6.3: Differentiable Functions

As we know, a function $f : E^{1} \rightarrow E\left(\text {on } E^{1}\right)$ is differentiable at $p \in E^{1}$ iff, with $\Delta f=f(x)-f(p)$ and $\Delta x=x-p$,

$$f^{\prime}(p)=\lim _{x \rightarrow p} \frac{\Delta f}{\Delta x} \text { exists } \quad(\text { finite }).$$

Setting $\Delta x=x-p=t, \Delta f=f(p+t)-f(p),$ and $f^{\prime}(p)=v,$ we may write this equation as

$$\lim _{t \rightarrow 0}\left|\frac{\Delta f}{t}-v\right|=0,$$

or

$$\lim _{t \rightarrow 0} \frac{1}{|t|}|f(p+t)-f(p)-v t|=0$$

Now define a map $\phi : E^{1} \rightarrow E$ by $\phi(t)=t v, v=f^{\prime}(p) \in E$.

Then $\phi$ is linear and continuous, i.e., $\phi \in L\left(E^{1}, E\right) ;$ so by Corollary 2 in §2, we may express $(1)$ as follows: there is a map $\phi \in L\left(E^{1}, E\right)$ such that

$$\lim _{t \rightarrow 0} \frac{1}{|t|}|\Delta f-\phi(t)|=0.$$

We adopt this as a definition in the general case, $f : E^{\prime} \rightarrow E,$ as well.

Definition: Differentiable at a Point

A function $f : E^{\prime} \rightarrow E$ where $E^{\prime}$ and $E$ are normed spaces over the same scalar field) is said to be differentiable at a point $\vec{p} \in E^{\prime}$ iff there is a map

$$\phi \in L\left(E^{\prime}, E\right)$$

such that

$$\lim _{\vec{t} \rightarrow \overrightarrow{0}} \frac{1}{|\vec{t}|}|\Delta f-\phi(\vec{t})|=0;$$

that is,

$$\lim _{\vec{t} \rightarrow \overrightarrow{0}} \frac{1}{|\vec{t}|}[f(\vec{p}+\vec{t})-f(\vec{p})-\phi(\vec{t})]=0.$$

As we show below, $\phi$ is unique (for a fixed $\vec{p} ),$ if it exists.

We call $\phi$ the differential of $f$ at $\vec{p},$ briefly denoted $d f .$ As it depends on $\vec{p},$ we also write $d f(\vec{p} ; \vec{t})$ for $d f(\vec{t})$ and $d f(\vec{p} ; \cdot)$ for $d f$.

Some authors write $f^{\prime}(\vec{p})$ for $d f(\vec{p} ; \cdot)$ and call it the derivative at $\vec{p},$ but we shall not do this (see Preface). Following M. Spivak, however, we shall use $"[f^{\prime}(\vec{p})] "$ for its matrix, as follows.

Definition: Jacobian matrix

If $E^{\prime}=E^{n}\left(C^{n}\right)$ and $E=E^{m}\left(C^{m}\right),$ and $f : E^{\prime} \rightarrow E$ is differentiable at $\vec{p},$ we set

$$\left[f^{\prime}(\vec{p})\right]=[d f(\vec{p} ; \cdot)]$$

and call it the Jacobian matrix of $f$ at $\vec{p}$.

Note 1. In Chapter 5, §6, we did not define $d f$ as a mapping. However, if $E^{\prime}=E^{1},$ the function value

$$d f(p ; t)=v t=f^{\prime}(p) \Delta x$$

is as in Chapter 5, §6.

Also, $\left[f^{\prime}(p)\right]$ is a $1 \times 1$ matrix with single term $f^{\prime}(p) .$ (Why?) This motivated Definition 2.

Theorem $\PageIndex{1}$

(uniqueness of $d f ) .$ If $f : E^{\prime} \rightarrow E$ is differentiable at $\vec{p},$ then the map $\phi$ described in Definition 1 is unique (dependent on $f$ and $\vec{p}$ only).

Proof

Suppose there is another linear map $g : E^{\prime} \rightarrow E$ such that

$$\lim _{\vec{t} \rightarrow \overrightarrow{0}} \frac{1}{|\vec{t}|}[f(\vec{p}+\vec{t})-f(\vec{p})-g(\vec{t})]=\lim _{\vec{t} \rightarrow \overrightarrow{0}} \frac{1}{|\vec{t}|}[\Delta f-g(\vec{t})]=0.$$

Let $h=\phi-g .$ By Corollary 1 in §2, $h$ is linear.

