# Prove-that-d-dx-e-x-e-x-Assume-that-you-do-not-know-that-the-above-statement-is-true-

Question Number 2980 by Filup last updated on 02/Dec/15
$$\mathrm{Prove}\:\mathrm{that}\:\frac{{d}}{{dx}}\left({e}^{{x}} \right)={e}^{{x}} \\$$$$\mathrm{Assume}\:\mathrm{that}\:\mathrm{you}\:\mathrm{do}\:\mathrm{not}\:\mathrm{know}\:\mathrm{that} \\$$$$\mathrm{the}\:\mathrm{above}\:\mathrm{statement}\:\mathrm{is}\:\mathrm{true}. \\$$
$${e}^{{x}} =\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+… \\$$$$\frac{{d}}{{dx}}\left({e}^{{x}} \right)=\frac{{d}}{{dx}}\left(\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+…\right) \\$$$$\:\:\:\:\:\:\:=\frac{{d}}{{dx}}\left(\mathrm{1}\right)+\frac{{d}}{{dx}}\left({x}\right)+\frac{\mathrm{1}}{\mathrm{2}!}\frac{{d}}{{dx}}\left({x}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{3}!}\frac{{d}}{{dx}}\left({x}^{\mathrm{3}} \right)+… \\$$$$\:\:\:=\mathrm{0}+\mathrm{1}+\frac{\mathrm{2}{x}}{\mathrm{2}!}+\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{3}!}+… \\$$$$\:\:\:\:\:\:\:=\mathrm{0}+\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+…={e}^{{x}} \\$$$$\\$$
Answered by 123456 last updated on 02/Dec/15
$$\frac{{d}}{{dx}}\left({e}^{{x}} \right)=\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{f}\left({x}+\Delta{x}\right)−{f}\left({x}\right)}{\Delta{x}} \\$$$$=\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}+\Delta{x}} −{e}^{{x}} }{\Delta{x}} \\$$$$=\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}} \left({e}^{\Delta{x}} −\mathrm{1}\right)}{\Delta{x}} \\$$$$={e}^{{x}} \underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{\Delta{x}} −\mathrm{1}}{\Delta{x}} \\$$$$={e}^{{x}} \\$$
Commented by 123456 last updated on 02/Dec/15
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{a}^{{x}} −\mathrm{1}}{{x}}=\mathrm{ln}\:{a} \\$$
Commented by Filup last updated on 02/Dec/15
$$\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{know}\:\:\:\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{\Delta{x}} −\mathrm{1}}{\Delta{x}}=\mathrm{1}? \\$$
Commented by Yozzi last updated on 03/Dec/15
$${Let}\:\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{\Delta{x}} −\mathrm{1}}{\Delta{x}}={l}.\:{Using}\:{the}\:{Maclaurin} \\$$$${series}\:{expansion}\:{for}\:{e}^{\Delta{x}\:} {we}\:{have} \\$$$${e}^{\Delta{x}} =\mathrm{1}+\Delta{x}+\frac{\left(\Delta{x}\right)^{\mathrm{2}} }{\mathrm{2}}+\frac{\left(\Delta{x}\right)^{\mathrm{3}} }{\mathrm{6}}+\frac{\left(\Delta{x}\right)^{\mathrm{4}} }{\mathrm{24}}+… \\$$$$\Rightarrow{e}^{\Delta{x}} −\mathrm{1}=\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\Delta{x}\right)^{{r}} }{{r}!} \\$$$$\Rightarrow\frac{{e}^{\Delta{x}} −\mathrm{1}}{\Delta{x}}=\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\Delta{x}\right)^{{r}−\mathrm{1}} }{{r}!} \\$$$${So},\:\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{\Delta{x}} −\mathrm{1}}{\Delta{x}}=\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\Delta{x}\right)^{{r}−\mathrm{1}} }{{r}!} \\$$$${l}=\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{1}}+\underset{{r}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(\Delta{x}\right)^{{r}−\mathrm{1}} }{{r}!} \\$$$$\\$$
Commented by Filup last updated on 03/Dec/15
$$\mathrm{Thats}\:\mathrm{true},\:{but}\:\mathrm{to}\:\mathrm{derive}\:\mathrm{thus}\:\mathrm{series}\:\mathrm{you} \\$$$$\mathrm{must}\:\mathrm{take}\:\mathrm{the}\:\mathrm{derivative}\:\mathrm{of}\:{e}^{{x}} …? \\$$
Commented by Yozzi last updated on 03/Dec/15
$${Yes}.\:{You}'{re}\:{right}.\:{Perhaps}\:{then}\:{another} \\$$$${a}\:{method}\:{of}\:{proof}\:{of}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}} −\mathrm{1}}{{x}}=\mathrm{1} \\$$$${is}\:{required}\:{that}\:{is}\:{devoid}\:{of}\: \\$$$${differentiating}\:{e}^{{x}} . \\$$$${It}\:{means}\:{then}\:{that}\:{the}\:{first}\:{answer} \\$$$${is}\:{not}\:{what}\:{you}\:{want}\:{since}\:{it}\:{is}\:{a}\:{series} \\$$$${based}\:{on}\:\frac{{d}}{{dx}}\left({e}^{{x}} \right).\: \\$$
