Let y=x2+2xx2−x ⇒(x2+2x)y=x2−x⇒x(x+2)y=x(x−1)⇒x[(x+2)y−(x−1)]=0∵x=0,∴(x+2)y−(x−1)=0⇒xy+2y−x+1=0⇒x(y−1)=−(2y+1)∴x=1−y2y+1⇒f−1(x)=1−x2x+1dxd(f−1(x))=(1−x)22(1−x)−(2x+1)(−1)=(1−x)22−2x+2x+1=(1−x)23
Let f(x)=x2+2xx2−xx=0,−2. Then dxd[f−1(x)] (wherever it is defined) is equal to:
Held on 9 Apr 2013 · Verified 6 Jul 2026.
(1−x)2−1
(1−x)23
(1−x)21
(1−x)2−3
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