Also, by the triangle law,

$$|h(\vec{t})|=|\phi(\vec{t})-g(\vec{t})| \leq|\Delta f-\phi(\vec{t})|+|\Delta f-g(\vec{t})|.$$

Hence, dividing by $|\vec{t}|$,

$$\left|h\left(\frac{\vec{t}}{|\vec{t}|}\right)\right|=\frac{1}{|\vec{t}|}|h(\vec{t})| \leq \frac{1}{|\vec{t}|}|\Delta f-\phi(\vec{t})|+\frac{1}{|\vec{t}|}|\Delta f-g(\vec{t})|.$$

By $(3)$ and $(2),$ the right side expressions tend to 0 as $\vec{t} \rightarrow \overrightarrow{0} .$ Thus

$$\lim _{\vec{t} \rightarrow \overrightarrow{0}} h\left(\frac{\vec{t}}{|\vec{t}|}\right)=0.$$

This remains valid also if $\vec{t} \rightarrow \overrightarrow{0}$ over any line through $\overrightarrow{0},$ so that $\vec{t} /|\vec{t}|$ remains constant, say $\vec{t} /|\vec{t}|=\vec{u},$ where $\vec{u}$ is an arbitrary (but fixed) unit vector.

Then

$$h\left(\frac{\vec{t}}{|\vec{t}|}\right)=h(\vec{u})$$

is constant; so it can tend to $0$ only if it equals $0,$ so $h(\vec{u})=0$ for any unit vector $\vec{u} .$

Since any $\vec{x} \in E^{\prime}$ can be written as $\vec{x}=|\vec{x}| \vec{u},$ linearity yields

$$h(\vec{x})=|\vec{x}| h(\vec{u})=0.$$

Thus $h=\phi-g=0$ on $E^{\prime},$ and so $\phi=g$ after all, proving the uniqueness of $\phi . \square$

Theorem $\PageIndex{2}$

If $f$ is differentiable at $\vec{p},$ then

(i) $f$ is continuous at $\vec{p}$;

(ii) for any $\vec{u} \neq \overrightarrow{0},$ has the $\vec{u}$-directed derivative

$$D_{\vec{u}} f(\vec{p})=d f(\vec{p} ; \vec{u}).$$

Proof

By assumption, formula $(2)$ holds for $\phi=d f(\vec{p} ; \cdot)$.

Thus, given $\varepsilon>0,$ there is $\delta>0$ such that, setting $\Delta f=f(\vec{p}+\vec{t})-f(\vec{p})$ we have

$$\frac{1}{|\vec{t}|} |\Delta f-\phi(\vec{t}) |<\varepsilon \text { whenever } 0<|\vec{t}|<\delta;$$

or, by the triangle law,

$$|\Delta f| \leq|\Delta f-\phi(\vec{t})|+|\phi(\vec{t})| \leq \varepsilon|\vec{t}|+|\phi(\vec{t})|, \quad 0<|\vec{t}|<\delta.$$

Now, by Definition $1, \phi$ is linear and continuous; so

$$\lim _{\vec{t} \rightarrow \overrightarrow{0}}|\phi(\vec{t})|=|\phi(\overrightarrow{0})|=0.$$

Thus, making $\vec{t} \rightarrow \overrightarrow{0}$ in $(5),$ with $\varepsilon$ fixed, we get

$$\lim _{\vec{t} \rightarrow \overrightarrow{0}}|\Delta f|=0.$$

As $\vec{t}$ is just another notation for $\Delta \vec{x}=\vec{x}-\vec{p},$ this proves assertion (i).

Next, fix any $\vec{u} \neq \overrightarrow{0}$ in $E^{\prime},$ and substitute $t \vec{u}$ for $\vec{t}$ in $(4)$.

In other words, $t$ is a real variable, $0<t<\delta /|\vec{u}|,$ so that $\vec{t}=t \vec{u}$ satisfies $0<|\vec{t}|<\delta$.

Multiplying by $|\vec{u}|,$ we use the linearity of $\phi$ to get

$$\varepsilon|\vec{u}|>\left|\frac{\Delta f}{t}-\frac{\phi(t \vec{u})}{t}\right|=\left|\frac{\Delta f}{t}-\phi(\vec{u})\right|=\left|\frac{f(\vec{p}+t \vec{u})-f(\vec{p})}{t}-\phi(\vec{u})\right|.$$

As $\varepsilon$ is arbitrary, we have

$$\phi(\vec{u})=\lim _{t \rightarrow 0} \frac{1}{t}[f(\vec{p}+t \vec{u})-f(\vec{p})].$$

But this is simply $D_{\vec{u}} f(\vec{p}),$ by Definition 1 in §1.

