# What is the derivative of 1/logx?

Aug 4, 2015

It is (-1)/(x(logx)^2(ln10) (Assuming that $\log x = {\log}_{10} x$).

#### Explanation:

The function $f \left(x\right) = \frac{1}{\log} x$

is of the form: $f \left(x\right) = \frac{1}{u} = {u}^{-} 1$

So we find its derivative using the power rule and the chain rule:

$f ' \left(x\right) = - 1 {u}^{-} 2 \cdot \frac{\mathrm{du}}{\mathrm{dx}} = \frac{- 1}{u} ^ 2 \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

In this case, $u = \log x$, which I take to mean ${\log}_{10} x$.

The derivative of ${\log}_{b} x = \frac{1}{x \ln b}$, so we have:

$\frac{\mathrm{du}}{\mathrm{dx}} \frac{d}{\mathrm{dx}} \left(\log x\right) = \frac{d}{\mathrm{dx}} \left({\log}_{10} x\right) = \frac{1}{x \ln 10}$.

Putting this together, we get:

f'(x) = (-1)/(logx)^2 * 1/(xln10) = (-1)/(x(logx)^2(ln10)

Note
If we are using $\log x$ for the natural logarithm (as is sometimes done), then,
that would be ${\log}_{e} x$ and natural log of $e$ is $1$, so we get

$f ' \left(x\right) = \frac{- 1}{\log x} ^ 2 \cdot \frac{1}{x} = \frac{- 1}{x {\left(\log x\right)}^{2}}$