How do you differentiate $y=ln(1/x)-1/lnx$ ?

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Answer 1 · 2018-04-01T18:08:08

$y'=-1/x+1/(x(lnx)^2)$

Let's rewrite the function a bit, mostly to get rid of $ln(1/x).$

$y=(ln1-lnx)-(lnx)^-1$

Since $ln1=0:$

$y=-lnx-(lnx)^-1$

$y'=-1/x-(-(lnx)^-2*d/dxlnx)$ (Chain Rule)

$y'=-1/x+1/(x(lnx)^2)$

How do you differentiate y=ln(1/x)-1/lnxy=ln(1/x)-1/lnx?