Thus $D_{\vec{u}} f(\vec{p})=\phi(\vec{u})=d f(\vec{p} ; \vec{u}),$ proving $(\mathrm{ii}) . \square$

Note 2. If $E^{\prime}=E^{n}(C^{n}),$ Theorem 2(iii) shows that if $f$ is differentiable at $\vec{p},$ it has the $n$ partials

$$D_{k} f(\vec{p})=d f\left(\vec{p} ; \vec{e}_{k}\right), \quad k=1, \ldots, n.$$

But the converse fails: the existence of the $D_{k} f(\vec{p})$ does not even imply continuity, let alone differentiability (see §1). Moreover, we have the following result.

Corollary $\PageIndex{1}$

If $E^{\prime}=E^{n}\left(C^{n}\right)$ and if $f : E^{\prime} \rightarrow E$ is differentiable at $\vec{p},$ then

$$d f(\vec{p} ; \vec{t})=\sum_{k=1}^{n} t_{k} D_{k} f(\vec{p})=\sum_{k=1}^{n} t_{k} \frac{\partial}{\partial x_{k}} f(\vec{p}),$$

where $\vec{t}=\left(t_{1}, \ldots, t_{n}\right)$.

Proof

By definition, $\phi=d f(\vec{p}; \cdot)$ is a linear map for a fixed $\vec{p}$.

If $E^{\prime}=E^{n}$ or $C^{n},$ we may use formula (3) of §2, replacing $f$ and $\vec{x}$ by $\phi$ and $\vec{t},$ and get

$$\phi(\vec{t})=d f(\vec{p}; \vec{t})=\sum_{k=1}^{n} t_{k} d f\left(\vec{p}; \vec{e}_{k}\right)=\sum_{k=1}^{n} t_{k} D_{k} f(\vec{p})$$

by Note 2. $\quad \square$

Note 3. In classical notation, one writes $\Delta x_{k}$ or $d x_{k}$ for $t_{k}$ in (6). Thus, omitting $\vec{p}$ and $\vec{t},$ formula (6) is often written as

$$d f=\frac{\partial f}{\partial x_{1}} d x_{1}+\frac{\partial f}{\partial x_{2}} d x_{2}+\cdots+\frac{\partial f}{\partial x_{n}} d x_{n}.$$

In particular, if $n=3,$ we write $x, y, z$ for $x_{1}, x_{2}, x_{3}.$ This yields

$$d f=\frac{\partial f}{\partial x} d x+\frac{\partial f}{\partial y} d y+\frac{\partial f}{\partial z} d z$$

(a familiar calculus formula).

Note 4. If the range space $E$ in Corollary 1 is $E^{1}(C),$ then the $D_{k} f(\vec{p})$ form an $n$-tuple of scalars, i.e., a vector in $E^{n}(C^{n}).$

In case $f : E^{n} \rightarrow E^{1},$ we denote it by

$$\nabla f(\vec{p})=\left(D_{1} f(\vec{p}), \ldots, D_{n} f(\vec{p})\right)=\sum_{k=1}^{n} \vec{e}_{k} D_{k} f(\vec{p}).$$

In case $f : C^{n} \rightarrow C,$ we replace the $D_{k} f(\vec{p})$ by their conjugates $\overline{D_{k} f(\vec{p})}$ and set

$$\nabla f(\vec{p})=\sum_{k=1}^{n} \vec{e}_{k} \overline{D_{k} f(\vec{p})}.$$

The vector $\nabla f(\vec{p})$ is called the gradient of $f$ ("grad $f$") at $\vec{p}$.

From (6) we obtain

$$d f(\vec{p};\vec{t})=\sum_{k=1}^{n} t_{k} D_{k} f(\vec{p})=\vec{t} \cdot \nabla f(\vec{p})$$

(dot product of $\vec{t}$ by $\nabla f(\vec{p}) ),$ provided $f : E^{n} \rightarrow E^{1}$ (or $f : C^{n} \rightarrow C )$ is differentiable at $\vec{p}$.

This leads us to the following result.

