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Question Number 2980 by Filup last updated on 02/Dec/15

Prove that (d/dx)(e^x )=e^x Assume that you do not know that the above statement is true.

$$\mathrm{Prove}\:\mathrm{that}\:\frac{{d}}{{dx}}\left({e}^{{x}} \right)={e}^{{x}} \\ $$$$\mathrm{Assume}\:\mathrm{that}\:\mathrm{you}\:\mathrm{do}\:\mathrm{not}\:\mathrm{know}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{above}\:\mathrm{statement}\:\mathrm{is}\:\mathrm{true}. \\ $$

Answered by RasheedAhmad last updated on 02/Dec/15

e^x =1+x+(x^2 /(2!))+(x^3 /(3!))+... (d/dx)(e^x )=(d/dx)(1+x+(x^2 /(2!))+(x^3 /(3!))+...) =(d/dx)(1)+(d/dx)(x)+(1/(2!))(d/dx)(x^2 )+(1/(3!))(d/dx)(x^3 )+... =0+1+((2x)/(2!))+((3x^2 )/(3!))+... =0+1+x+(x^2 /(2!))+(x^3 /(3!))+...=e^x

$${e}^{{x}} =\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+... \\ $$$$\frac{{d}}{{dx}}\left({e}^{{x}} \right)=\frac{{d}}{{dx}}\left(\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+...\right) \\ $$$$\:\:\:\:\:\:\:=\frac{{d}}{{dx}}\left(\mathrm{1}\right)+\frac{{d}}{{dx}}\left({x}\right)+\frac{\mathrm{1}}{\mathrm{2}!}\frac{{d}}{{dx}}\left({x}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{3}!}\frac{{d}}{{dx}}\left({x}^{\mathrm{3}} \right)+... \\ $$$$\:\:\:=\mathrm{0}+\mathrm{1}+\frac{\mathrm{2}{x}}{\mathrm{2}!}+\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{3}!}+... \\ $$$$\:\:\:\:\:\:\:=\mathrm{0}+\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+...={e}^{{x}} \\ $$$$ \\ $$

Answered by 123456 last updated on 02/Dec/15

(d/dx)(e^x )=lim_(Δx→0) ((f(x+Δx)−f(x))/(Δx)) =lim_(Δx→0) ((e^(x+Δx) −e^x )/(Δx)) =lim_(Δx→0) ((e^x (e^(Δx) −1))/(Δx)) =e^x lim_(Δx→0) ((e^(Δx) −1)/(Δx)) =e^x

$$\frac{{d}}{{dx}}\left({e}^{{x}} \right)=\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{f}\left({x}+\Delta{x}\right)−{f}\left({x}\right)}{\Delta{x}} \\ $$$$=\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}+\Delta{x}} −{e}^{{x}} }{\Delta{x}} \\ $$$$=\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}} \left({e}^{\Delta{x}} −\mathrm{1}\right)}{\Delta{x}} \\ $$$$={e}^{{x}} \underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{\Delta{x}} −\mathrm{1}}{\Delta{x}} \\ $$$$={e}^{{x}} \\ $$

Commented by 123456 last updated on 02/Dec/15

lim_(x→0) ((a^x −1)/x)=ln a

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{a}^{{x}} −\mathrm{1}}{{x}}=\mathrm{ln}\:{a} \\ $$

Commented by Filup last updated on 02/Dec/15

How do you know lim_(Δx→0) ((e^(Δx) −1)/(Δx))=1?

$$\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{know}\:\:\:\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{\Delta{x}} −\mathrm{1}}{\Delta{x}}=\mathrm{1}? \\ $$

Commented by Yozzi last updated on 03/Dec/15

Let lim_(Δx→0) ((e^(Δx) −1)/(Δx))=l. Using the Maclaurin series expansion for e^(Δx ) we have e^(Δx) =1+Δx+(((Δx)^2 )/2)+(((Δx)^3 )/6)+(((Δx)^4 )/(24))+... ⇒e^(Δx) −1=Σ_(r=1) ^∞ (((Δx)^r )/(r!)) ⇒((e^(Δx) −1)/(Δx))=Σ_(r=1) ^∞ (((Δx)^(r−1) )/(r!)) So, lim_(Δx→0) ((e^(Δx) −1)/(Δx))=lim_(Δx→0) Σ_(r=1) ^∞ (((Δx)^(r−1) )/(r!)) l=lim_(Δx→0) (1/1)+Σ_(r=2) ^∞ (((Δx)^(r−1) )/(r!))

