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@@ -160,7 +160,6 @@ $=0$
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如$\lim\limits_{x\to 1}\dfrac{f(x)}{x-1}=2$且$f(x)$连续,可以推出$f(1)=0$与$f'(1)=2$。
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\section{函数求导法则}
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\subsection{四则运算}
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