已知y=³√［(x+1)(x+2)/(x+3)(4-x)］，求y′
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y = ³√［(x+1)(x+2)/(x+3)(4-x)］
lny = (1/3)[ln(x+1)+ln(x+2)-ln(x+3)-ln(4-x)]
y'/y = (1/3)[1/(x+1)+1/(x+2)-1/(x+3)+1/(4-x)]
y' = (1/3) ³√［(x+1)(x+2)/(x+3)(4-x)] [1/(x+1)+1/(x+2)-1/(x+3)+1/(4-x)]
lny = (1/3)[ln(x+1)+ln(x+2)-ln(x+3)-ln(4-x)]
y'/y = (1/3)[1/(x+1)+1/(x+2)-1/(x+3)+1/(4-x)]
y' = (1/3) ³√［(x+1)(x+2)/(x+3)(4-x)] [1/(x+1)+1/(x+2)-1/(x+3)+1/(4-x)]
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