2009
232
J & K CETJ & K CET 2014Continuity and Differentiability
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Solution:
We have y=xe2y Taking log on both sides, we get logy=log(xe2y) ⇒logy=logx+2yloge ⇒logy=logx+2y
On differentiating w. r. t. x, we get y1dxdy=x1+2dxdy ⇒dxdy(y1−2)=x1 ⇒dxdy=x1×(1−2y)y ⇒dxdy=x(1−2y)y