Corollary $\PageIndex{2}$

A function $f : E^{n} \rightarrow E^{1}$ (or $f : C^{n} \rightarrow C$) is differentiable at $\vec{p}$ iff

$$\lim _{\vec{t} \rightarrow \overline{0}} \frac{1}{|\vec{t}|}|f(\vec{p}+\vec{t})-f(\vec{p})-\vec{t} \cdot \vec{v}|=0$$

for some $\vec{v} \in E^{n}(C^{n})$.

In this case, necessarily $\vec{v}=\nabla f(\vec{p})$ and $\vec{t} \cdot \vec{v}=d f(\vec{p} ; \vec{t}), \vec{t} \in E^{n}(C^{n})$.

Proof

If $f$ is differentiable at $\vec{p},$ we may set $\phi=d f(\vec{p} ; \cdot)$ and $\vec{v}=\nabla f(\vec{p})$

Then by (7),

$$\phi(\vec{t})=d f(\vec{p} ; \vec{t})=\vec{t} \cdot \vec{v};$$

so by Definition 1, (8) results.

Conversely, if some $\vec{v}$ satisfies (8), set $\phi(\vec{t})=\vec{t} \cdot \vec{v}.$ Then (8) implies (2), and $\phi$ is linear and continuous.

Thus by definition, $f$ is differentiable at $\vec{p};$ so (7) holds.

Also, $\phi$ is a linear functional on $E^{n}(C^{n}).$ By Theorem 2(ii) in §2, the $\vec{v}$ in $\phi(\vec{t})=\vec{t} \cdot \vec{v}$ is unique, as is $\phi.$

Thus by (7), $\vec{v}=\nabla f(\vec{p})$ necessarily. $\quad \square$

Corollary $\PageIndex{3}$ (law of the mean)

If $f : E^{n} \rightarrow E^{1}$ (real) is relatively continuous on a closed segment $L[\vec{p}, \vec{q}], \vec{p} \neq \vec{q},$ and differentiable on $L(\vec{p}, \vec{q}),$ then

$$f(\vec{q})-f(\vec{p})=(\vec{q}-\vec{p}) \cdot \nabla f(\vec{x}_{0})$$

for some $\vec{x}_{0} \in L(\vec{p}, \vec{q})$.

Proof

Let

$$r=|\vec{q}-\vec{p}|, \quad \vec{v}=\frac{1}{r}(\vec{q}-\vec{p}), \text { and } r \vec{v}=(\vec{q}-\vec{p}).$$

By (7) and Theorem 2(ii),

$$D_{\vec{v}} f(\vec{x})=d f(\vec{x} ; \vec{v})=\vec{v} \cdot \nabla f(\vec{x})$$

for $\vec{x} \in L(\vec{p}, \vec{q}).$ Thus by formula (3') of Corollary 2 in §1,

$$f(\vec{q})-f(\vec{p})=r D_{\vec{v}} f\left(\vec{x}_{0}\right)=r \vec{v} \cdot \nabla f(\vec{x}_{0})=(\vec{q}-\vec{p}) \cdot \nabla f(\vec{x}_{0})$$

for some $\vec{x}_{0} \in L(\vec{p}, \vec{q}). \quad \square$

As we know, the mere existence of partials does not imply differentiability. But the existence of continuous partials does. Indeed, we have the following theorem.

Theorem $\PageIndex{3}$

Let $E^{\prime}=E^{n}(C^{n})$.

If $f : E^{\prime} \rightarrow E$ has the partial derivatives $D_{k} f(k=1, \ldots, n)$ on all of an open set $A \subseteq E^{\prime},$ and if the $D_{k} f$ are continuous at some $\vec{p} \in A,$ then $f$ is differentiable at $\vec{p}$.

Proof

With $\vec{p}$ as above, let

$$\phi(\vec{t})=\sum_{k=1}^{n} t_{k} D_{k} f(\vec{p}) \text { with } \vec{t}=\sum_{k=1}^{n} t_{k} \vec{e}_{k} \in E^{\prime}.$$

Then $\phi$ is continuous (a polynomial!) and linear (Corollary 2 in §2).

