f′(x)=limh→0(x+h)2+2(x+h)−x2−2xh=limh→02xh+2h+h2h=limh→0(2x+2+h)=2x+2f'(x)=lim_{h\to 0}\frac{(x+h)^2+2(x+h)-x^2-2x}{h}=lim_{h\to 0}\frac{2xh+2h+h^2}{h}=lim_{h\to 0}(2x+2+h)=2x+2f′(x)=limh→0h(x+h)2+2(x+h)−x2−2x=limh→0h2xh+2h+h2=limh→0(2x+2+h)=2x+2
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