分化gydF4y2Ba

在这一页中，我们将遇到一些微分规则的证明，这些证明用于大多数微分问题。在证明这些规则时，将使用标准的“PEMDAS”(括号，指数，乘法，除法，加法，减法)。gydF4y2Ba

让gydF4y2Ba $一个gydF4y2Ba$是任意的实数。常数法则指出gydF4y2Ba

$\frac{d\ (a)}{dx} = 0。gydF4y2Ba$

证明:gydF4y2Ba

回忆一下，对于任意的函数gydF4y2Ba $f (x)gydF4y2Ba$，gydF4y2Ba

$\压裂{d \ \大(f (x) \大)}{dx} = \ lim_ {h \ \ \ 0} \压裂{f (x + h) - f (x)} {h}。gydF4y2Ba$

让gydF4y2Ba $F (x) = agydF4y2Ba$．替代gydF4y2Ba $x + hgydF4y2Ba$为gydF4y2Ba $xgydF4y2Ba$．自gydF4y2Ba $F (x) = agydF4y2Ba$是常数函数，gydF4y2Ba $F (x+h) = agydF4y2Ba$．gydF4y2Ba

$\ \压裂{d (a)} {dx} = \ lim_ {h \ \ \ 0} \压裂{一}- {h} = \ lim_ {h \ \ \ 0} \压裂{0}{h} = \ lim_ h \ \ \{0} 0 = 0。\ _ \广场gydF4y2Ba$

让gydF4y2Ba $一个gydF4y2Ba$是任意的实数，并且gydF4y2Ba $g (x)gydF4y2Ba$任意可微函数。gydF4y2Ba

$\压裂{d \ \大(g (x) \ \ cdot大)}{dx} = \ cdot g ' (x)。gydF4y2Ba$

证明:gydF4y2Ba

回忆一下，对于任意的函数gydF4y2Ba $f (x)gydF4y2Ba$，gydF4y2Ba

$\压裂{d \ \大(f (x) \大)}{dx} = \ lim_ {h \ \ \ 0} \压裂{f (x + h) - f (x)} {h}。gydF4y2Ba$

让gydF4y2Ba $F (x) = a \ c导g(x)gydF4y2Ba$．替代gydF4y2Ba $x + hgydF4y2Ba$为gydF4y2Ba $xgydF4y2Ba$得到gydF4y2Ba $F (x+h) = a \cdot g(x+h)gydF4y2Ba$．gydF4y2Ba

$\开始{对齐}\压裂{d \ \大(g (x) \ \ cdot大)}{dx} & = \ lim_ {h \ \ \ 0} \压裂{\ cdot g (x + h) - \ cdot g (x)} {h} \ \ & = \ lim_ {h \ \ \ 0} \压裂{\大(g (x + h) - g (x) \大)}{h} \ \ & = \离开(\ lim_ h \ \ \ {0} \) \ lim_ {h \ \ \ 0} \压裂{g (x + h) - g (x)} {h}。结束\{对齐}gydF4y2Ba$

根据极限和导数的定义，gydF4y2Ba

$\压裂{d \ \大(g (x) \ \ cdot大)}{dx} = \ cdot g ' (x)。\ _ \广场gydF4y2Ba$

让gydF4y2Ba $ngydF4y2Ba$而且gydF4y2Ba $xgydF4y2Ba$是实数，用gydF4y2Ba $N \neq 0gydF4y2Ba$而且gydF4y2Ba $X \neq 0gydF4y2Ba$．gydF4y2Ba

幂法则指出gydF4y2Ba

$\frac{d\ (x^n)}{dx} = nx^{n-1}。gydF4y2Ba$

证据1:gydF4y2Ba

回忆一下，对于任意的函数gydF4y2Ba $f (x)gydF4y2Ba$，gydF4y2Ba

$\压裂{d \ \大(f (x) \大)}{dx} = \ lim_ {h \ \ \ 0} \压裂{f (x + h) - f (x)} {h}。gydF4y2Ba$

让gydF4y2Ba $F (x) = x^ngydF4y2Ba$．替代gydF4y2Ba $x + hgydF4y2Ba$为gydF4y2Ba $xgydF4y2Ba$得到gydF4y2Ba $F (x+h) = (x+h)^ngydF4y2Ba$．gydF4y2Ba

$\ \压裂{d (x ^ n)} {dx} = \ lim_ {h \ \ \ 0} \压裂{(x + h) ^ n - x ^ n} {h}。gydF4y2Ba$

