x2x−2xxcoty−1=0 Now x=1, 1−2coty−1=0⇒coty=0⇒y=2π Now differentiating eq. (1) w.r.t. ' x ' 2x2x(1+logx)−2[xx(−cosec2y)dxdy+cotyxx(1+logx)]=0 Now at (1,2π) 2(1+log1)−2(1(−1)(dxdy)(1,2π)+0)=0⇒2+2(dxdy)(1,2π)=0⇒(dxdy)(1,2π)=−1
JEE Main 2009 — Mathematics Calculus
Let y be an implicit function of x defined by x2x−2xxcoty−1=0. Then y′(1) equals
Held on 30 Apr 2009 · Verified 6 Jul 2026.
−1
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log2
−log2
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