$${Let}\:\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{\Delta{x}} −\mathrm{1}}{\Delta{x}}={l}.\:{Using}\:{the}\:{Maclaurin} \\ $$$${series}\:{expansion}\:{for}\:{e}^{\Delta{x}\:} {we}\:{have} \\ $$$${e}^{\Delta{x}} =\mathrm{1}+\Delta{x}+\frac{\left(\Delta{x}\right)^{\mathrm{2}} }{\mathrm{2}}+\frac{\left(\Delta{x}\right)^{\mathrm{3}} }{\mathrm{6}}+\frac{\left(\Delta{x}\right)^{\mathrm{4}} }{\mathrm{24}}+... \\ $$$$\Rightarrow{e}^{\Delta{x}} −\mathrm{1}=\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\Delta{x}\right)^{{r}} }{{r}!} \\ $$$$\Rightarrow\frac{{e}^{\Delta{x}} −\mathrm{1}}{\Delta{x}}=\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\Delta{x}\right)^{{r}−\mathrm{1}} }{{r}!} \\ $$$${So},\:\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{\Delta{x}} −\mathrm{1}}{\Delta{x}}=\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\Delta{x}\right)^{{r}−\mathrm{1}} }{{r}!} \\ $$$${l}=\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{1}}+\underset{{r}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(\Delta{x}\right)^{{r}−\mathrm{1}} }{{r}!} \\ $$$$ \\ $$

Commented by Filup last updated on 03/Dec/15

Thats true, but to derive thus series you must take the derivative of e^x ...?

$$\mathrm{Thats}\:\mathrm{true},\:{but}\:\mathrm{to}\:\mathrm{derive}\:\mathrm{thus}\:\mathrm{series}\:\mathrm{you} \\ $$$$\mathrm{must}\:\mathrm{take}\:\mathrm{the}\:\mathrm{derivative}\:\mathrm{of}\:{e}^{{x}} ...? \\ $$

Commented by Yozzi last updated on 03/Dec/15

Yes. You′re right. Perhaps then another a method of proof of lim_(x→0) ((e^x −1)/x)=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 (d/dx)(e^x ).

$${Yes}.\:{You}'{re}\:{right}.\:{Perhaps}\:{then}\:{another} \\ $$$${a}\:{method}\:{of}\:{proof}\:{of}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}} −\mathrm{1}}{{x}}=\mathrm{1} \\ $$$${is}\:{required}\:{that}\:{is}\:{devoid}\:{of}\: \\ $$$${differentiating}\:{e}^{{x}} . \\ $$$${It}\:{means}\:{then}\:{that}\:{the}\:{first}\:{answer} \\ $$$${is}\:{not}\:{what}\:{you}\:{want}\:{since}\:{it}\:{is}\:{a}\:{series} \\ $$$${based}\:{on}\:\frac{{d}}{{dx}}\left({e}^{{x}} \right).\: \\ $$

Commented by Yozzi last updated on 03/Dec/15

Consider the limit l of the form l=lim_(x→0) ((e^(ax) −1)/x) (a∈N). Since e=lim_(x→0) (1+x)^(1/x) ⇒e^(ax) =(lim_(x→0) (1+x)^(1/x) )^(ax) e^(ax) =lim_(x→0) {(1+x)^(1/x) }^(ax) e^(ax) =lim_(x→0) (1+x)^a e^(ax) −1=lim_(x→0) {(1+x)^a −1} ((e^(ax) −1)/x)=(1/x)lim_(x→0) {(1+x)^a −1} ⇒lim_(x→0) ((e^(ax) −1)/x)=lim_(x→0) (1/x)lim_(x→0) {(x+1)^a −1} The repeated limits on the right hand side absorb into one limit. ∴lim_(x→0) ((e^(ax) −1)/x)=lim_(x→0) (((x+1)^a −1)/x). ∴l=lim_(x→0) (((1+x)^a −1)/x) By the Binomial Theorem , for a∈N, (1+x)^a =Σ_(n=0) ^a ((a),(n) )x^n . ∴ l=lim_(x→0) ((1+ ((a),(1) )x+ ((a),(2) )x^2 +...+ ((a),(a) )x^a −1)/x) l=lim_(x→0) { ((a),(1) )+ ((a),(2) )x+...+x^(a−1) } l= ((a),(1) ) Now, let a=1. ∴ l= ((1),(1) )=1 □

