Question Number 1369 by 123456 last updated on 25/Jul/15
$$\mathrm{lets}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{continuous}\:\mathrm{and}\:\mathrm{differentiable} \\ $$$$\mathrm{compute} \\ $$$${g}\left({x}\right)=\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{f}\left({x}+\Delta{x}\right)} −{e}^{{f}\left({x}\right)} }{{f}\left({x}+\Delta{x}\right)−{f}\left({x}\right)} \\ $$
Answered by prakash jain last updated on 27/Jul/15
$${e}^{{f}\left({x}+\Delta{x}\right)} =\mathrm{1}+{f}\left({x}+\Delta{x}\right)+\frac{{f}\left({x}+\Delta{x}\right)^{\mathrm{2}} }{\mathrm{2}!}+\frac{{f}\left({x}+\Delta{x}\right)^{\mathrm{3}} }{\mathrm{3}!}+ \\ $$$${e}^{{f}\left({x}\right)} =\mathrm{1}+{f}\left({x}\right)+\frac{{f}\left({x}\right)^{\mathrm{2}} }{\mathrm{2}!}+\frac{{f}\left({x}\right)^{\mathrm{3}} }{\mathrm{3}!}+ \\ $$$$\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{f}\left({x}+\Delta{x}\right)} −{e}^{{f}\left({x}\right)} }{{f}\left({x}+\Delta{x}\right)−{f}\left({x}\right)}=\mathrm{1}+\frac{\mathrm{2}{f}\left({x}\right)}{\mathrm{2}!}+\frac{\mathrm{3}{f}\left({x}\right)^{\mathrm{2}} }{\mathrm{3}!}+.. \\ $$$$={e}^{{f}\left({x}\right)} \\ $$