(a)OR(b)

If $\displaystyle \mathrm{x}=\mathrm{e}^{\mathrm{x} / \mathrm{y}}$, prove that $\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\log \mathrm{x}-1}{(\log \mathrm{x})^{2}}$

$$\begin{aligned} & \mathrm{x}=\mathrm{e}^{\frac{\mathrm{x}}{\mathrm{y}}} \Rightarrow \log \mathrm{x}=\frac{\mathrm{x}}{\mathrm{y}} \Rightarrow \mathrm{y}=\frac{\mathrm{x}}{\log \mathrm{x}} & \Rightarrow \frac{d \mathrm{y}}{d \mathrm{x}}=\frac{(\log \mathrm{x})(1)-\mathrm{x}\left(\dfrac{1}{\mathrm{x}}\right)}{(\log \mathrm{x})^{2}}=\frac{\log \mathrm{x}-1}{(\log \mathrm{x})^{2}} \end{aligned}LHD at $\displaystyle \mathrm{x}=1$=\lim _{h \rightarrow 0} \frac{\mathrm{f}(1-h)-\mathrm{f}(1)}{-h}=\lim _{h \rightarrow 0} \frac{\left[(1-h)^{2}+1\right]-2}{-h}=2RHD at $\displaystyle \mathrm{x}=1$=\lim _{h \rightarrow 0} \frac{\mathrm{f}(1+h)-\mathrm{f}(1)}{h}=\lim _{h \rightarrow 0} \frac{[3-(1+h)]-2}{h}=-1as $\displaystyle \mathrm{LHD} \neq \mathrm{RHD}$, so $\displaystyle \mathrm{f}(\mathrm{x})$ is not differentiable at $\displaystyle \mathrm{x}=1$