2(1−x)2
−2(1−x)2
−1√1−x
1√1−x
Correct answer is A
y=1+x1−x
Using quotient rule, vdudx−udvdxv2, we have
dydx=(1−x)(1)−(1+x)(−1)(1−x)2=(1−x+1+x)(1−x)2
= 2(1−x)2.
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