procedure and say that the differential gives us for every real number p another real number, $a_p$. With respect to your example, we write in short $$ f(x) = x^3 $$ then $$ f'(p) = 3 \; p^2 $$ which means for every real number p we get the linear transformation $x \mapsto 3 p^2 x$.

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In two more ways, too. The derivative is often called the 'best linear approximation' - in the sense that you can use it Newton-style to approximate functions. This is Ross's answer (which appeared just now to me).

But the act of taking a derivative is also linear, in the sense that if f and g are both differentiable and $\alpha$ and $\beta$ are scalars, then $\frac{d}{dx} (\alpha f (x) + \beta g(x) ) = \alpha \frac{df}{dx} (x) + \beta \frac{dg}{dx} (x) $.

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