Commented by Yozzi last updated on 03/Dec/15
$${Consider}\:{the}\:{limit}\:{l}\:{of}\:{the}\:{form} \\$$$${l}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{ax}} −\mathrm{1}}{{x}}\:\:\:\left({a}\in\mathbb{N}\right). \\$$$${Since}\:{e}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+{x}\right)^{\mathrm{1}/{x}} \\$$$$\Rightarrow{e}^{{ax}} =\left(\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+{x}\right)^{\mathrm{1}/{x}} \right)^{{ax}} \\$$$${e}^{{ax}} =\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\left(\mathrm{1}+{x}\right)^{\mathrm{1}/{x}} \right\}^{{ax}} \\$$$${e}^{{ax}} =\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+{x}\right)^{{a}} \\$$$${e}^{{ax}} −\mathrm{1}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\left(\mathrm{1}+{x}\right)^{{a}} −\mathrm{1}\right\} \\$$$$\frac{{e}^{{ax}} −\mathrm{1}}{{x}}=\frac{\mathrm{1}}{{x}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\left(\mathrm{1}+{x}\right)^{{a}} −\mathrm{1}\right\} \\$$$$\Rightarrow\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{ax}} −\mathrm{1}}{{x}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{x}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\left({x}+\mathrm{1}\right)^{{a}} −\mathrm{1}\right\} \\$$$${The}\:{repeated}\:{limits}\:{on}\:{the}\:{right}\:{hand} \\$$$${side}\:{absorb}\:{into}\:{one}\:{limit}. \\$$$$\therefore\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{ax}} −\mathrm{1}}{{x}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({x}+\mathrm{1}\right)^{{a}} −\mathrm{1}}{{x}}. \\$$$$\therefore{l}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+{x}\right)^{{a}} −\mathrm{1}}{{x}} \\$$$${By}\:{the}\:{Binomial}\:{Theorem}\:,\:{for}\:{a}\in\mathbb{N}, \\$$$$\left(\mathrm{1}+{x}\right)^{{a}} =\underset{{n}=\mathrm{0}} {\overset{{a}} {\sum}}\begin{pmatrix}{{a}}\\{{n}}\end{pmatrix}{x}^{{n}} . \\$$$$\therefore\:{l}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}+\begin{pmatrix}{{a}}\\{\mathrm{1}}\end{pmatrix}{x}+\begin{pmatrix}{{a}}\\{\mathrm{2}}\end{pmatrix}{x}^{\mathrm{2}} +…+\begin{pmatrix}{{a}}\\{{a}}\end{pmatrix}{x}^{{a}} −\mathrm{1}}{{x}} \\$$$${l}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\begin{pmatrix}{{a}}\\{\mathrm{1}}\end{pmatrix}+\begin{pmatrix}{{a}}\\{\mathrm{2}}\end{pmatrix}{x}+…+{x}^{{a}−\mathrm{1}} \right\} \\$$$${l}=\begin{pmatrix}{{a}}\\{\mathrm{1}}\end{pmatrix} \\$$$${Now},\:{let}\:{a}=\mathrm{1}.\:\therefore\:{l}=\begin{pmatrix}{\mathrm{1}}\\{\mathrm{1}}\end{pmatrix}=\mathrm{1}\:\Box \\$$$$\\$$
Commented by Yozzi last updated on 03/Dec/15
$${Alternatively},\:{consider}\:{the}\:{limit} \\$$$${l}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{ln}\left({x}+\mathrm{1}\right)}{{x}}.\:\:{By}\:{the}\:{power}\:{rule} \\$$$${of}\:{logarithms}\:{we}\:{obtain}\: \\$$$$\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}={ln}\left(\mathrm{1}+{x}\right)^{\mathrm{1}/{x}} . \\$$$$\Rightarrow{l}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{ln}\left(\mathrm{1}+{x}\right)^{\mathrm{1}/{x}} \\$$$${l}={ln}\left\{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+{x}\right)^{\mathrm{1}/{x}} \right\} \\$$$${Since}\:{e}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+{x}\right)^{\mathrm{1}/{x}} \\$$$$\Rightarrow{l}={lne}=\mathrm{1} \\$$$${Define}\:{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}} −\mathrm{1}}{{x}}\:{and}\:{let}\:{u}={e}^{{x}} −\mathrm{1}. \\$$$$\therefore\:{u}\rightarrow\mathrm{0}\Leftrightarrow{x}\rightarrow\mathrm{0}.\:\:{Also},\:{x}={ln}\left({u}+\mathrm{1}\right). \\$$$$\therefore\:{L}=\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{u}}{{ln}\left({u}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{ln}\left({u}+\mathrm{1}\right)}{{u}}} \\$$$${L}=\frac{\mathrm{1}}{\mathrm{1}}=\mathrm{1}\:. \\$$$$\\$$$$\\$$$$\\$$$$\\$$
Commented by Rasheed Soomro last updated on 03/Dec/15
$$\mathcal{T}{here}\:{is}\:{also}\:{a}\:{proof}\:{using}\:{Sandwitch}\:{theorm}. \\$$
Commented by Yozzi last updated on 03/Dec/15
$${Certainly}.\: \\$$