一阶微分公式的存在性

若：
n∈N,n≥1,n \in \mathbb N, n \ge 1,
x=(x1,⋯xn)⊺ \mathbf{x} = (x_1, \cdots x _n)^\intercal
f(x),g(x)f( \mathbf{x} ), g( \mathbf{x}) 可微，
则： kf,f+g,f⋅g,fg(g(x)≠0)kf, f + g, f \cdot g, \frac {f} {g} (g( \mathbf{x}) \ne 0) 都可微。

证明：

定义内积： ∀x=(x1,⋯xn)⊺,∀y=(y1,⋯yn)⊺,\forall {\mathbf{x}} = (x_1, \cdots x _n)^\intercal, \forall {\mathbf{y}} = (y_1, \cdots y _n)^\intercal,
⟨x,y⟩=∑ni=1xiyi \langle {\mathbf{x}}, {\mathbf{y}} \rangle = \sum _{i = 1}^{n} x_i y_i
令 a=(∂f(x)∂x1,⋯,∂f(x)∂xn),b=(∂g(x)∂x1,⋯,∂g(x)∂xn),\mathbf{a} = \left (\dfrac{\partial f(\mathbf{x})}{\partial x_1}, \cdots, \dfrac{\partial f(\mathbf{x})}{\partial x_n} \right ), \mathbf{b} = \left (\dfrac{\partial g(\mathbf{x})}{\partial x_1}, \cdots, \dfrac{\partial g(\mathbf{x})}{\partial x_n} \right ),
Δx=(Δx1,⋯,Δxn)⊺,\mathbf{\Delta x} = (\mathbf{\Delta x}_1, \cdots, \mathbf{\Delta x}_n)^\intercal,
r=⟨Δx,Δx⟩−−−−−−−−√,r = \sqrt {\langle \mathbf{\Delta x}, \mathbf{\Delta x} \rangle} ,
则：
f(x+Δx)−f(x)=⟨a,Δx⟩+rα,limΔx→0α=0,f(\mathbf{x} + \mathbf{\Delta x}) - f(\mathbf{x}) = \langle \mathbf{a}, \mathbf{\Delta x} \rangle + r \alpha, \lim _{\mathbf{\Delta x} \rightarrow \mathbf{0}} \alpha = 0,
g(x+Δx)−g(x)=⟨b,Δx⟩+rβ,limΔx→0β=0,g(\mathbf{x} + \mathbf{\Delta x}) - g(\mathbf{x}) = \langle \mathbf{b}, \mathbf{\Delta x} \rangle + r \beta, \lim _{\mathbf{\Delta x} \rightarrow \mathbf{0}} \beta = 0,
于是：

