How do you find the derivative of $\frac{1}{log x}$ $1/logx$ ?

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Answer 1 · 2016-06-05T19:46:29

It is $= - \frac{1}{x {log (x)}^{2}}$ $=-1/(xlog(x)^2)$ .

$\frac{1}{log (x)} = {log (x)}^{- 1}$ $1/log(x)=log(x)^-1$

So you apply the chain rule to the power and then to the log:

$\frac{d}{d x} {log (x)}^{- 1} = - 1 \cdot {log (x)}^{- 2} \cdot \frac{d}{d x} log (x)$ $d/dxlog(x)^-1=-1*log(x)^-2*d/dxlog(x)$

$= - \frac{1}{{log (x)}^{2}} \frac{1}{x}$ $=-1/(log(x)^2) 1/x$

$= - \frac{1}{x {log (x)}^{2}}$ $=-1/(xlog(x)^2)$ .

How do you find the derivative of 1logx1/logx?