Question

how to differntiate (1/logx)

Answer 1

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}

Answer 2

Answer:

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}