Answers

2015-01-20T01:36:28+05:30
Y = 1/logx
y = 1/(lnx/ln10)                                             as lnx = ln10*logx
y = ln10*(lnx)^-1
let lnx = u
we know d(u^n)/dx = nu^(n-1)du/dx
d(lnx)/dx = 1/x
so
dy/dx = ln10{-1*(u)^-2}du/dx
put u = lnx
dy/dx = ln10{-(lnx)^-2}*1/x
hence
dy/dx = -ln10/{x(lnx)²}
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