二项式定理说明了gydF4y2Ba

$\{对齐}开始(x + h) ^ n & = \ \ limits_和c \ = \ {0} ^ n \ _nC_c \ x ^ {c} \ h c ^ \ \ & = \ _nC_0 \ x ^ {n-0} \ h ^ 0 + \ _nC_1 \ x ^ {n} \ h ^ 1 + \ _nC_2 \ x ^ {2} \ h ^ 2 + \ cdots + \ _nC_ {n} \ x ^ 1 \ h ^ {n} + \ _nC_n \ x ^ {n n} \ h ^ n \ \ & = x ^ n + nx ^ {n} h + \压裂{n (n - 1)} {2} x ^ {2} h ^ 2 + \ cdots + h ^ n。结束\{对齐}gydF4y2Ba$

代入gydF4y2Ba $(x + h) ^ ngydF4y2Ba$得到gydF4y2Ba

$\ \压裂{d (x ^ n)} {dx} = \ lim_ {h \ \ \ 0} \压裂{\离开(x ^ n + nx ^ {n} h + \压裂{n (n - 1)} {2} x ^ {2} h ^ 2 + \ cdots + h ^ n \右)- x ^ n} {h}。gydF4y2Ba$

取消gydF4y2Ba $x ^ ngydF4y2Ba$而且gydF4y2Ba $hgydF4y2Ba$得到gydF4y2Ba

${对齐}\ \开始压裂{d \ (x ^ n)} {dx} & = \ lim_ {h \ \ \ 0} \压裂{\离开(nx ^ {n} h + \压裂{n (n - 1)} {2} x ^ {2} h ^ 2 + \ cdots + h ^ n \右)}{h } \\\\ &= \ lim_ {h \ \ \ 0} \压裂{h \ \离开(nx ^ {n} + \压裂{n (n - 1)} {2} x ^ {2} h + \ cdots + h ^ {n} \右)}{h } \\\\ &= \ lim_ {h \ \ \ 0} \离开(nx ^ {n} + \压裂{n (n - 1)} {2} x ^ {2} h + \ cdots + h ^ {n} \右)。结束\{对齐}gydF4y2Ba$

根据极限的定义，gydF4y2Ba

$\ \压裂{d (x ^ n)} {dx} = nx ^ {n} + 0 + \ cdots + 0 = \压裂{d \ (x ^ n)} {dx} = nx ^ {n}。\ _ \广场gydF4y2Ba$

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让gydF4y2Ba $ngydF4y2Ba$而且gydF4y2Ba $xgydF4y2Ba$是实数，用gydF4y2Ba $N \neq 0gydF4y2Ba$而且gydF4y2Ba $X \neq 0gydF4y2Ba$．gydF4y2Ba

根据求导的幂法则，我们知道gydF4y2Ba

$\frac{\delta(x^n)}{\delta x} = n\cdot x^{n-1}。gydF4y2Ba$

(希腊语单词"gydF4y2Ba $\δgydF4y2Ba$意思是“…的小变化”。)gydF4y2Ba

证据2:gydF4y2Ba

让gydF4y2Ba $Y = x^ngydF4y2Ba$．gydF4y2Ba

我们将看看我们会得到什么样的小变化gydF4y2Ba $ygydF4y2Ba$如果我们改变gydF4y2Ba $xgydF4y2Ba$加入少量的gydF4y2Ba $xgydF4y2Ba$(用gydF4y2Ba $\δxgydF4y2Ba$)：gydF4y2Ba

$Y = x^n \意味着Y + Y = (x + x)^n。gydF4y2Ba$

扩大gydF4y2Ba $(x + \ x) ^ngydF4y2Ba$用二项式展开得到gydF4y2Ba

$y + \δy = x ^ n + n \ cdot x ^ {n}(\δx) + \压裂{n (n - 1) \ cdot x ^{2} \δx ^ 2} {2 !} + (\text{具有}\delta x更高升序幂的项)。gydF4y2Ba$

替代gydF4y2Ba $x ^ ngydF4y2Ba$为gydF4y2Ba $ygydF4y2Ba$方程两边同时除以gydF4y2Ba $\δxgydF4y2Ba$得到gydF4y2Ba

