check if $\sqrt{x} \cdot \log(x-1)$ is differentiable $\forall x \in (2,4)$

Question

let

$f:[2,4] \to \mathbb{R}, f(x)= \sqrt{x} \cdot \log(x-1)$.

How can I check if $f$ is differentiable $\forall x \in (2,4)$ ?

The function is valid $\forall x \in (2,4)$ because $x+h > 0 \forall x \in (2,4),\quad h \to 0$ and $x+h-1 > 1 \forall x \in (2,4),\quad h \to 0$

$\begin{align} \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} &= \lim_{h \to 0} \frac{\sqrt{x+h} \cdot \log(x+h-1)- \sqrt{x} \cdot \log(x-1)}{h} \\ \end{align}$

Question: How can I go on ? If I use L'Hopital's rule $h$ turns into $0$ in the denominator. So that step does not make any sense.

Answer 1

$$\frac{\sqrt x\log(x-1)-\sqrt{x_0}\log(x_0-1)}{x-x_0}=$$

$$=\frac{\sqrt x\log(x-1)-\sqrt{x_0}\log(x-1)+\sqrt{x_0}\log(x-1)-\sqrt{x_0}\log(x_0-1)}{x-x_0}=$$

$$\frac{\left(\sqrt x-\sqrt{x_0}\right)\log(x-1)+\sqrt{x_0}\log\frac{x-1}{x_0-1}}{x-x_0}=$$

$$=\left(\sqrt x+\sqrt{x_0}\right)\log(x-1)+\sqrt{x_0}\frac{\log\frac{x-1}{x_0-1}}{x-x_0}\xrightarrow[x\to x_0]{}2\sqrt{x_0}\log(x_0-1)+\frac{\sqrt{x_0}}{x_0-1}\frac{x_0-1}{x_0-1}=$$

$$=2\sqrt{x_0}\log(x_0-1)+\frac{\sqrt{x_0}}{x_0-1}$$

Answer 2

Just $$(\sqrt{x}\ln(x-1))'=\frac{\ln(x-1)}{2\sqrt{x}}+\frac{\sqrt{x}}{x-1},$$ which exists for all $x\in(2,4)$.

Answer 3

You can continue by adding and subtracting $$\frac{\sqrt{x+h}\log(x+h)-\sqrt{x}\log(x)}{h}=\frac{\sqrt{x+h}-\sqrt{x}}{h}\log(x+h)+\sqrt{x}\frac{\log(x+h)-\log(x)}{h}$$

Then multiplying and dividing by $\sqrt{x+h}+\sqrt{x}$

$=\frac{1}{\sqrt{x+h}+\sqrt{x}}\log(x+h)+\sqrt{x}\log(1+h/x)^{1/h}$

and taking limits and using the fundamental exponential limit $(1+h)^{1/h}\to e$ as $h\to 0$.

This tends to $\frac{1}{2\sqrt{x}}\log(x)+\sqrt{x}\log(e^{1/x})$