$$\begin{aligned} \text{We have,}\quad & f(x)=|x-5| \\ \therefore\quad & f(x)=\left\{\begin{array}{l} -(x-5) \text {, if } x<5 \\ x-5, \quad \text { if } x \geq 5 \end{array}\right. \end{aligned}$$

For continuity at $x=5$,

$$\begin{aligned} \mathrm{LHL} & =\lim _{x \rightarrow 5^{-}}(-x+5) \\ & =\lim _{h \rightarrow 0}[-(5-h)+5]=\lim _{h \rightarrow 0} h=0 \\ \mathrm{RHL} & =\lim _{x \rightarrow 5^{+}}(x-5) \\ & =\lim _{h \rightarrow 0}(5+h-5)=\lim _{h \rightarrow 0} h=0 \\ \therefore \quad \mathrm{f}(5) & =5-5=0 \\ \Rightarrow \quad \mathrm{LHL} & =\mathrm{RHL}=f(5) \end{aligned}$$

Hence, $f(x)$ is continuous at $x=5$.

Now,

$$\begin{aligned} L f^{\prime}(5) & =\lim _{x \rightarrow 5^{-}} \frac{f(x)-f(5)}{x-5} \\ & =\lim _{x \rightarrow 5^{-}} \frac{-x+5-0}{x-5}=-1 \\ R f^{\prime}(5) & =\lim _{x \rightarrow 5^{+}} \frac{f(x)-f(5)}{x-5} \\ & =\lim _{x \rightarrow 5^{+}} \frac{x-5-0}{x-5}=1 \end{aligned}$$

$$\therefore \quad L f^{\prime}(5) \neq R f^{\prime}(5)$$

So, $f(x)=|x-5|$ is not differentiable at $x=5$.

Let $f: R \rightarrow R$ satisfies the equation $f(x+y)=f(x) \cdot f(y), \forall x, y \in R, f(x) \neq 0$.

Let $f(x)$ is differentiable at $x=0$ and $f^{\prime}(0)=2$.

$$\begin{array}{ll} \Rightarrow & f^{\prime}(0)=\lim _\limits{x \rightarrow 0} \frac{f(x)-f(0)}{x-0} \\ \Rightarrow & 2=\lim _\limits{x \rightarrow 0} \frac{f(x)-f(0)}{x} \\ \Rightarrow & 2=\lim _\limits{h \rightarrow 0} \frac{f(0+h)-f(0)}{0+h} \\ \Rightarrow & 2=\lim _\limits{h \rightarrow 0} \frac{f(0) \cdot f(h)-f(0)}{h} \\ \Rightarrow & 2=\lim _\limits{h \rightarrow 0} \frac{f(0)[f(h)-1]}{h}\quad [\because f(0)=f(h)] \ldots(\mathrm{i}) \end{array}$$

Also, $$f^{\prime}(x)=\lim _\limits{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$

$$\begin{aligned} & =\lim _{h \rightarrow 0} \frac{f(x) \cdot f(h)-f(x)}{h} & {[\because f(x+y)=f(x) \cdot f(y)] } \\ & =\lim _{h \rightarrow 0} \frac{f(x)[f(h)-1]}{h}=2 f(x) & \quad \text { [using Eq. (i)] } \\ \therefore\quad f^{\prime}(x) & =2 f(x) & \end{aligned}$$

$$\begin{aligned} \text{Let}\quad y & =2^{\cos ^2 x} \\ \therefore\quad \log y & =\log 2^{\cos ^2 x}=\cos ^2 x \cdot \log 2 \end{aligned}$$

On differentiating w.r.t. $x$, we get

$$\begin{aligned} \frac{d}{d y} \log y \cdot \frac{d y}{d x} & =\frac{d}{d x} \log 2 \cdot \cos ^2 x \\ \Rightarrow \quad \frac{1}{y} \cdot \frac{d y}{d x} & =\log 2 \frac{d}{d x}(\cos x)^2 \\ \Rightarrow \quad \frac{1}{y} \cdot \frac{d y}{d x} & =\log 2 \cdot[2 \cos x] \cdot \frac{d}{d x} \cos x \\ & =\log 2 \cdot 2 \cos x \cdot(-\sin x) \\ & =\log 2 \cdot[-(\sin 2 x)] \\ \therefore \quad \frac{d y}{d x} & =-y \cdot \log 2(\sin 2 x) \\ & =-2^{\cos ^2 x} \cdot \log 2(\sin 2 x) \end{aligned}$$

Let $$y=\frac{8^x}{x^8} \Rightarrow \log y=\log \frac{8^x}{x^8}$$

$$\begin{array}{ll} \Rightarrow & \frac{d}{d y} \log y \cdot \frac{d y}{d x}=\frac{d}{d x}\left[\log 8^x-\log x^8\right] \\ \Rightarrow & \frac{1}{y} \cdot \frac{d y}{d x}=\frac{d}{d x}[x \cdot \log 8-8 \cdot \log x] \end{array}$$

On differentiating w.r.t. $x$, we get

$$\begin{aligned} & \frac{1}{y} \cdot \frac{d y}{d x} =\log 8 \cdot 1-8 \cdot \frac{1}{x} \\ \Rightarrow \quad & \frac{1}{y} \cdot \frac{d y}{d x} =\log 8-\frac{8}{x} \\ \therefore \quad & \frac{d y}{d x} =y\left(\log 8-\frac{8}{x}\right)=\frac{8^x}{x^8}\left(\log 8-\frac{8}{x}\right) \end{aligned} $$

$$\begin{aligned} \text { Let } \quad & =\log \left(x+\sqrt{x^2+a}\right) \\ \therefore \quad \frac{d y}{d x} & =\frac{d}{d x} \log \left(x+\sqrt{x^2+a}\right) \\ & =\frac{1}{\left(x+\sqrt{x^2+a}\right)} \cdot \frac{d}{d x}\left[x+\sqrt{x^2+a}\right] \\ & =\frac{1}{\left(x+\sqrt{x^2+a}\right)}\left[1+\frac{1}{2}\left(x^2+a\right)^{-1 / 2} \cdot 2 x\right] \\ & =\frac{1}{\left(x+\sqrt{x^2+a}\right)} \cdot\left(1+\frac{x}{\sqrt{x^2+a}}\right) \\ & =\frac{\left(\sqrt{x^2+a}+x\right)}{\left(x+\sqrt{x^2+a}\right)\left(\sqrt{x^2+a}\right)}=\frac{1}{\left(\sqrt{x^2+a}\right)} \end{aligned}$$