函数求导简介
函数相乘求导
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[f(x)\cdot g(x)]'=f'(x)g(x)+f(x)g'(x)
[f(x)⋅g(x)]′=f′(x)g(x)+f(x)g′(x)
证明：
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\qquad [f(x)\cdot g(x)]'=\lim\limits_{\Delta x\rightarrow0}\dfrac{f(x+\Delta x)\cdot g(x+\Delta x)-f(x)\cdot g(x)}{\Delta x}
[f(x)⋅g(x)]′=Δx→0limΔxf(x+Δx)⋅g(x+Δx)−f(x)⋅g(x)
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\qquad \qquad \qquad \qquad =\lim\limits_{\Delta x\rightarrow0}\dfrac{f(x+\Delta x)\cdot g(x+\Delta x)-f(x)\cdot g(x+\Delta x)+f(x)\cdot g(x+\Delta x)-f(x)\cdot g(x)}{\Delta x}
=Δx→0limΔxf(x+Δx)⋅g(x+Δx)−f(x)⋅g(x+Δx)+f(x)⋅g(x+Δx)−f(x)⋅g(x)
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\qquad \qquad \qquad \qquad =\lim\limits_{\Delta x\rightarrow0}\dfrac{f'(x)\Delta x\cdot g(x)+g'(x)\Delta x\cdot f(x)}{\Delta x}
=Δx→0limΔxf′(x)Δx⋅g(x)+g′(x)Δx⋅f(x)
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\qquad \qquad \qquad \qquad =f'(x)g(x)+f(x)g'(x)
=f′(x)g(x)+f(x)g′(x)
函数相除求导
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[\dfrac{f(x)}{g(x)}]'=\dfrac{f'(x)g(x)-f(x)g'(x)}{g^2(x)}
[g(x)f(x)]′=g2(x)f′(x)g(x)−f(x)g′(x)
证明：
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\qquad [\dfrac{f(x)}{g(x)}]'=\lim\limits_{\Delta x\rightarrow0}\dfrac{\frac{f(x+\Delta x)}{g(x+\Delta x)}-\frac{f(x)}{g(x)}}{\Delta x}
[g(x)f(x)]′=Δx→0limΔxg(x+Δx)f(x+Δx)−g(x)f(x)
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\qquad \qquad \qquad \qquad =\lim\limits_{\Delta x\rightarrow0}\dfrac{f(x+\Delta x)g(x)-f(x)g(x+\Delta x)}{g(x+\Delta x)g(x)}\cdot\dfrac{1}{\Delta x}
=Δx→0limg(x+Δx)g(x)f(x+Δx)g(x)−f(x)g(x+Δx)⋅Δx1
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\qquad \qquad \qquad \qquad =\lim\limits_{\Delta x\rightarrow0}\dfrac{f(x+\Delta x)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+\Delta x)}{g(x+\Delta x)g(x)}\cdot\dfrac{1}{\Delta x}
=Δx→0limg(x+Δx)g(x)f(x+Δx)g(x)−f(x)g(x)+f(x)g(x)−f(x)g(x+Δx)⋅Δx1
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\qquad \qquad \qquad \qquad =\lim\limits_{\Delta x\rightarrow0}\dfrac{f'(x)\Delta x\cdot g(x)-f(x)\cdot g'(x)\Delta x}{g^2(x)\cdot \Delta x}
=Δx→0limg2(x)⋅Δxf′(x)Δx⋅g(x)−f(x)⋅g′(x)Δx
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\qquad \qquad \qquad \qquad =\dfrac{f'(x)g(x)-f(x)g'(x)}{g^2(x)}
=g2(x)f′(x)g(x)−f(x)g′(x)