What is the derivative of $y=1/log_2(x)$ ?

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Answer 1 · 2015-08-06T21:46:09

$y'=-log(2)/(x(logx)^2)$

Note that:

$1/log_2x=log(2)/log(x)$

We then have using the chain rule:

$y=kf(g(x))$
$y'=kf'(g(x))g'(x)$
$f(u)=1/u$
$g(x)=logx$

These have derivatives:
$f'(u)=-1/u^2$
$g'(x)=1/x$

The full derivative is therefore:
$y'=-log(2)/(x(logx)^2)$

What is the derivative of y=1/log_2(x)y=1/log_2(x)?