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y=【(2x+1)(2-3x)^1/3】/(x-3)^2/3求导
人气:142 ℃ 时间:2019-12-01 12:37:12
解答
y=(2x+1)(2-3x)^(1/3)(x-3)^(-2/3)
y'=(2x+1)'*(2-3x)^(1/3)(x-3)^(-2/3)+(2x+1)[(2-3x)^(1/3)]'(x-3)^(-2/3)+(2x+1)(2-3x)^(1/3)[(x-3)^(-2/3)]'
=2(2-3x)^(1/3)(x-3)^(-2/3)+1/3*(-3)*(2x+1)(2-3x)^(-2/3)(x-3)^(-2/3)-(2/3)(2x+1)(2-3x)^(1/3)(x-3)^(-5/3)
=2(2-3x)^(1/3)(x-3)^(-2/3)-(2x+1)(2-3x)^(-2/3)(x-3)^(-2/3)-(2/3)(2x+1)(2-3x)^(1/3)(x-3)^(-5/3)
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