$\{对齐}开始x ^ n + \δy & = x ^ n + n \ cdot x ^ {n}(\δx) + \压裂{n (n - 1) \ cdot x ^{2} \δx ^ 2} {2 !}+ (\text{{\delta x升幂较高的项)\\ \\ \frac{\delta y}{\delta x} &= n \cdot x^{n-1}+ \frac{n(n-1) \cdot x^{n-2}(\delta x)}{2!} + (\text{具有较高升序幂的}\delta x的项). \end{aligned}gydF4y2Ba$

作为gydF4y2Ba $x \到0，gydF4y2Ba$的幂项gydF4y2Ba $\δxgydF4y2Ba$大于gydF4y2Ba $0gydF4y2Ba$变得小到可以被忽略:gydF4y2Ba

$\displaystyle \lim_{\ x \to 0} \frac{\ y}{\ x} = n \cdot x^{n-1}。\ _ \广场gydF4y2Ba$

让gydF4y2Ba $xgydF4y2Ba$那么，它是一个任意变量gydF4y2Ba

$\frac{d\ (x)}{dx} = 1。gydF4y2Ba$

证明:gydF4y2Ba

让gydF4y2Ba $ngydF4y2Ba$而且gydF4y2Ba $xgydF4y2Ba$是实数，用gydF4y2Ba $N \neq 0gydF4y2Ba$而且gydF4y2Ba $X \neq 0gydF4y2Ba$．gydF4y2Ba

幂法则指出gydF4y2Ba

$\frac{d\ (x^n)}{dx} = nx^{n-1}。gydF4y2Ba$

让gydF4y2Ba $X ^n = X = X ^1gydF4y2Ba$．gydF4y2Ba

$\frac{d\ (x)}{dx} = 1 \cdot x^{1-1} = x^0 = 1。\ _ \广场gydF4y2Ba$

让gydF4y2Ba $f (x)gydF4y2Ba$而且gydF4y2Ba $g (x)gydF4y2Ba$是任意可微函数。gydF4y2Ba

回忆一下，对于任意的函数gydF4y2Ba $(x)gydF4y2Ba$，gydF4y2Ba

$\压裂{d \ \大(大)(x) \} {dx} = ' (x) = \ lim_ {h \ \ \ 0} \压裂{(x + h)——(x)} {h}。gydF4y2Ba$

的gydF4y2Ba求和规则gydF4y2Ba州gydF4y2Ba

$\压裂{d \ \大(f (x) + g (x) \大)}{dx} = f (x) + g的(x)。gydF4y2Ba$

证明:gydF4y2Ba

让gydF4y2Ba $A (x) = f(x) + g(x)gydF4y2Ba$．替代gydF4y2Ba $x + hgydF4y2Ba$为gydF4y2Ba $xgydF4y2Ba$得到gydF4y2Ba $A (x+h) = f(x+h) + g(x+h)gydF4y2Ba$：gydF4y2Ba

${对齐}\ \开始压裂{d \ \大(f (x) + g (x) \大)}{dx} & = \ lim_ {h \ \ \ 0} \压裂{f (x + h) + g (x + h) - \大(f (x) + g (x) \大)}{h} \ \ & = \ lim_ {h \ \ \ 0} \压裂{f (x + h) + g (x + h) - f (x) g (x)} {h} \ \ & = \ lim_ {h \ \ \ 0} \压裂{f (x + h) - f (x)} {h} + \ lim_ {h \ \ \ 0} \压裂{g (x + h) - g (x)} {h}。结束\{对齐}gydF4y2Ba$

根据导数的定义，gydF4y2Ba

$\压裂{d \ \大(f (x) + g (x) \大)}{dx} = f (x) + g的(x)。\ _ \广场gydF4y2Ba$

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的gydF4y2Ba不同的规则gydF4y2Ba州gydF4y2Ba

$\压裂{d \ \大(f (x), g (x) \大)}{dx} = f (x) - g的(x)。gydF4y2Ba$

证明:gydF4y2Ba

让gydF4y2Ba $A (x) = f(x) - g(x)gydF4y2Ba$．替代gydF4y2Ba $x + hgydF4y2Ba$为gydF4y2Ba $xgydF4y2Ba$得到gydF4y2Ba $A (x+h) = f(x+h) - g(x+h)gydF4y2Ba$：gydF4y2Ba