kf(x+Δx)−kf(x)=k[f(x+Δx)−f(x)]k f(\mathbf{x} + \mathbf{\Delta x}) - k f(x) = k \left [ f(\mathbf{x} + \mathbf{\Delta x}) - f(x) \right ]
=k(⟨a,Δx⟩+rα) = k ( \langle \mathbf{a}, \mathbf{\Delta x} \rangle + r \alpha )
=⟨ka,Δx⟩+r⋅kα = \langle k \mathbf{a}, \mathbf{\Delta x} \rangle + r \cdot k \alpha
=⟨ka,Δx⟩+r⋅o(1)= \langle k \mathbf{a}, \mathbf{\Delta x} \rangle + r \cdot o(1)
[f(x+Δx)+g(x+Δx)]−[f(x)+g(x)]\left [f(\mathbf{x} + \mathbf{\Delta x}) + g(\mathbf{x} + \mathbf{\Delta x})\right ] - \left [f(\mathbf{x}) + g(\mathbf{x})\right ]
=(f(x+Δx)−f(x))+(g(x+Δx)−g(x))= (f(\mathbf{x} + \mathbf{\Delta x}) - f(\mathbf{x}) ) + (g(\mathbf{x} + \mathbf{\Delta x}) - g(\mathbf{x}))
=⟨a,Δx⟩+rα+⟨b,Δx⟩+rβ= \langle \mathbf{a}, \mathbf{\Delta x} \rangle + r \alpha + \langle \mathbf{b}, \mathbf{\Delta x} \rangle + r \beta
=⟨a+b,Δx⟩+r(α+β)= \langle \mathbf{a + b}, \mathbf{\Delta x} \rangle + r (\alpha + \beta)
=⟨a+b,Δx⟩+r⋅o(1)= \langle \mathbf{a + b}, \mathbf{\Delta x} \rangle + r \cdot o(1)
f(x+Δx)⋅g(x+Δx)−f(x)⋅g(x)f(\mathbf{x} + \mathbf{\Delta x}) \cdot g(\mathbf{x} + \mathbf{\Delta x}) - f(\mathbf{x}) \cdot g(\mathbf{x})
=[f(x+Δx)−f(x)]⋅g(x+Δx)+f(x)⋅[g(x+Δx)−g(x)]= \left [f(\mathbf{x} + \mathbf{\Delta x}) - f(\mathbf {x})\right ] \cdot g(\mathbf{x} + \mathbf{\Delta x}) + f(\mathbf{x}) \cdot \left [g(\mathbf{x} + \mathbf{\Delta x}) - g(\mathbf{x})\right ]
=(⟨a,Δx⟩+rα)⋅g(x+Δx)+f(x)⋅(⟨b,Δx⟩+rβ)= (\langle \mathbf{a}, \mathbf{\Delta x} \rangle + r \alpha) \cdot g(\mathbf{x} + \mathbf{\Delta x}) + f(\mathbf{x}) \cdot ( \langle \mathbf{b}, \mathbf{\Delta x} \rangle + r \beta)
=⟨a,Δx⟩⋅g(x+Δx)+f(x)⋅⟨b,Δx⟩+rα⋅g(x+Δx)+rβf(x)= \langle \mathbf{a}, \mathbf{\Delta x} \rangle \cdot g(\mathbf{x} + \mathbf{\Delta x}) + f(\mathbf{x}) \cdot \langle \mathbf{b}, \mathbf{\Delta x} \rangle + r \alpha \cdot g(\mathbf{x} + \mathbf{\Delta x}) + r \beta f(\mathbf{x})
=⟨a,Δx⟩⋅g(x+Δx)+f(x)⋅⟨b,Δx⟩+r[α⋅g(x+Δx)+βf(x)]= \langle \mathbf{a}, \mathbf{\Delta x} \rangle \cdot g(\mathbf{x} + \mathbf{\Delta x}) + f(\mathbf{x}) \cdot \langle \mathbf{b}, \mathbf{\Delta x} \rangle + r \left [ \alpha \cdot g(\mathbf{x} + \mathbf{\Delta x}) +\beta f(\mathbf{x}) \right ]
=⟨a,Δx⟩⋅g(x+Δx)+f(x)⋅⟨b,Δx⟩+r⋅o(1)= \langle \mathbf{a}, \mathbf{\Delta x} \rangle \cdot g(\mathbf{x} + \mathbf{\Delta x}) + f(\mathbf{x}) \cdot \langle \mathbf{b}, \mathbf{\Delta x} \rangle + r \cdot o(1)
=⟨a⋅g(x+Δx)+f(x)⋅b,Δx⟩+r⋅o(1)= \langle \mathbf{a} \cdot g(\mathbf{x} + \mathbf{\Delta x}) + f(\mathbf{x}) \cdot \mathbf{b} , \mathbf{\Delta x} \rangle + r \cdot o(1)
=⟨a⋅g(x)+f(x)⋅b,Δx⟩+r⋅o(1)+⟨a⋅[g(x+Δx)−g(x)],Δx⟩= \langle \mathbf{a} \cdot g(\mathbf{x}) + f(\mathbf{x}) \cdot \mathbf{b} , \mathbf{\Delta x} \rangle + r \cdot o(1) + \langle \mathbf{a} \cdot \left [ g(\mathbf{x} + \mathbf{\Delta x}) - g(\mathbf{x}) \right ], \mathbf{\Delta x} \rangle

