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