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I am now studying total derivative of a function. The defination of the differentiability of the vector-valued function $\mathbf{f}$ is given in my book as follows :

Let $\mathbf{f} : S \longrightarrow \mathbb R^{m}$ be a function defined on a set $S \subset \mathbb R^{n}$ with values in $\mathbb R^{m}$.Let $\mathbf{c}$ be an interior point of $S$, and let $B(\mathbf{c}; r)$ be an $n$-ball lying in $S$. Let $\mathbf{v}$ be a point in $\mathbb R^{n}$ with $||\mathbf{v}|| < r$, so that $\mathbf{c} + \mathbf{v} \in B(\mathbf{c}; r)$.Then the function $\mathbf{f}$ is said to be differentiable at $\mathbf{c}$ if there exists a linear operator $T_{\mathbf{c}} : \mathbb R^{n} \longrightarrow \mathbb R^{m}$ such that

$\mathbf{f}(\mathbf{c} + \mathbf{v}) = \mathbf{f}(\mathbf{c}) + T_{\mathbf{c}}(\mathbf{v}) + ||\mathbf{v}|| E_{\mathbf{c}}(\mathbf{v})$ , where $E_{\mathbf{c}}(\mathbf{v}) \rightarrow \mathbf{0}$ as $\mathbf{v} \rightarrow \mathbf{0}$.The linear function $T_{\mathbf{c}}$ is called the total derivative of $\mathbf{f}$ at $\mathbf{c}$.

But my question is ''how can the total derivative be linear operator?''In particular if $f$ be a real valued function of real variable then if $f$ is differentiable at $c$ then I have a question.

Is $f'(c)$ the total derivative of $f$ at $c$?

If the answer is affirmative then how is the real number $f'(c)$ considered to be linear operator i.e. to be function.Please help me in understanding this concept.

Thank you in advance.

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  • $\begingroup$ Think Taylor series. $T_c(v) = f'(c) v$ is linear. $\endgroup$ – Ken Duna Dec 27 '16 at 16:11
  • $\begingroup$ the real question is how on earth did we get away with the derivative being a number before this point in the development... $\endgroup$ – James S. Cook Dec 27 '16 at 16:11
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In the case where $f:\mathbb R \to \mathbb R$, $T_c$ would be the linear operator defined by $T_c(v) = f'(c) v$.

Further comment: When $f:\mathbb R^n \to \mathbb R^m$ is differentiable at $c$, I like to define $f'(c)$ to be the $m \times n$ matrix that represents $T_c$ (with respect to the standard bases), so that: $$ \underbrace{T_c(v)}_{m \times 1} = \underbrace{f'(c)}_{m \times n} \underbrace{v}_{n \times1}. $$ In the case where $m = n = 1$, $f'(c)$ is just a $1 \times 1$ matrix, which is consistent with the fact that in introductory calculus courses $f'(c)$ is a scalar.

But note that some authors actually define $f'(c)$ to be the linear transformation itself, rather than the matrix that represents it.

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  • $\begingroup$ Do you mean @Olittle $T_{c} = f'(c)$? $\endgroup$ – Arnab Chatterjee. Dec 27 '16 at 16:14
  • $\begingroup$ What will be the linear operator if $f : \mathbb R^2 \longrightarrow \mathbb R$ is differentiable at $\mathbf{c} \in \mathbb R^{2}$?Is it defined as $T_{\mathbf{c}} (h,k) = Ah +Bk$,where $A$ and $B$ are the partial derivatives $D_{x}f(\mathbf{c})$ and $D_{y}f(\mathbf{c})$ of $f$ at $\mathbf{c}$ w.r.t. $x$ and $y$ respectively.If the answer is "yes" then what is $T_{\mathbf{c}}$ here? Because $T_{\mathbf{c}} (h,k)$ is not same as $T_{\mathbf{c}}$. $\endgroup$ – Arnab Chatterjee. Dec 27 '16 at 16:38
  • $\begingroup$ @ArnabChatterjee. Yes, that is what $T_c$ is in that case. In other words, $T_c$ is the linear operator defined by $T_c(h,k) = D_xf(c) h + D_y f(c) k$ for all $h,k$. That is the answer to your question "What is $T_c$ here?" There is no more explicit way to describe $T_c$ than that. By the way, using my preferred definition of $f'(c)$, in this case we would have $f'(c) = \begin{bmatrix} D_x f(c) & D_y f(c) \end{bmatrix}$. So $f'(c)$ is the matrix that represents the linear operator $T_c$. $\endgroup$ – littleO Dec 27 '16 at 23:49
  • $\begingroup$ I have understood the fact that $T_{c}$ is the total derivative or the differential coefficient of $f$ at $c$.But what does $T_{c} (h,k)$ signify?If $(h,k) \rightarrow (0,0)$ then can we say that $T_{c} (h,k) \rightarrow df$ where $df$ represents the total differential of $f$.If the answer is 'yes' then my next question "Is there any other significance of the value of $T_{c}$ at some point $(h,k)$? $\endgroup$ – Arnab Chatterjee. Dec 28 '16 at 5:13
  • $\begingroup$ Can we say that "If $f$ is differentiable at $c$ then $T_{c}$ is bounded on the domain $S \subset \mathbb R^{n}$ of $f$?" $\endgroup$ – Arnab Chatterjee. Dec 28 '16 at 5:39

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