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http://www.mathcenter.net/forum/showthread.php?t=1566



ѵҡ¹ŧ¢ͧ

f(x) ¹ҡ x x + h hf(x+h)−f(x) ö¹աẺ˹
ѵҡ¹ŧ¢ͧ f(x) ¹ҡ x1 x2 x2−x1f(x2)−f(x1) ҧ 1) s(t) = 2t - 3
ѵҡ¹ŧ t ¹ҡ 2 2.2 2.2−2s(2.2)−s(2) s(2) = 2(2) - 3 s(2.2) = 2(2.2) - 3 ᷹ҡ稺Ѻ

ǹѵҡ¹з x ͧ¹ Ҩҡ
limh→0hf(x+h)−f(x)
¹仡͡ٵá͹ؾѹ ҧ



dxdx=1
dxd(cx)=c



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