x+xy+y=2y=(2−x)(x+1)y′=((2−x)(x+1))′=(2−x)′∗(x+1)−(2−x)∗(x+1)′(x+1)2==(−1)∗(x+1)−(2−x)∗1(x+1)2=−3(x+1)2x+xy+y=2\\ y={\frac {(2-x)} {(x+1)}}\\ y'=({\frac {(2-x)} {(x+1)}})'={\frac {(2-x)'*(x+1)-(2-x)*(x+1)'}{(x+1)^2}}=\\ ={\frac {(-1)*(x+1)-(2-x)*1}{(x+1)^2}}={\frac{-3}{(x+1)^2}}x+xy+y=2y=(x+1)(2−x)y′=((x+1)(2−x))′=(x+1)2(2−x)′∗(x+1)−(2−x)∗(x+1)′==(x+1)2(−1)∗(x+1)−(2−x)∗1=(x+1)2−3
answer: −3(x+1)2{\frac{-3}{(x+1)^2}}(x+1)2−3
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