证明微分乘法律 $ d(\lambda \mu)=\lambda d\mu + \mu d\lambda $
对微分乘法法则的推导,即证明: $\quad d(\lambda \mu)=\lambda d\mu + \mu d\lambda $
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\[若y=\mu \lambda ,\quad \lambda = f(x),\quad \mu = g(x), 二者均以x为自变量
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\[y+\Delta y=(\mu +\Delta \mu) (\lambda +\Delta \lambda )
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\[=\mu \lambda +\lambda \Delta \mu +\mu \Delta \lambda +\Delta \lambda \Delta \mu
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\[y+\Delta y=y+\lambda \Delta \mu +\mu \Delta \lambda +\Delta \lambda \Delta \mu
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\[\Delta y=\lambda \Delta \mu +\mu \Delta \lambda +\Delta \lambda \Delta \mu
\]
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\[同时除以 \Delta x: \quad
\frac{\Delta y}{\Delta x}=\frac{\lambda \Delta \mu}{\Delta x}
+\frac{\mu \Delta \lambda }{\Delta x}
+\frac{\Delta \lambda \Delta \mu}{\Delta x}
\]
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\[引入极限,设 \Delta x \to 0: \\
\lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x}=
\lambda \lim_{\Delta x \to 0} \frac{\Delta \mu}{\Delta x}
+\mu \lim_{\Delta x \to 0} \frac{\Delta \lambda }{\Delta x}
+\lim_{\Delta x \to 0} \frac{\Delta \lambda \Delta \mu}{\Delta x}
\]
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\[\lim_{\Delta x \to 0} \frac{\Delta \lambda \Delta \mu}{\Delta x}
= \lim_{\Delta x \to 0} \frac{\Delta \lambda}{\Delta x} \cdot
\lim_{\Delta x \to 0} \Delta \mu
\]
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\[而\lim_{\Delta x \to 0} \Delta \mu =\lim_{\Delta x \to 0} g(x+\Delta x)-g(x)
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\[\because \Delta x \to 0,\quad
\therefore \lim_{\Delta x \to 0} \Delta \mu =0
\]
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\[得: \quad \frac{dy}{dx}= \lambda \frac{d\mu }{dx}
+\mu\frac{d\lambda }{dx} +\frac{d\lambda }{dx} \cdot 0
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\[\frac{dy}{dx}= \frac{\lambda d\mu +\mu d\lambda }{dx}
\]
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\[约去dx: \quad dy=\lambda d\mu + \mu d\lambda
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\[证明成立
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