$\开始{对齐}\压裂{d \ \大(f (x), g (x) \大)}{dx} & = \ lim_ {h \ \ \ 0} \压裂{f (x + h) - g (x + h) - \大(f (x), g (x) \大)}{h} \ \ & = \ lim_ {h \ \ \ 0} \压裂{f (x + h) - g (x + h) - f (x) + g (x)} {h} \ \ & = \ lim_ {h \ \ \ 0} \压裂{f (x + h) - f (x)} {h} - \ lim_ {h \ \ \ 0} \压裂{g (x + h) - g (x)} {h}。结束\{对齐}gydF4y2Ba$

根据导数的定义，gydF4y2Ba

$\压裂{d \ \大(f (x), g (x) \大)}{dx} = f (x) - g的(x)。\ _ \广场gydF4y2Ba$

让gydF4y2Ba $f (x)gydF4y2Ba$而且gydF4y2Ba $g (x)gydF4y2Ba$是任意可微函数。gydF4y2Ba

乘法法则指出gydF4y2Ba

$\压裂{d \ \大(f (x) \ cdot g (x) \大)}{dx} = f (x)的\ cdot g (x) + f (x) \ cdot g’(x)。gydF4y2Ba$

证明:gydF4y2Ba

回忆一下，对于任意的函数gydF4y2Ba $(x)gydF4y2Ba$，gydF4y2Ba

$\压裂{d \ \大(大)(x) \} {dx} = ' (x) = \ lim_ {h \ \ \ 0} \压裂{(x + h)——(x)} {h}。gydF4y2Ba$

让gydF4y2Ba $A (x) = f(x) \cdot g(x)gydF4y2Ba$．替代gydF4y2Ba $x + hgydF4y2Ba$为gydF4y2Ba $xgydF4y2Ba$得到gydF4y2Ba $A (x+h) = f(x+h) \cdot g(x+h)gydF4y2Ba$：gydF4y2Ba

$\压裂{d \ \大(f (x) \ cdot g (x) \大)}{dx} = \ lim_ {h \ \ \ 0} \压裂{f (x + h) \ cdot g (x + h) - f (x) \ cdot g (x)} {h} = \ lim_ {h \ \ \ 0} \压裂{- f (x) \ cdot g (x) + f (x + h) \ cdot g (x + h)} {h}。gydF4y2Ba$

加减法gydF4y2Ba $F (x+h) \cdot g(x)gydF4y2Ba$在分子上gydF4y2Ba

${对齐}\ \开始压裂{d \ \大(f (x) \ cdot g (x) \大)}{dx} & = \ lim_ {h \ \ \ 0} \压裂{f (x + h) \ cdot g (x) - f (x) \ cdot g (x) + f (x + h) \ cdot g (x + h) - f (x + h) \ cdot g (x)} {h } \\ \\ &= \ lim_ {h \ \ \ 0} \压裂{f (x + h) \ cdot g (x) - f (x) \ cdot g (x)} {h} + \ lim_ {h \ \ \ 0} \压裂{f (x + h) \ cdot g (x + h) - f (x + h) \ cdot g (x)} {h } \\ \\ &= \ lim_ {h \ \ \ 0} \离开(g (x) \ cdot \压裂{f (x + h) - f (x)} {h} \右)+ \ lim_ {h \ \ \ 0} \离开(f (x + h) \ cdot \压裂{g (x + h) - g (x)} {h} \ ) \\ \\ &= \ 左(\ lim_ h \ \ \ {0} g (x) \) \ cdoth \ \ \ \ lim_{0} \压裂{f (x + h) - f (x)} {h} + \离开(\ lim_ \ \ \ 0} {h f (x + h) \) \ cdot \ lim_ {h \ \ \ 0} \压裂{g (x + h) - g (x)} {h}。结束\{对齐}gydF4y2Ba$

根据极限和导数的定义，gydF4y2Ba

$\压裂{d \ \大(f (x) \ cdot g (x) \大)}{dx} = g (x) \ cdot f (x) + f (x) \ cdot g f (x) = ' (x) \ cdot g (x) + f (x) \ cdot g’(x)。\ _ \广场gydF4y2Ba$

让gydF4y2Ba $f (x)gydF4y2Ba$而且gydF4y2Ba $g (x)gydF4y2Ba$是可微函数gydF4y2Ba $F (x) \neq 0gydF4y2Ba$而且gydF4y2Ba $G (x) \neq 0gydF4y2Ba$．gydF4y2Ba

除法法则指出gydF4y2Ba

$左(\ \ \ \压裂{d压裂{f (x)} {g (x)} \右)}{dx} = \压裂{f (x) \ cdot g的(x) - f (x) \ cdot g’(x)}{\大(g (x) \大)^ 2}。gydF4y2Ba$

