(a)OR(b)

If $\displaystyle x=\mathrm{e}^{\frac{x}{\mathrm{y}}}$, then prove that $\displaystyle \frac{\mathrm{dy}}{\mathrm{d} x}=\frac{x-\mathrm{y}}{x \log x}$.

$\displaystyle x=\mathrm{e}^{\frac{x}{\mathrm{y}}}$ $\displaystyle \Rightarrow \log x=\frac{x}{\mathrm{y}}$ $\displaystyle \Rightarrow \mathrm{y} \log x=x$ Differentiating both sides w.r.to x , we get $\displaystyle \frac{\mathrm{y}}{x}+\log x \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}=1$ $\displaystyle \Rightarrow \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}=\frac{x-\mathrm{y}}{x \log x}$ $\displaystyle L \mathrm{f}^{\prime}(-2)=\lim _{h \rightarrow 0} \frac{\mathrm{f}(-2-h)-\mathrm{f}(-2)}{-h} \quad(h>0)$ $\displaystyle =\lim _{h \rightarrow 0} \frac{2(-2-h)-3-(-7)}{-h}$ $\displaystyle =\lim _{h \rightarrow 0} 2=2$ $\displaystyle R \mathrm{f}^{\prime}(-2)=\lim _{h \rightarrow 0} \frac{\mathrm{f}(-2+h)-\mathrm{f}(-2)}{h} \quad(h>0)$ $\displaystyle =\lim _{h \rightarrow 0} \frac{-2+h+1-(-7)}{h}$ $\displaystyle =\lim _{h \rightarrow 0} \frac{6+h}{h}$, which does not exist, i.e., RHD does not exist.