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f(x) = (x3+1)ex ÀÏ ¶§, f′(0) ÀÇ °ªÀº? [3Á¡]

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1) f′(x) = (x3+1)′ex+(x3+1)(ex)′ = 3x2ex+(x3+1)ex
2) f′(0) = 0+e0 = 1 ^^

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1) (ax)′ = axlogea, (ex)′ = ex
2) (uv)′ =u′v+uv′, (uvw)′ = u′vw+uv′w+uvw′

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