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Mathematics LibreTexts

6.3: Differentiable Functions

  • Page ID
    19199
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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.