Questions
If
a
yx(1n xlogx−1)+2 1n x 1n (1nx)
b
yx(logx)log(logx)(2log(logx)+1)
c
yx 1nx[(1nx)2+21n(1nx)]
d
yxlogylogx[2log(logx)+1]
detailed solution
Correct option is B
y=x(logx)log(logx)⇒log y=(logx)(logx)log(logx)→(1)Taking log of both sides, we get ⇒log (logy)=log(logx)+log(logx)log(logx)Diff. w.r.t x, we get 1logy.1ydydx=1xlogx+2log(logx)logx1x=2log(logx)+1xlogx⇒dydx=yx.logylogx(2log(logx)+1)Substituting the value of log y from (1), we get yx(logx)log(logx)(2log(logx)+1)Talk to our academic expert!
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