证明:gydF4y2Ba

回忆一下，对于任意的函数gydF4y2Ba $(x)gydF4y2Ba$，gydF4y2Ba

$\压裂{d \ \大(大)(x) \} {dx} = ' (x) = \ lim_ {h \ \ \ 0} \压裂{(x + h)——(x)} {h}。gydF4y2Ba$

让gydF4y2Ba $A (x) = \frac{f(x)}{g(x)}gydF4y2Ba$．替代gydF4y2Ba $x + hgydF4y2Ba$为gydF4y2Ba $xgydF4y2Ba$得到gydF4y2Ba $A (x+h) = \frac{f(x+h)}{g(x+h)}:gydF4y2Ba$

${对齐}\ \开始压裂{d \ \离开(\压裂{f (x)} {g (x)} \右)}{dx} & = \ lim_ {h \ \ \ 0} \压裂{\压裂{f (x + h)} {g (x + h)} - \压裂{f (x)} {g (x)}} {h } \\ \\ &= \ lim_ {h \ \ \ 0} \压裂{\压裂{f (x + h) \ \ cdot \ g (x)} {g (x + h) \ \ cdot \ g (x)} - \压裂{f (x) \ \ cdot \ g (x + h)} {g (x) \ \ cdot \ g (x + h)}} {h } \\ \\ &= \ lim_ h \ \ \{0}左(\ \压裂{1}{h} \ cdot \压裂{f (x + h) \ cdot g (x) - f (x) \ cdot g (x + h)} {g (x + h) \ cdot g (x)} \ ) \\ \\ &= \ lim_ {h \ \ \ 0} \压裂{f (x + h) \ cdot g (x) - f (x) \ cdot g (x + h)} {h (x + h) \ \ cdot g cdot g (x)}。结束\{对齐}gydF4y2Ba$

加减法gydF4y2Ba $F (x) \cdot g(x)gydF4y2Ba$在分子上gydF4y2Ba

${对齐}\ \开始压裂{d \ \离开(\压裂{f (x)} {g (x)} \右)}{dx} & = \ lim_ {h \ \ \ 0} \压裂{f (x + h) \ cdot g (x) - f (x) \ cdot g (x) - f (x) \ cdot g (x + h) + f (x) \ cdot g (x)} {h \ cdot g (x + h) \ cdot g (x )} \\ &= \ lim_ {h \ \ \ 0} \压裂{\压裂{f (x + h) \ \ cdot \ g (x) - f (x) \ \ cdot \ g (x) - f (x) \ \ cdot \ g (x + h) + f (x) \ \ cdot \ g (x)} {h}} {g (x + h) \ cdot g (x )} \\ \\ &= \ lim_ {h \ \ \ 0} \压裂{\压裂{f (x + h) \ \ cdot \ g (x) - f (x) \ \ cdot \ g (x)} {h} - \压裂{f (x) \ \ cdot \ g (x + h) - f (x) \ \ cdot \ g (x)} {h}} {g (x + h) \ cdot g (x )} \\ \\ &=h \ \ \ \ lim_{0} \压裂{g (x) \ cdot \压裂{f (x + h) - f (x)} {h} - f (x) \ cdot \压裂{g (x + h) - g (x)} {h}} {g (x + h) \ cdot g (x )} \\ \\ &= \ 压裂{\ displaystyle \ lim_ h \ \ \ {0} g (x) \ cdot \ lim_ {h \ \ \ 0} \压裂{f (x + h) - f (x)} {h} - \ lim_ \ \ \ 0} {h f (x) \ cdot \ lim_ {h \ \ \ 0} \压裂{g (x + h) - g (x)} {h}} {\ displaystyle \ lim_ h \ \ \ {0} g (x + h) \ cdot \ lim_ h \ \ \ {0} g (x)}。结束\{对齐}gydF4y2Ba$

根据极限和导数的定义，gydF4y2Ba

$左(\ \ \ \压裂{d压裂{f (x)} {g (x)} \右)}{dx} = \压裂{g (x) \ cdot f (x) - f (x) \ cdot g’(x)} {g (x) \ cdot g (x)} = \压裂{f (x) \ cdot g的(x) - f (x) \ cdot g’(x)}{\大(g (x) \大)^ 2}。\ _ \广场gydF4y2Ba$