⟨a⋅[g(x+Δx)−g(x)],Δx⟩ \langle \mathbf{a} \cdot \left [ g(\mathbf{x} + \mathbf{\Delta x}) - g(\mathbf{x}) \right ], \mathbf{\Delta x} \rangle
=[g(x+Δx)−g(x)]⋅⟨a,Δx⟩= \left [ g(\mathbf{x} + \mathbf{\Delta x}) - g(\mathbf{x}) \right ] \cdot \langle \mathbf{a} , \mathbf{\Delta x} \rangle
=⟨a,Δx⟩⋅o(1)= \langle \mathbf{a} , \mathbf{\Delta x} \rangle \cdot o(1)
=1r⋅⟨a,Δx⟩⋅r⋅o(1)= \frac {1 }{ r} \cdot \langle \mathbf{a} , \mathbf{\Delta x} \rangle \cdot r \cdot o(1)
（注： |⟨a,Δx⟩|≤⟨a,a⟩⟨Δx,Δx⟩−−−−−−−−−−−−√=r⋅⟨a,a⟩−−−−−√| \langle \mathbf{a}, \mathbf{\Delta x} \rangle | \le \sqrt {\langle \mathbf{a}, \mathbf {a} \rangle \langle \mathbf{\Delta x}, \mathbf {\mathbf{\Delta x}} \rangle} = r \cdot \sqrt {\langle \mathbf{a}, \mathbf {a} \rangle } ）
=r⋅o(1)= r \cdot o(1)
f(x+Δx)g(x+Δx)−f(x)g(x)\frac {f \left (\mathbf{x} + \mathbf{\Delta x}\right )}{g \left (\mathbf{x} + \mathbf{\Delta x}\right )} - \frac {f \left (\mathbf{x}\right )}{g \left (\mathbf{x}\right )}
=f(x+Δx)g(x)−f(x)g(x+Δx)g(x+Δx)g(x)= \frac {f \left (\mathbf{x} + \mathbf{\Delta x}\right )g \left (\mathbf{x}\right ) - f \left (\mathbf{x}\right )g \left (\mathbf{x} + \mathbf{\Delta x}\right )}{g \left (\mathbf{x} + \mathbf{\Delta x}\right ) g \left (\mathbf{x}\right )}
=1[g(x)]2⋅g(x)g(x+Δx)⋅{[f(x+Δx)−f(x)]g(x)−f(x)[g(x+Δx)−g(x)]}= \frac{1}{{ \left [g \left (\mathbf{x}\right )\right ]}^2} \cdot \frac{g \left (\mathbf{x}\right )}{g \left (\mathbf{x} + \mathbf{\Delta x}\right )} \cdot \left \{ \left [f \left (\mathbf{x} + \mathbf{\Delta x}\right ) - f \left (\mathbf{x}\right )\right ] g \left (\mathbf{x}\right ) - f \left (\mathbf{x}\right ) \left [g \left (\mathbf{x} + \mathbf{\Delta x}\right ) - g \left (\mathbf{x}\right )\right ] \right \}
=1[g(x)]2⋅[1+o(1)]⋅[(⟨a,Δx⟩+rα)g(x)−f(x)(⟨b,Δx⟩+rβ)]= \frac{1}{{ \left [g \left (\mathbf{x}\right )\right ]}^2} \cdot \left [ 1 + o \left (1\right )\right ] \cdot \left [ \left ( \langle \mathbf{a}, \mathbf{\Delta x} \rangle + r \alpha \right ) g \left (\mathbf{x}\right ) - f \left (\mathbf{x}\right ) \left (\langle \mathbf{b}, \mathbf{\Delta x} \rangle + r \beta\right )\right ]
=[1[g(x)]2+o(1)]⋅{⟨a⋅g(x)−f(x)⋅b,Δx⟩+r⋅[αg(x)−βf(x)]}= \left [ \frac{1}{{ \left [g \left (\mathbf{x}\right )\right ]}^2} + o \left (1\right )\right ] \cdot \left \{ \langle \mathbf{a} \cdot g \left (\mathbf{x}\right ) - f \left (\mathbf{x}\right ) \cdot \mathbf{b}, \mathbf{\Delta x} \rangle + r \cdot \left [ \alpha g \left (\mathbf{x}\right ) - \beta f \left (\mathbf{x}\right )\right ] \right \}
=⟨1[g(x)]2⋅[a⋅g(x)−f(x)⋅b],Δx⟩+r⋅o(1)+r⋅o(1)⋅[1r⋅⟨a⋅g(x)−f(x)⋅b,Δx⟩+o(1)]= \langle \frac{1}{{ \left [g \left (\mathbf{x}\right )\right ]}^2} \cdot \left [ \mathbf{a} \cdot g \left (\mathbf{x}\right ) - f \left (\mathbf{x}\right ) \cdot \mathbf{b}\right ] , \mathbf{\Delta x} \rangle + r \cdot o \left (1\right ) + r \cdot o \left (1\right ) \cdot \left [\frac{1}{r} \cdot \langle \mathbf{a} \cdot g \left (\mathbf{x}\right ) - f \left (\mathbf{x}\right ) \cdot \mathbf{b}, \mathbf{\Delta x} \rangle + o \left (1\right )\right ]
=⟨1[g(x)]2⋅[a⋅g(x)−f(x)⋅b],Δx⟩+r⋅o(1)= \langle \frac{1}{{ \left [g \left (\mathbf{x}\right )\right ]}^2} \cdot \left [ \mathbf{a} \cdot g \left (\mathbf{x}\right ) - f \left (\mathbf{x}\right ) \cdot \mathbf{b}\right ] , \mathbf{\Delta x} \rangle + r \cdot o \left (1\right )