Thus by Definition 1, it remains to show that

$$\lim _{\vec{t} \rightarrow \overrightarrow{0}|\vec{t}|}|\Delta f-\phi(\vec{t})|=0;$$

that is;

$$\lim _{\vec{t} \in \overrightarrow{0}} \frac{1}{|\vec{t}|}\left|f(\vec{p}+\vec{t})-f(\vec{p})-\sum_{k=1}^{n} t_{k} D_{k} f(\vec{p})\right|=0.$$

To do this, fix $\varepsilon>0.$ As $A$ is open and the $D_{k} f$ are continuous at $\vec{p} \in A$ there is a $\delta>0$ such that $G_{\vec{p}}(\delta) \subseteq A$ and simultaneously (explain this!)

$$(\forall \vec{x} \in G_{\vec{p}}(\delta)) \quad\left|D_{k} f(\vec{x})-D_{k} f(\vec{p})\right|<\frac{\varepsilon}{n}, k=1, \ldots, n.$$

Hence for any set $I \subseteq G_{\vec{p}}(\delta)$

$$\sup _{\vec{x} \in I}\left|D_{k} f(\vec{x})-D_{k} f(\vec{p})\right| \leq \frac{\varepsilon}{n}. \quad (Why?)$$

Now fix any $\vec{t} \in E^{\prime}, 0<|\vec{t}|<\delta,$ and let $\vec{p}_{0}=\vec{p}$,

$$\vec{p}_{k}=\vec{p}+\sum_{i=1}^{k} t_{i} e_{i}, \quad k=1, \ldots, n.$$

Then

$$\vec{p}_{n}=\vec{p}+\sum_{i=1}^{n} t_{i} \vec{e}_{i}=\vec{p}+\vec{t},$$

$\left|\vec{p}_{k}-\vec{p}_{k-1}\right|=\left|t_{k}\right|,$ and all $\vec{p}_{k}$ lie in $G_{\vec{p}}(\delta),$ for

$$\left|\vec{p}_{k}-\vec{p}\right|=\left|\sum_{i=1}^{k} t_{i} e_{i}\right|=\sqrt{\sum_{i=1}^{k}\left|t_{i}\right|^{2}} \leq \sqrt{\sum_{i=1}^{n}\left|t_{i}\right|^{2}}=|\vec{t}|<\delta,$$

as required.

As $G_{p}(\delta)$ is convex (Chapter 4, §9), the segments $I_{k}=L\left[\vec{p}_{k-1}, \vec{p}_{k}\right]$ all lie in $G_{\vec{p}}(\delta) \subseteq A;$ and by assumption, $f$ has all partials there.

Hence by Theorem 1 in §1, $f$ is relatively continuous on all $I_{k}$.

All this also applies to the functions $g_{k},$ defined by

$$\left(\forall \vec{x} \in E^{\prime}\right) \quad g_{k}(\vec{x})=f(\vec{x})-x_{k} D_{k} f(\vec{p}), \quad k=1, \ldots, n.$$

(Why?) Here

$$D_{k} g_{k}(\vec{x})=D_{k} f(\vec{x})-D_{k} f(\vec{p}).$$

(Why?)

Thus by Corollary 2 in §1, and (11) above,

$$\begin{aligned}\left|g_{k}\left(\vec{p}_{k}\right)-g_{k}\left(\vec{p}_{k-1}\right)\right| & \leq\left|\vec{p}_{k}-\vec{p}_{k-1}\right| \sup _{x \in I_{k}}\left|D_{k} f(\vec{x})-D_{k} f(\vec{p})\right| \\ & \leq \frac{\varepsilon}{n}\left|t_{k}\right| \leq \frac{\varepsilon}{n}|\vec{t}|, \end{aligned}$$

since

$$\left|\vec{p}_{k}-\vec{p}_{k-1}\right|=\left|t_{k} \vec{e}_{k}\right| \leq|\vec{t}|,$$

by construction.

Combine with (12), recalling that the $k$th coordinates $x_{k},$ for $\vec{p}_{k}$ and $\vec{p}_{k-1}$ differ by $t_{k};$ so we obtain

$$\begin{aligned}\left|g_{k}\left(\vec{p}_{k}\right)-g_{k}\left(\vec{p}_{k-1}\right)\right| &=\left|f\left(\vec{p}_{k}\right)-f\left(\vec{p}_{k-1}\right)-t_{k} D_{k} f(\vec{p})\right| \\ & \leq \frac{\varepsilon}{n}|\vec{t}|. \end{aligned}$$

Also,

$$\begin{aligned} \sum_{k=1}^{n}\left[f\left(\vec{p}_{k}\right)-f\left(\vec{p}_{k-1}\right)\right] &=f\left(\vec{p}_{n}\right)-f\left(\vec{p}_{0}\right) \\ &=f(\vec{p}+\vec{t})-f(\vec{p})=\Delta f(\text {see above}). \end{aligned}$$

