If $f(x)=|\cos x-\sin x|$, then $f^{\prime}\left(\frac{\pi}{3}\right)$ is equal to ............. .
$$\begin{aligned} \because\quad f(x) & =|\cos x-\sin x|, \\ \therefore\quad f^{\prime}\left(\frac{\pi}{3}\right) & =\frac{\sqrt{3}+1}{2} \end{aligned}$$
$$\begin{aligned} &\text { We know that, } \frac{\pi}{4}< x<\frac{\pi}{2}, \sin x>\cos x\\ &\therefore \cos x-\sin x \leq 0 \text { i.e., } \quad f(x)=-(\cos x-\sin x) \end{aligned}$$
$$\begin{aligned} f^{\prime}(x) & =-[-\sin x-\cos x] \\ \therefore\quad f^{\prime}\left(\frac{\pi}{3}\right) & =-\left(\frac{-\sqrt{3}}{2}-\frac{1}{2}\right)=\left(\frac{\sqrt{3}+1}{2}\right) \end{aligned}$$
For the curve $\sqrt{x}+\sqrt{y}=1, \frac{d y}{d x}$ at $\left(\frac{1}{4}, \frac{1}{4}\right)$ is ............. .
For the curve $\sqrt{x}+\sqrt{y}=1, \frac{d y}{d x}$ at $\left(\frac{1}{4}, \frac{1}{4}\right)$ is $-1$.
$$\begin{array}{lrl} \text { We have, } & \sqrt{x}+\sqrt{y} & =1 \\ \Rightarrow & \frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \frac{d y}{d x} & =0 \\ \Rightarrow & \frac{d y}{d x} & =-\frac{\sqrt{y}}{\sqrt{x}} \\ \therefore & \left(\frac{d y}{d x}\right)_{\left(\frac{1}{4} \cdot \frac{1}{4}\right)} & =\frac{-\frac{1}{2}}{\frac{1}{2}}=-1 \end{array}$$
Rolle's theorem is applicable for the function $f(x)=|x-1|$ in $[0,2]$.
If $f$ is continuous on its domain $D$, then $|f|$ is also continuous on $D$.
The composition of two continuous function is a continuous function.