f(x)=x³,0∆x→1
Δf(x)Δx=(x+1)3−x3=(x+1)2+x(x+1)+x2=3x2+3x+1\frac{\Delta f(x)}{\Delta x}=(x+1)^3-x^3=(x+1)^2+x(x+1)+x^2=3x^2+3x+1ΔxΔf(x)=(x+1)3−x3=(x+1)2+x(x+1)+x2=3x2+3x+1
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