一阶微分公式

d(kf)\mathrm{d} \left (k f \right )
=∑ni=1∂(kf)∂xidxi = \sum _{i = 1}^ {n} \frac {\partial \left (kf \right )}{\partial x_i} \mathrm{d} x_i
=∑ni=1k∂(f)∂xidxi = \sum _{i = 1}^ {n} k \frac {\partial \left (f \right )}{\partial x_i} \mathrm{d} x_i
=k∑ni=1∂(f)∂xidxi = k \sum _{i = 1}^ {n} \frac {\partial \left (f \right )}{\partial x_i} \mathrm{d} x_i
=kdf = k \mathrm{d} f
d(f+g)\mathrm{d} \left (f + g\right )
=∑ni=1∂(f+g)∂xidxi= \sum _{i = 1}^ {n} \frac {\partial \left (f + g\right )}{\partial x_i} \mathrm{d} x_i
=∑ni=1∂f∂xidxi+∑ni=1∂g∂xidxi=df+dg= \sum _{i = 1}^ {n} \frac {\partial f}{\partial x_i} \mathrm{d} x_i + \sum _{i = 1}^ {n} \frac {\partial g}{\partial x_i} \mathrm{d} x_i = \mathrm{d} f + \mathrm{d} g
d(f⋅g)\mathrm{d} \left (f \cdot g\right )
=∑ni=1∂(f⋅g)∂xidxi= \sum _{i = 1}^ {n} \frac {\partial \left (f \cdot g\right )}{\partial x_i} \mathrm{d} x_i
=∑ni=1(∂f∂xi⋅g+f⋅∂g∂xi)dxi= \sum _{i = 1}^ {n} \left (\frac {\partial f}{\partial x_i} \cdot g + f \cdot \frac {\partial g}{\partial x_i}\right ) \mathrm{d} x_i
=∑ni=1(∂f∂xidxi⋅g+f⋅∂g∂xidxi)= \sum _{i = 1}^ {n} \left (\frac {\partial f}{\partial x_i} \mathrm{d} x_i \cdot g + f \cdot \frac {\partial g}{\partial x_i} \mathrm{d} x_i\right )
=df⋅g+f⋅dg= \mathrm{d} f \cdot g + f \cdot \mathrm{d} g
dfg\mathrm{d} \frac {f}{g}
=∑ni=1∂fg∂xidxi= \sum _{i = 1}^ {n} \frac {\partial \frac {f}{g}}{\partial x_i} \mathrm{d} x_i
=∑ni=11g2(∂f∂xi⋅g−f⋅∂g∂xi)dxi= \sum _{i = 1}^ {n} \frac{1}{g^2} \left (\frac {\partial f}{\partial x_i} \cdot g - f \cdot \frac {\partial g}{\partial x_i}\right ) \mathrm{d} x_i
=1g2∑ni=1(∂f∂xi⋅g−f⋅∂g∂xi)dxi= \frac{1}{g^2} \sum _{i = 1}^ {n} \left (\frac {\partial f}{\partial x_i} \cdot g - f \cdot \frac {\partial g}{\partial x_i}\right ) \mathrm{d} x_i
=1g2∑ni=1(∂f∂xidxi⋅g−f⋅∂g∂xidxi)= \frac{1}{g^2} \sum _{i = 1}^ {n} \left (\frac {\partial f}{\partial x_i}\mathrm{d} x_i \cdot g - f \cdot \frac {\partial g}{\partial x_i} \mathrm{d} x_i\right )
=df⋅g−f⋅dgg2 = \frac {\mathrm{d} f \cdot g - f \cdot \mathrm{d} g} {g^2}

推论

定义 f′(x)=(∂f(x)xj)1×n,g′(x)=(∂g(x)xj)1×n,f'(\mathbf{x}) = \left ( \dfrac {\partial f(\mathbf{x})} {x_j} \right )_ { 1 \times n }, g'(\mathbf{x}) = \left ( \dfrac {\partial g(\mathbf{x})} {x_j} \right )_ { 1 \times n }, 则

(kf)′(x)=kf′(x)(kf) '(\mathbf{x}) = k f'(\mathbf{x})
(f+g)′(x)=f′(x)+g′(x)(f + g) '(\mathbf{x}) = f'(\mathbf{x}) + g'(\mathbf{x})
(f⋅g)′(x)=f′(x)g(x)+f(x)g′(x)(f \cdot g) '(\mathbf{x}) = f'(\mathbf{x}) g(\mathbf{x}) + f(\mathbf{x}) g'(\mathbf{x})
(fg)′(x)=f′(x)⋅g(x)−f(x)⋅g′(x)g(x)2\left ( \frac {f}{g} \right) '(\mathbf{x}) = \frac { f' (\mathbf{x}) \cdot g (\mathbf{x}) - f (\mathbf{x})\cdot g ' (\mathbf{x})} {g (\mathbf{x})^2}

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多元函数四则运算的一阶微分公式的存在性与性质

一阶微分公式的存在性

证明：

一阶微分公式

推论

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