# How do you find where f(x)=1/ln(x) is continuous and differentiable on which interval(s)?

Aug 6, 2015

It's continuous and differentiable on the union $\left(0 , 1\right) \setminus \cup \left(1 , \infty\right)$. That is, for all $x > 0$ except $x = 1$.

#### Explanation:

The function $\ln \left(x\right)$ is continuous and differentiable for all $x > 0$. Therefore, $\frac{1}{\ln \left(x\right)}$ will be continuous and differentiable for all such values of $x$ as well, except for those values of $x$ where $\ln \left(x\right) = 0$. The only such value of $x$ where the logarithm is zero is $x = 1$.

Therefore, $\frac{1}{\ln \left(x\right)}$ is continuous and differentiable on the union $\left(0 , 1\right) \setminus \cup \left(1 , \infty\right)$. That is, for all $x > 0$ except $x = 1$.

Moreover, if you are interested, the Quotient Rule gives

$\frac{d}{\mathrm{dx}} \left(\frac{1}{\ln \left(x\right)}\right) = \frac{\ln \left(x\right) \cdot 0 - 1 \cdot \frac{1}{x}}{\ln \left(x\right)} ^ 2 = - \frac{1}{x {\left(\ln \left(x\right)\right)}^{2}}$ for $x > 0$ and $x \ne 1$.