高数同济第六版第157页例1的导数是怎么求的？
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f(x) = (x-4)(x+1)^(2/3)
f'(x) = (x+1)^(2/3) + (2/3)(x-4)(x+1)^(-1/3)
= (3/3)(x+1)(x+1)^(-1/3) + (2/3)(x-4)(x+1)^(-1/3)
= (1/3)[ 3x+3 + 2x - 8 ](x+1)^(-1/3)
= (1/3)[ 5x - 5 ](x+1)^(-1/3)
= (5/3)[ x - 1 ](x+1)^(-1/3)
f'(x) = (x+1)^(2/3) + (2/3)(x-4)(x+1)^(-1/3)
= (3/3)(x+1)(x+1)^(-1/3) + (2/3)(x-4)(x+1)^(-1/3)
= (1/3)[ 3x+3 + 2x - 8 ](x+1)^(-1/3)
= (1/3)[ 5x - 5 ](x+1)^(-1/3)
= (5/3)[ x - 1 ](x+1)^(-1/3)
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