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# 6.3: Differentiable Functions

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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.