(a) Check the differentiability of function f ( x )=[ x ] at x =-3, where [·] denotes greatest integer function.

$$\begin{aligned} & \mathrm{f}(\mathrm{x})=[\mathrm{x}] \text { at } \mathrm{x}=-3 \\ & \begin{aligned} \text { RHD } & =\lim _{h \rightarrow 0} \frac{\mathrm{f}(-3+h)-\mathrm{f}(-3)}{h} \\ & =\lim _{h \rightarrow 0} \frac{-3-(-3)}{h}=0 \\ \text { LHD } & =\lim _{h \rightarrow 0} \frac{\mathrm{f}(-3-h)-\mathrm{f}(-3)}{-h} \\ & =\lim _{h \rightarrow 0} \frac{-4-(-3)}{h}=\lim _{h \rightarrow 0}\left(\frac{-1}{h}\right) \\ & =\text { not defined } \end{aligned} \\ & \because \text { LHD } \neq \text { RHD } \end{aligned} $$ So f is not differentiable at $\displaystyle \mathrm{x}=-3$ $$\begin{aligned} & \frac{1}{3} \mathrm{x}^{\frac{-2}{3}}+\frac{1}{3} y^{\frac{-2}{3}} \frac{d y}{d \mathrm{x}}=0 \\ & \frac{d y}{d \mathrm{x}}=\frac{-\mathrm{x}^{\frac{-2}{3}}}{y^{\frac{-2}{3}}} \\ & \left(\frac{d y}{d \mathrm{x}}\right)_{\left(\frac{1}{8}, \frac{1}{8}\right)}=\frac{-4}{4}=-1 \end{aligned} $$