$${Consider}\:{the}\:{limit}\:{l}\:{of}\:{the}\:{form} \\ $$$${l}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{ax}} −\mathrm{1}}{{x}}\:\:\:\left({a}\in\mathbb{N}\right). \\ $$$${Since}\:{e}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+{x}\right)^{\mathrm{1}/{x}} \\ $$$$\Rightarrow{e}^{{ax}} =\left(\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+{x}\right)^{\mathrm{1}/{x}} \right)^{{ax}} \\ $$$${e}^{{ax}} =\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\left(\mathrm{1}+{x}\right)^{\mathrm{1}/{x}} \right\}^{{ax}} \\ $$$${e}^{{ax}} =\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+{x}\right)^{{a}} \\ $$$${e}^{{ax}} −\mathrm{1}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\left(\mathrm{1}+{x}\right)^{{a}} −\mathrm{1}\right\} \\ $$$$\frac{{e}^{{ax}} −\mathrm{1}}{{x}}=\frac{\mathrm{1}}{{x}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\left(\mathrm{1}+{x}\right)^{{a}} −\mathrm{1}\right\} \\ $$$$\Rightarrow\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{ax}} −\mathrm{1}}{{x}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{x}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\left({x}+\mathrm{1}\right)^{{a}} −\mathrm{1}\right\} \\ $$$${The}\:{repeated}\:{limits}\:{on}\:{the}\:{right}\:{hand} \\ $$$${side}\:{absorb}\:{into}\:{one}\:{limit}. \\ $$$$\therefore\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{ax}} −\mathrm{1}}{{x}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({x}+\mathrm{1}\right)^{{a}} −\mathrm{1}}{{x}}. \\ $$$$\therefore{l}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+{x}\right)^{{a}} −\mathrm{1}}{{x}} \\ $$$${By}\:{the}\:{Binomial}\:{Theorem}\:,\:{for}\:{a}\in\mathbb{N}, \\ $$$$\left(\mathrm{1}+{x}\right)^{{a}} =\underset{{n}=\mathrm{0}} {\overset{{a}} {\sum}}\begin{pmatrix}{{a}}\\{{n}}\end{pmatrix}{x}^{{n}} . \\ $$$$\therefore\:{l}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}+\begin{pmatrix}{{a}}\\{\mathrm{1}}\end{pmatrix}{x}+\begin{pmatrix}{{a}}\\{\mathrm{2}}\end{pmatrix}{x}^{\mathrm{2}} +...+\begin{pmatrix}{{a}}\\{{a}}\end{pmatrix}{x}^{{a}} −\mathrm{1}}{{x}} \\ $$$${l}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\begin{pmatrix}{{a}}\\{\mathrm{1}}\end{pmatrix}+\begin{pmatrix}{{a}}\\{\mathrm{2}}\end{pmatrix}{x}+...+{x}^{{a}−\mathrm{1}} \right\} \\ $$$${l}=\begin{pmatrix}{{a}}\\{\mathrm{1}}\end{pmatrix} \\ $$$${Now},\:{let}\:{a}=\mathrm{1}.\:\therefore\:{l}=\begin{pmatrix}{\mathrm{1}}\\{\mathrm{1}}\end{pmatrix}=\mathrm{1}\:\Box \\ $$$$ \\ $$

Commented by Yozzi last updated on 03/Dec/15

Alternatively, consider the limit l=lim_(x→0) ((ln(x+1))/x). By the power rule of logarithms we obtain ((ln(1+x))/x)=ln(1+x)^(1/x) . ⇒l=lim_(x→0) ln(1+x)^(1/x) l=ln{lim_(x→0) (1+x)^(1/x) } Since e=lim_(x→0) (1+x)^(1/x) ⇒l=lne=1 Define L=lim_(x→0) ((e^x −1)/x) and let u=e^x −1. ∴ u→0⇔x→0. Also, x=ln(u+1). ∴ L=lim_(u→0) (u/(ln(u+1)))=(1/(lim_(x→0) ((ln(u+1))/u))) L=(1/1)=1 .

$${Alternatively},\:{consider}\:{the}\:{limit} \\ $$$${l}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{ln}\left({x}+\mathrm{1}\right)}{{x}}.\:\:{By}\:{the}\:{power}\:{rule} \\ $$$${of}\:{logarithms}\:{we}\:{obtain}\: \\ $$$$\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}={ln}\left(\mathrm{1}+{x}\right)^{\mathrm{1}/{x}} . \\ $$$$\Rightarrow{l}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{ln}\left(\mathrm{1}+{x}\right)^{\mathrm{1}/{x}} \\ $$$${l}={ln}\left\{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+{x}\right)^{\mathrm{1}/{x}} \right\} \\ $$$${Since}\:{e}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+{x}\right)^{\mathrm{1}/{x}} \\ $$$$\Rightarrow{l}={lne}=\mathrm{1} \\ $$$${Define}\:{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}} −\mathrm{1}}{{x}}\:{and}\:{let}\:{u}={e}^{{x}} −\mathrm{1}. \\ $$$$\therefore\:{u}\rightarrow\mathrm{0}\Leftrightarrow{x}\rightarrow\mathrm{0}.\:\:{Also},\:{x}={ln}\left({u}+\mathrm{1}\right). \\ $$$$\therefore\:{L}=\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{u}}{{ln}\left({u}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{ln}\left({u}+\mathrm{1}\right)}{{u}}} \\ $$$${L}=\frac{\mathrm{1}}{\mathrm{1}}=\mathrm{1}\:. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by Rasheed Soomro last updated on 03/Dec/15

There is also a proof using Sandwitch theorm.

$$\mathcal{T}{here}\:{is}\:{also}\:{a}\:{proof}\:{using}\:{Sandwitch}\:{theorm}. \\ $$

Commented by Yozzi last updated on 03/Dec/15

Certainly.

$${Certainly}.\: \\ $$

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