$|x|+|x-1|$ is continuous everywhere but fails to be differentiable exactly at two points $x=0$ and $x=1$.

So, there can be more such examples of functions.

Derivative of $x^2$ w.r.t. $x^3$ is $\frac{2}{3 x}$.

Let $$u=x^2 \text { and } v=x^3$$

$$\begin{array}{ll} \therefore & \frac{d u}{d x}=2 x \text { and } \frac{d v}{d x}=3 x^2 \\ \Rightarrow & \frac{d u}{d v}=\frac{2 x}{3 x^2}=\frac{2}{3 x} \end{array}$$

If $f(x)=|\cos x|$, then $f^{\prime}\left(\frac{\pi}{4}\right)$

$\because \quad 0< x<\frac{\pi}{2}, \cos x>0$.

$$\begin{array}{l} \therefore & f(x) & =+\cos x & \\ \Rightarrow & f^{\prime}(x) & =(-\sin x) \\ \Rightarrow & f^{\prime}\left(\frac{\pi}{4}\right) & =-\sin \frac{\pi}{4}=\frac{-1}{\sqrt{2}} & {\left[\because \sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right]} \end{array}$$

$$\begin{aligned} \because\quad f(x) & =|\cos x-\sin x|, \\ \therefore\quad f^{\prime}\left(\frac{\pi}{3}\right) & =\frac{\sqrt{3}+1}{2} \end{aligned}$$

$$\begin{aligned} &\text { We know that, } \frac{\pi}{4}< x<\frac{\pi}{2}, \sin x>\cos x\\ &\therefore \cos x-\sin x \leq 0 \text { i.e., } \quad f(x)=-(\cos x-\sin x) \end{aligned}$$

$$\begin{aligned} f^{\prime}(x) & =-[-\sin x-\cos x] \\ \therefore\quad f^{\prime}\left(\frac{\pi}{3}\right) & =-\left(\frac{-\sqrt{3}}{2}-\frac{1}{2}\right)=\left(\frac{\sqrt{3}+1}{2}\right) \end{aligned}$$

For the curve $\sqrt{x}+\sqrt{y}=1, \frac{d y}{d x}$ at $\left(\frac{1}{4}, \frac{1}{4}\right)$ is $-1$.

$$\begin{array}{lrl} \text { We have, } & \sqrt{x}+\sqrt{y} & =1 \\ \Rightarrow & \frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \frac{d y}{d x} & =0 \\ \Rightarrow & \frac{d y}{d x} & =-\frac{\sqrt{y}}{\sqrt{x}} \\ \therefore & \left(\frac{d y}{d x}\right)_{\left(\frac{1}{4} \cdot \frac{1}{4}\right)} & =\frac{-\frac{1}{2}}{\frac{1}{2}}=-1 \end{array}$$