Asked by p
how to calculate dy/dx for the curve y=x^(ln x)^2
Answers
Answered by
Reiny
take ln of both sides
lny = ln[x^(lnx)^2]
= (lnx)^2(lnx)
= (lnx)^3
(dy/dx)/y = 3(lnx)^2 (1/x+
dy/dx = 3y(lnx)^2 /x
or
3x^(lnx)^2 (lnx)^2 /x
lny = ln[x^(lnx)^2]
= (lnx)^2(lnx)
= (lnx)^3
(dy/dx)/y = 3(lnx)^2 (1/x+
dy/dx = 3y(lnx)^2 /x
or
3x^(lnx)^2 (lnx)^2 /x
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