Thus,

$$\begin{aligned}\left|\Delta f-\sum_{k=1}^{n} t_{k} D_{k} f(\vec{p})\right| &=\left|\sum_{k=1}^{n}\left[f\left(\vec{p}_{k}\right)-f\left(\vec{p}_{k-1}\right)-t_{k} D_{k} f(\vec{p})\right]\right| \\ & \leq n \cdot \frac{\varepsilon}{n}|\vec{t}|=\varepsilon|\vec{t}|. \end{aligned}$$

As $\varepsilon$ is arbitrary, (10) follows, and all is proved. $\quad \square$

Theorem $\PageIndex{4}$

If $f : E^{n} \rightarrow E^{m}$ (or $f : C^{n} \rightarrow C^{m})$ is differentiable at $\vec{p},$ with $f=\left(f_{1}, \ldots, f_{m}\right),$ then $\left[f^{\prime}(\vec{p})\right]$ is an $m \times n$ matrix,

$$\left[f^{\prime}(\vec{p})\right]=\left[D_{k} f_{i}(\vec{p})\right], \quad i=1, \ldots, m, k=1, \ldots, n.$$

Proof

By definition, $\left[f^{\prime}(\vec{p})\right]$ is the matrix of the linear map $\phi=d f(\vec{p} ; \cdot),$ $\phi=\left(\phi_{1}, \ldots, \phi_{m}\right).$ Here

$$\phi(\vec{t})=\sum_{k=1}^{n} t_{k} D_{k} f(\vec{p})$$

by Corollary 1.

As $f=\left(f_{1}, \ldots, f_{m}\right),$ we can compute $D_{k} f(\vec{p})$ componentwise by Theorem 5 of Chapter 5, §1, and Note 2 in §1 to get

$$\begin{aligned} D_{k} f(\vec{p}) &=\left(D_{k} f_{1}(\vec{p}), \ldots, D_{k} f_{m}(\vec{p})\right) \\ &=\sum_{i=1}^{m} e_{i}^{\prime} D_{k} f_{i}(\vec{p}), \quad k=1,2, \ldots, n, \end{aligned}$$

where the $e_{i}^{\prime}$ are the basic vectors in $E^{m}\left(C^{m}\right).$ (Recall that the $\vec{e}_{k}$ are the basic vectors in $E^{n}\left(C^{n}\right).)$

Thus

$$\phi(\vec{t})=\sum_{i=1}^{m} e_{i}^{\prime} \phi_{i}(\vec{t}).$$

Also,

$$\phi(\vec{t})=\sum_{k=1}^{n} t_{k} \sum_{i=1}^{m} e_{i}^{\prime} D_{k} f_{i}(\vec{p})=\sum_{i=1}^{m} e_{i}^{\prime} \sum_{k=1}^{n} t_{k} D_{k} f_{i}(\vec{p}).$$

The uniqueness of the decomposition (Theorem 2 in Chapter 3, §§1-3) now yields

$$\phi_{i}(\vec{t})=\sum_{k=1}^{n} t_{k} D_{k} f_{i}(\vec{p}), \quad i=1, \ldots, m, \quad \vec{t} \in E^{n}\left(C^{n}\right).$$

If here $\vec{t}=\vec{e}_{k},$ then $t_{k}=1,$ while $t_{j}=0$ for $j \neq k.$ Thus we obtain

$$\phi_{i}\left(\vec{e}_{k}\right)=D_{k} f_{i}(\vec{p}), \quad i=1, \ldots, m, k=1, \ldots, n.$$

Hence,

$$\phi\left(\vec{e}_{k}\right)=\left(v_{1 k}, v_{2 k}, \ldots, v_{m k}\right),$$

where

$$v_{i k}=\phi_{i}\left(\vec{e}_{k}\right)=D_{k} f_{i}(\vec{p}).$$

But by Note 3 of §2, $v_{1 k}, \ldots, v_{m k}$ (written vertically) is the $k$th column of the $m \times n$ matrix $[\phi]=\left[f^{\prime}(\vec{p})\right].$ Thus formula (14) results indeed. $\quad \square$

In conclusion, let us stress again that while $D_{\vec{u}} f(\vec{p})$ is a constant, for a fixed $\vec{p}, d f(\vec{p} ; \cdot)$ is a mapping

$$\phi \in L\left(E^{\prime}, E\right),$$

especially "tailored" for $\vec{p}$.

The reader should carefully study at least the "arrowed" problems below.