How do you differentiate $y = 1 / log_2 x$ ?

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Answer 1 · 2018-03-31T08:54:58

$(dy)/(dx)=-ln2/(x(lnx)^2)$

$log_2x=lnx/ln2$

Therefore $y=ln2/lnx=ln2*1/lnx$

and $(dy)/(dx)=ln2*(-1/(lnx)^2)*1/x$

= $-ln2/(x(lnx)^2)$

How do you differentiate y = 1 / log_2 xy = 1 / log_2 x?