# What is the derivative of f(x)=x/ln(1/x)?

Nov 7, 2015

$\frac{d}{\mathrm{dx}} \frac{x}{\ln \left(\frac{1}{x}\right)} = \frac{\ln \left(\frac{1}{x}\right) + 1}{{\ln}^{2} \left(\frac{1}{x}\right)}$

#### Explanation:

To solve this, we will be using the quotient rule
$\frac{d}{\mathrm{dx}} f \frac{x}{g} \left(x\right) = \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{g} ^ 2 \left(x\right)$

and the chain rule
$\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = f ' \left(g \left(x\right)\right) g ' \left(x\right)$

For convenience, let's let ${f}_{1} \left(x\right) = x$ and ${f}_{2} \left(x\right) = \ln \left(\frac{1}{x}\right)$

First, we apply the quotient rule to get
$\frac{d}{\mathrm{dx}} \frac{x}{\ln \left(\frac{1}{x}\right)} = \frac{{f}_{1} ' \left(x\right) {f}_{2} \left(x\right) - {f}_{1} \left(x\right) {f}_{2} ' \left(x\right)}{f} _ {2}^{2} \left(x\right)$

We can easily see that ${f}_{1} ' \left(x\right) = 1$, but we will need to do a little more work to calculate ${f}_{2} ' \left(x\right)$

Before moving on, let's let ${f}_{3} \left(x\right) = \ln \left(x\right)$ and ${f}_{4} \left(x\right) = \frac{1}{x}$

Then ${f}_{2} \left(x\right) = {f}_{3} \left({f}_{4} \left(x\right)\right)$ and we can apply the chain rule to obtain
${f}_{2} ' \left(x\right) = {f}_{3} ' \left({f}_{4} \left(x\right)\right) {f}_{4} ' \left(x\right)$

${f}_{3} ' \left(x\right) = \frac{1}{x}$ and ${f}_{4} ' \left(x\right) = - \frac{1}{x} ^ 2$
so, substituting those in,

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

And now we can substitute back into the original equation to obtain

$\frac{d}{\mathrm{dx}} \frac{x}{\ln \left(\frac{1}{x}\right)} = \frac{1 \cdot \ln \left(\frac{1}{x}\right) - x \cdot \left(- \frac{1}{x}\right)}{{\ln}^{2} \left(\frac{1}{x}\right)} = \frac{\ln \left(\frac{1}{x}\right) + 1}{{\ln}^{2} \left(\frac{1}{x}\right)}$

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While the above shows a general method, it is worth noting that in this case, we can save some time and effort by observing that
$\ln \left(\frac{1}{x}\right) = \ln \left({x}^{-} 1\right) = - \ln \left(x\right)$

This removes the necessity of using the chain rule, as we can directly calculate the derivative
$\frac{d}{\mathrm{dx}} \ln \left(\frac{1}{x}\right) = \frac{d}{\mathrm{dx}} \left(- \ln \left(x\right)\right) = - \frac{1}{x}$