How do you differentiate $y=log_x2$ ?

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Answer 1 · 2017-07-11T13:29:11

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

As $y=log_x2$ , we have $y=ln2/lnx$

Hence assuming $g(f(x))=1/(f(x))$ , where $f(x)=lnx$

$(dy)/(dx)=ln2xx(dg(x))/(df(x))xx(df)/(dx)$

$=ln2xx(-1/(lnx)^2)xx1/x$

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

How do you differentiate y=log_x2 y=log_x2 ?