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设y=(1/6)ln[(x+1)²/(x²-x+1)]+(1/√3)arctan[(2x-1)/√3]，x≠-1，求y′．

高老师2年前 (2024-03-25)高等数学(一)(00020)11

y=(1/3)ln(x+1)-(1/6)ln(x²-x+1)+(1/√3)arctan[(2x-1)/√3]y'=1/3(x+1)-(x²-x+1)′/6(x²-x+1)+1/√3{1/[1+(2x-1)²/3]}[(2x-1)/√3]=1/[3(x+1)]-(2x-1)/[6(x²-x+1)]+1/(2x²-2x+2)=1/(1+x³)

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设y=(1/6)ln[(x+1)2/(x2-x+1)]+(1/√3)arctan[(2x-1)/√3]，x≠-1，求y′．

设y=(1/6)ln[(x+1)²/(x²-x+1)]+(1/√3)arctan[(2x-1)/√3]，x≠-1，求y′．